Find sum of series $ sum_{n=1}^{infty} (ncdot ln frac{2n+1}{2n-1} - 1) $












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how can I find sum of series $ sum_{n=1}^{infty} (ncdot ln frac{2n+1}{2n-1} - 1) $?
It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...



Let consider sum no to infinity, but to n
$$ sum_{k=1}^{n} (kcdot ln frac{2k+1}{2k-1} - 1) =$$
$$ ln frac{3}{1}cdot left(frac{5}{3}right)^2 cdot...cdot left(frac{2n+1}{2n-1}right)^n - n = ln frac{1}{1}cdot frac{1}{3}cdot frac{1}{5}cdot ... frac{1}{2n-1} - n $$
but $$ n = ln e^n $$
so
it will be $$lnfrac{1}{e^n} cdot frac{1}{1}cdot frac{1}{3}cdot frac{1}{5}cdot ... frac{1}{2n-1}$$



So the limit of it is $-infty$
Have I done this well or I missed sth?






real-analysis sequences-and-series







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  • 2




    $begingroup$
    it doesn't converge because the summand is around $nln(1+frac{2}{2n-1})-1 approx nfrac{2}{2n-1}-1 = frac{1}{2n-1}$.
    $endgroup$
    – mathworker21
    Nov 29 '18 at 19:08








  • 1




    $begingroup$
    It certainly converges, since the $n$'th term is asymptotic to $1/(12 n^2)$.
    $endgroup$
    – Robert Israel
    Nov 29 '18 at 19:09










  • $begingroup$
    @RobertIsrael where did I go wrong
    $endgroup$
    – mathworker21
    Nov 29 '18 at 19:14










  • $begingroup$
    @mathworker21 seems correct to me.
    $endgroup$
    – Connor Harris
    Nov 29 '18 at 19:16










  • $begingroup$
    where should it be?
    $endgroup$
    – mvxxx
    Nov 29 '18 at 19:28
















3












$begingroup$


how can I find sum of series $ sum_{n=1}^{infty} (ncdot ln frac{2n+1}{2n-1} - 1) $?
It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...



Let consider sum no to infinity, but to n
$$ sum_{k=1}^{n} (kcdot ln frac{2k+1}{2k-1} - 1) =$$
$$ ln frac{3}{1}cdot left(frac{5}{3}right)^2 cdot...cdot left(frac{2n+1}{2n-1}right)^n - n = ln frac{1}{1}cdot frac{1}{3}cdot frac{1}{5}cdot ... frac{1}{2n-1} - n $$
but $$ n = ln e^n $$
so
it will be $$lnfrac{1}{e^n} cdot frac{1}{1}cdot frac{1}{3}cdot frac{1}{5}cdot ... frac{1}{2n-1}$$



So the limit of it is $-infty$
Have I done this well or I missed sth?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    it doesn't converge because the summand is around $nln(1+frac{2}{2n-1})-1 approx nfrac{2}{2n-1}-1 = frac{1}{2n-1}$.
    $endgroup$
    – mathworker21
    Nov 29 '18 at 19:08








  • 1




    $begingroup$
    It certainly converges, since the $n$'th term is asymptotic to $1/(12 n^2)$.
    $endgroup$
    – Robert Israel
    Nov 29 '18 at 19:09










  • $begingroup$
    @RobertIsrael where did I go wrong
    $endgroup$
    – mathworker21
    Nov 29 '18 at 19:14










  • $begingroup$
    @mathworker21 seems correct to me.
    $endgroup$
    – Connor Harris
    Nov 29 '18 at 19:16










  • $begingroup$
    where should it be?
    $endgroup$
    – mvxxx
    Nov 29 '18 at 19:28














3












3








3





$begingroup$


how can I find sum of series $ sum_{n=1}^{infty} (ncdot ln frac{2n+1}{2n-1} - 1) $?
It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...



Let consider sum no to infinity, but to n
$$ sum_{k=1}^{n} (kcdot ln frac{2k+1}{2k-1} - 1) =$$
$$ ln frac{3}{1}cdot left(frac{5}{3}right)^2 cdot...cdot left(frac{2n+1}{2n-1}right)^n - n = ln frac{1}{1}cdot frac{1}{3}cdot frac{1}{5}cdot ... frac{1}{2n-1} - n $$
but $$ n = ln e^n $$
so
it will be $$lnfrac{1}{e^n} cdot frac{1}{1}cdot frac{1}{3}cdot frac{1}{5}cdot ... frac{1}{2n-1}$$



So the limit of it is $-infty$
Have I done this well or I missed sth?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$




how can I find sum of series $ sum_{n=1}^{infty} (ncdot ln frac{2n+1}{2n-1} - 1) $?
It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...



Let consider sum no to infinity, but to n
$$ sum_{k=1}^{n} (kcdot ln frac{2k+1}{2k-1} - 1) =$$
$$ ln frac{3}{1}cdot left(frac{5}{3}right)^2 cdot...cdot left(frac{2n+1}{2n-1}right)^n - n = ln frac{1}{1}cdot frac{1}{3}cdot frac{1}{5}cdot ... frac{1}{2n-1} - n $$
but $$ n = ln e^n $$
so
it will be $$lnfrac{1}{e^n} cdot frac{1}{1}cdot frac{1}{3}cdot frac{1}{5}cdot ... frac{1}{2n-1}$$



So the limit of it is $-infty$
Have I done this well or I missed sth?






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 19:25







mvxxx

















asked Nov 29 '18 at 19:06









mvxxxmvxxx

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  • 2




    $begingroup$
    it doesn't converge because the summand is around $nln(1+frac{2}{2n-1})-1 approx nfrac{2}{2n-1}-1 = frac{1}{2n-1}$.
    $endgroup$
    – mathworker21
    Nov 29 '18 at 19:08








  • 1




    $begingroup$
    It certainly converges, since the $n$'th term is asymptotic to $1/(12 n^2)$.
    $endgroup$
    – Robert Israel
    Nov 29 '18 at 19:09










  • $begingroup$
    @RobertIsrael where did I go wrong
    $endgroup$
    – mathworker21
    Nov 29 '18 at 19:14










  • $begingroup$
    @mathworker21 seems correct to me.
    $endgroup$
    – Connor Harris
    Nov 29 '18 at 19:16










  • $begingroup$
    where should it be?
    $endgroup$
    – mvxxx
    Nov 29 '18 at 19:28














  • 2




    $begingroup$
    it doesn't converge because the summand is around $nln(1+frac{2}{2n-1})-1 approx nfrac{2}{2n-1}-1 = frac{1}{2n-1}$.
    $endgroup$
    – mathworker21
    Nov 29 '18 at 19:08








  • 1




    $begingroup$
    It certainly converges, since the $n$'th term is asymptotic to $1/(12 n^2)$.
    $endgroup$
    – Robert Israel
    Nov 29 '18 at 19:09










  • $begingroup$
    @RobertIsrael where did I go wrong
    $endgroup$
    – mathworker21
    Nov 29 '18 at 19:14










  • $begingroup$
    @mathworker21 seems correct to me.
    $endgroup$
    – Connor Harris
    Nov 29 '18 at 19:16










  • $begingroup$
    where should it be?
    $endgroup$
    – mvxxx
    Nov 29 '18 at 19:28








2




2




$begingroup$
it doesn't converge because the summand is around $nln(1+frac{2}{2n-1})-1 approx nfrac{2}{2n-1}-1 = frac{1}{2n-1}$.
$endgroup$
– mathworker21
Nov 29 '18 at 19:08






$begingroup$
it doesn't converge because the summand is around $nln(1+frac{2}{2n-1})-1 approx nfrac{2}{2n-1}-1 = frac{1}{2n-1}$.
$endgroup$
– mathworker21
Nov 29 '18 at 19:08






1




1




$begingroup$
It certainly converges, since the $n$'th term is asymptotic to $1/(12 n^2)$.
$endgroup$
– Robert Israel
Nov 29 '18 at 19:09




$begingroup$
It certainly converges, since the $n$'th term is asymptotic to $1/(12 n^2)$.
$endgroup$
– Robert Israel
Nov 29 '18 at 19:09












$begingroup$
@RobertIsrael where did I go wrong
$endgroup$
– mathworker21
Nov 29 '18 at 19:14




$begingroup$
@RobertIsrael where did I go wrong
$endgroup$
– mathworker21
Nov 29 '18 at 19:14












$begingroup$
@mathworker21 seems correct to me.
$endgroup$
– Connor Harris
Nov 29 '18 at 19:16




$begingroup$
@mathworker21 seems correct to me.
$endgroup$
– Connor Harris
Nov 29 '18 at 19:16












$begingroup$
where should it be?
$endgroup$
– mvxxx
Nov 29 '18 at 19:28




$begingroup$
where should it be?
$endgroup$
– mvxxx
Nov 29 '18 at 19:28










4 Answers
4






active

oldest

votes


















2












$begingroup$

We have that



$$sum_{n=1}^{N} left(ncdot ln frac{2n+1}{2n-1} - 1right)=sum_{n=1}^{N} left(ncdot ln (2n+1)-nln (2n-1) - 1right)=$$



$$=(1cdot ln 3-1cdot ln1-1)+(2cdot ln 5-2cdot ln3-1)+(3cdot ln 7-3cdot ln5-1)+ldots=$$



$$=-ln(3cdot 5cdot 7cdot ldotscdot (2N-1))+Ncdotln(2N+1)-N=$$$$=-lnleft(frac{(2N)!}{2^NN!}right)+Ncdotln(2N+1)-N=$$



$$=lnleft(frac{(2^N)^2N!N^N}{(2N)!e^N}right)+Ncdotlnleft(1+frac1{2N}right)$$



and by Stirling's approximation $N!sim sqrt{2pi N}left(frac{N}{e}right)^N$



$$frac{(2^N)^2N!N^N}{(2N!)e^N}simfrac{(2^N)^2N^N}{e^N}frac{sqrt{2pi N}}{sqrt{4pi N}}frac{N^Ne^{2N}}{e^N4^NN^{2N}}=frac{1}{sqrt 2}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I edited my post, can you look at this?
    $endgroup$
    – mvxxx
    Nov 29 '18 at 19:25










  • $begingroup$
    As noticed the limit exists since the term is asymthotic to $1/(12n^2)$. Starting by the first step indicated you see that the sum telescopes.
    $endgroup$
    – gimusi
    Nov 29 '18 at 19:28










  • $begingroup$
    @gimusi: Just was curious how did you precisely say the term is asymptotic to $1/(12n^2)$?
    $endgroup$
    – Yadati Kiran
    Nov 29 '18 at 19:36










  • $begingroup$
    @YadatiKiran It is not difficult to see by $ln frac{2n+1}{2n-1}=ln (1+1/2n)-ln (1-1/2n)$ and then using Taylor's series for log.
    $endgroup$
    – gimusi
    Nov 29 '18 at 19:46










  • $begingroup$
    @gimusi: Got it! Thanks.
    $endgroup$
    – Yadati Kiran
    Nov 29 '18 at 19:55



















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If I'm not mistaken, the actual sum is
$$ frac{1 - ln(2)}{2}$$






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  • $begingroup$
    Yes that's agree with my result also.
    $endgroup$
    – gimusi
    Nov 29 '18 at 19:56



















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Let $f(x)=displaystylesum_{n=1}^{infty}left(nlnfrac{n+x}{n-x}-2xright)$ for $xin(-1,1)$. Then
$$f'(x)=displaystylesum_{n=1}^{infty}frac{2x^2}{n^2-x^2}=1-pi xcotpi x$$
(termwise differentiation is admissible because of uniform convergence of the latter series in $[-a,a]$ for any $0<a<1$; the second equality is known). Thus,
$$f(x)=x-frac{1}{pi}int_{0}^{pi x}tcot t,dt=x(1-lnsinpi x)+frac{1}{pi}int_{0}^{pi x}lnsin t,dt.$$
Your sum is $f(1/2)=(1-ln2)/2$.






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    1












    $begingroup$

    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
    newcommand{dd}{mathrm{d}}
    newcommand{ds}[1]{displaystyle{#1}}
    newcommand{expo}[1]{,mathrm{e}^{#1},}
    newcommand{ic}{mathrm{i}}
    newcommand{mc}[1]{mathcal{#1}}
    newcommand{mrm}[1]{mathrm{#1}}
    newcommand{pars}[1]{left(,{#1},right)}
    newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$




    With $ds{N in mathbb{N}_{geq 1}}$:




    begin{align}
    &bbox[#ffd,10px]{sum_{n = 1}^{N}bracks{nlnpars{2n + 1 over 2n - 1} - 1}}
    \[5mm] = &
    sum_{n = 1}^{N}nlnpars{2n + 1} - sum_{n = 1}^{N}nlnpars{2n - 1} - N
    \[5mm] = &
    -N + sum_{n = 0}^{N}nlnpars{2n + 1} -
    sum_{n = 0}^{N - 1}pars{n + 1}lnpars{2n + 1}
    \[5mm] = &
    -N + Nlnpars{2N + 1} -Nlnpars{2} -
    sum_{n = 0}^{N - 1}lnpars{n + {1 over 2}}
    \[5mm] = &
    -N + Nlnpars{N + {1 over 2}} -
    lnpars{prod_{n = 0}^{N - 1}bracks{n + {1 over 2}}}
    \[5mm] = &
    -N + Nlnpars{N + {1 over 2}} -
    lnpars{bracks{N - 1/2}! over Gammapars{1/2}}
    \[5mm] stackrel{mrm{as} N to infty}{sim}&
    -N + Nlnpars{N + {1 over 2}} -
    lnpars{root{2pi}bracks{N - 1/2}^{N}expo{-N + 1/2} over root{pi}}
    \[5mm] = &
    -N + Nlnpars{N + {1 over 2}} -
    lnpars{2^{1/2}N^{N}bracks{1 - {1/2 over N}}^{N}
    expo{-N + 1/2}}
    \[5mm] stackrel{mrm{as} N to infty}{sim} &
    -N + Nlnpars{N + {1 over 2}} -
    bracks{{1 over 2},lnpars{2} + Nlnpars{N} - N}
    \[5mm] = &
    underbrace{Nlnpars{1 + {1 over 2N}}}
    _{ds{stackrel{mrm{as} N to infty}{to} {1 over 2}}}
    - {1 over 2},lnpars{2}label{1}tag{1}
    \[5mm] stackrel{mrm{as} N to infty}{to} &
    bbx{1 - lnpars{2} over 2} approx 0.1534
    end{align}






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How have you moved from one before last to last line?
      $endgroup$
      – mvxxx
      Nov 29 '18 at 21:16










    • $begingroup$
      @mvxxx I add one more line ( line (1) ) where you can see that the first $displaystyle -N$ cancels with the $displaystyle -left(-Nright)$ at the far right and $displaystyle Nlnleft(N + {1 over 2}right) - Nlnleft(Nright) = Nlnleft(1 + {1 over 2N}right) to {large{1 over 2}}$
      $endgroup$
      – Felix Marin
      Nov 29 '18 at 21:34













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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    We have that



    $$sum_{n=1}^{N} left(ncdot ln frac{2n+1}{2n-1} - 1right)=sum_{n=1}^{N} left(ncdot ln (2n+1)-nln (2n-1) - 1right)=$$



    $$=(1cdot ln 3-1cdot ln1-1)+(2cdot ln 5-2cdot ln3-1)+(3cdot ln 7-3cdot ln5-1)+ldots=$$



    $$=-ln(3cdot 5cdot 7cdot ldotscdot (2N-1))+Ncdotln(2N+1)-N=$$$$=-lnleft(frac{(2N)!}{2^NN!}right)+Ncdotln(2N+1)-N=$$



    $$=lnleft(frac{(2^N)^2N!N^N}{(2N)!e^N}right)+Ncdotlnleft(1+frac1{2N}right)$$



    and by Stirling's approximation $N!sim sqrt{2pi N}left(frac{N}{e}right)^N$



    $$frac{(2^N)^2N!N^N}{(2N!)e^N}simfrac{(2^N)^2N^N}{e^N}frac{sqrt{2pi N}}{sqrt{4pi N}}frac{N^Ne^{2N}}{e^N4^NN^{2N}}=frac{1}{sqrt 2}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I edited my post, can you look at this?
      $endgroup$
      – mvxxx
      Nov 29 '18 at 19:25










    • $begingroup$
      As noticed the limit exists since the term is asymthotic to $1/(12n^2)$. Starting by the first step indicated you see that the sum telescopes.
      $endgroup$
      – gimusi
      Nov 29 '18 at 19:28










    • $begingroup$
      @gimusi: Just was curious how did you precisely say the term is asymptotic to $1/(12n^2)$?
      $endgroup$
      – Yadati Kiran
      Nov 29 '18 at 19:36










    • $begingroup$
      @YadatiKiran It is not difficult to see by $ln frac{2n+1}{2n-1}=ln (1+1/2n)-ln (1-1/2n)$ and then using Taylor's series for log.
      $endgroup$
      – gimusi
      Nov 29 '18 at 19:46










    • $begingroup$
      @gimusi: Got it! Thanks.
      $endgroup$
      – Yadati Kiran
      Nov 29 '18 at 19:55
















    2












    $begingroup$

    We have that



    $$sum_{n=1}^{N} left(ncdot ln frac{2n+1}{2n-1} - 1right)=sum_{n=1}^{N} left(ncdot ln (2n+1)-nln (2n-1) - 1right)=$$



    $$=(1cdot ln 3-1cdot ln1-1)+(2cdot ln 5-2cdot ln3-1)+(3cdot ln 7-3cdot ln5-1)+ldots=$$



    $$=-ln(3cdot 5cdot 7cdot ldotscdot (2N-1))+Ncdotln(2N+1)-N=$$$$=-lnleft(frac{(2N)!}{2^NN!}right)+Ncdotln(2N+1)-N=$$



    $$=lnleft(frac{(2^N)^2N!N^N}{(2N)!e^N}right)+Ncdotlnleft(1+frac1{2N}right)$$



    and by Stirling's approximation $N!sim sqrt{2pi N}left(frac{N}{e}right)^N$



    $$frac{(2^N)^2N!N^N}{(2N!)e^N}simfrac{(2^N)^2N^N}{e^N}frac{sqrt{2pi N}}{sqrt{4pi N}}frac{N^Ne^{2N}}{e^N4^NN^{2N}}=frac{1}{sqrt 2}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I edited my post, can you look at this?
      $endgroup$
      – mvxxx
      Nov 29 '18 at 19:25










    • $begingroup$
      As noticed the limit exists since the term is asymthotic to $1/(12n^2)$. Starting by the first step indicated you see that the sum telescopes.
      $endgroup$
      – gimusi
      Nov 29 '18 at 19:28










    • $begingroup$
      @gimusi: Just was curious how did you precisely say the term is asymptotic to $1/(12n^2)$?
      $endgroup$
      – Yadati Kiran
      Nov 29 '18 at 19:36










    • $begingroup$
      @YadatiKiran It is not difficult to see by $ln frac{2n+1}{2n-1}=ln (1+1/2n)-ln (1-1/2n)$ and then using Taylor's series for log.
      $endgroup$
      – gimusi
      Nov 29 '18 at 19:46










    • $begingroup$
      @gimusi: Got it! Thanks.
      $endgroup$
      – Yadati Kiran
      Nov 29 '18 at 19:55














    2












    2








    2





    $begingroup$

    We have that



    $$sum_{n=1}^{N} left(ncdot ln frac{2n+1}{2n-1} - 1right)=sum_{n=1}^{N} left(ncdot ln (2n+1)-nln (2n-1) - 1right)=$$



    $$=(1cdot ln 3-1cdot ln1-1)+(2cdot ln 5-2cdot ln3-1)+(3cdot ln 7-3cdot ln5-1)+ldots=$$



    $$=-ln(3cdot 5cdot 7cdot ldotscdot (2N-1))+Ncdotln(2N+1)-N=$$$$=-lnleft(frac{(2N)!}{2^NN!}right)+Ncdotln(2N+1)-N=$$



    $$=lnleft(frac{(2^N)^2N!N^N}{(2N)!e^N}right)+Ncdotlnleft(1+frac1{2N}right)$$



    and by Stirling's approximation $N!sim sqrt{2pi N}left(frac{N}{e}right)^N$



    $$frac{(2^N)^2N!N^N}{(2N!)e^N}simfrac{(2^N)^2N^N}{e^N}frac{sqrt{2pi N}}{sqrt{4pi N}}frac{N^Ne^{2N}}{e^N4^NN^{2N}}=frac{1}{sqrt 2}$$






    share|cite|improve this answer











    $endgroup$



    We have that



    $$sum_{n=1}^{N} left(ncdot ln frac{2n+1}{2n-1} - 1right)=sum_{n=1}^{N} left(ncdot ln (2n+1)-nln (2n-1) - 1right)=$$



    $$=(1cdot ln 3-1cdot ln1-1)+(2cdot ln 5-2cdot ln3-1)+(3cdot ln 7-3cdot ln5-1)+ldots=$$



    $$=-ln(3cdot 5cdot 7cdot ldotscdot (2N-1))+Ncdotln(2N+1)-N=$$$$=-lnleft(frac{(2N)!}{2^NN!}right)+Ncdotln(2N+1)-N=$$



    $$=lnleft(frac{(2^N)^2N!N^N}{(2N)!e^N}right)+Ncdotlnleft(1+frac1{2N}right)$$



    and by Stirling's approximation $N!sim sqrt{2pi N}left(frac{N}{e}right)^N$



    $$frac{(2^N)^2N!N^N}{(2N!)e^N}simfrac{(2^N)^2N^N}{e^N}frac{sqrt{2pi N}}{sqrt{4pi N}}frac{N^Ne^{2N}}{e^N4^NN^{2N}}=frac{1}{sqrt 2}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 29 '18 at 21:30

























    answered Nov 29 '18 at 19:21









    gimusigimusi

    92.8k84494




    92.8k84494












    • $begingroup$
      I edited my post, can you look at this?
      $endgroup$
      – mvxxx
      Nov 29 '18 at 19:25










    • $begingroup$
      As noticed the limit exists since the term is asymthotic to $1/(12n^2)$. Starting by the first step indicated you see that the sum telescopes.
      $endgroup$
      – gimusi
      Nov 29 '18 at 19:28










    • $begingroup$
      @gimusi: Just was curious how did you precisely say the term is asymptotic to $1/(12n^2)$?
      $endgroup$
      – Yadati Kiran
      Nov 29 '18 at 19:36










    • $begingroup$
      @YadatiKiran It is not difficult to see by $ln frac{2n+1}{2n-1}=ln (1+1/2n)-ln (1-1/2n)$ and then using Taylor's series for log.
      $endgroup$
      – gimusi
      Nov 29 '18 at 19:46










    • $begingroup$
      @gimusi: Got it! Thanks.
      $endgroup$
      – Yadati Kiran
      Nov 29 '18 at 19:55


















    • $begingroup$
      I edited my post, can you look at this?
      $endgroup$
      – mvxxx
      Nov 29 '18 at 19:25










    • $begingroup$
      As noticed the limit exists since the term is asymthotic to $1/(12n^2)$. Starting by the first step indicated you see that the sum telescopes.
      $endgroup$
      – gimusi
      Nov 29 '18 at 19:28










    • $begingroup$
      @gimusi: Just was curious how did you precisely say the term is asymptotic to $1/(12n^2)$?
      $endgroup$
      – Yadati Kiran
      Nov 29 '18 at 19:36










    • $begingroup$
      @YadatiKiran It is not difficult to see by $ln frac{2n+1}{2n-1}=ln (1+1/2n)-ln (1-1/2n)$ and then using Taylor's series for log.
      $endgroup$
      – gimusi
      Nov 29 '18 at 19:46










    • $begingroup$
      @gimusi: Got it! Thanks.
      $endgroup$
      – Yadati Kiran
      Nov 29 '18 at 19:55
















    $begingroup$
    I edited my post, can you look at this?
    $endgroup$
    – mvxxx
    Nov 29 '18 at 19:25




    $begingroup$
    I edited my post, can you look at this?
    $endgroup$
    – mvxxx
    Nov 29 '18 at 19:25












    $begingroup$
    As noticed the limit exists since the term is asymthotic to $1/(12n^2)$. Starting by the first step indicated you see that the sum telescopes.
    $endgroup$
    – gimusi
    Nov 29 '18 at 19:28




    $begingroup$
    As noticed the limit exists since the term is asymthotic to $1/(12n^2)$. Starting by the first step indicated you see that the sum telescopes.
    $endgroup$
    – gimusi
    Nov 29 '18 at 19:28












    $begingroup$
    @gimusi: Just was curious how did you precisely say the term is asymptotic to $1/(12n^2)$?
    $endgroup$
    – Yadati Kiran
    Nov 29 '18 at 19:36




    $begingroup$
    @gimusi: Just was curious how did you precisely say the term is asymptotic to $1/(12n^2)$?
    $endgroup$
    – Yadati Kiran
    Nov 29 '18 at 19:36












    $begingroup$
    @YadatiKiran It is not difficult to see by $ln frac{2n+1}{2n-1}=ln (1+1/2n)-ln (1-1/2n)$ and then using Taylor's series for log.
    $endgroup$
    – gimusi
    Nov 29 '18 at 19:46




    $begingroup$
    @YadatiKiran It is not difficult to see by $ln frac{2n+1}{2n-1}=ln (1+1/2n)-ln (1-1/2n)$ and then using Taylor's series for log.
    $endgroup$
    – gimusi
    Nov 29 '18 at 19:46












    $begingroup$
    @gimusi: Got it! Thanks.
    $endgroup$
    – Yadati Kiran
    Nov 29 '18 at 19:55




    $begingroup$
    @gimusi: Got it! Thanks.
    $endgroup$
    – Yadati Kiran
    Nov 29 '18 at 19:55











    2












    $begingroup$

    If I'm not mistaken, the actual sum is
    $$ frac{1 - ln(2)}{2}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Yes that's agree with my result also.
      $endgroup$
      – gimusi
      Nov 29 '18 at 19:56
















    2












    $begingroup$

    If I'm not mistaken, the actual sum is
    $$ frac{1 - ln(2)}{2}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Yes that's agree with my result also.
      $endgroup$
      – gimusi
      Nov 29 '18 at 19:56














    2












    2








    2





    $begingroup$

    If I'm not mistaken, the actual sum is
    $$ frac{1 - ln(2)}{2}$$






    share|cite|improve this answer









    $endgroup$



    If I'm not mistaken, the actual sum is
    $$ frac{1 - ln(2)}{2}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 29 '18 at 19:42









    Robert IsraelRobert Israel

    322k23212465




    322k23212465












    • $begingroup$
      Yes that's agree with my result also.
      $endgroup$
      – gimusi
      Nov 29 '18 at 19:56


















    • $begingroup$
      Yes that's agree with my result also.
      $endgroup$
      – gimusi
      Nov 29 '18 at 19:56
















    $begingroup$
    Yes that's agree with my result also.
    $endgroup$
    – gimusi
    Nov 29 '18 at 19:56




    $begingroup$
    Yes that's agree with my result also.
    $endgroup$
    – gimusi
    Nov 29 '18 at 19:56











    2












    $begingroup$

    Let $f(x)=displaystylesum_{n=1}^{infty}left(nlnfrac{n+x}{n-x}-2xright)$ for $xin(-1,1)$. Then
    $$f'(x)=displaystylesum_{n=1}^{infty}frac{2x^2}{n^2-x^2}=1-pi xcotpi x$$
    (termwise differentiation is admissible because of uniform convergence of the latter series in $[-a,a]$ for any $0<a<1$; the second equality is known). Thus,
    $$f(x)=x-frac{1}{pi}int_{0}^{pi x}tcot t,dt=x(1-lnsinpi x)+frac{1}{pi}int_{0}^{pi x}lnsin t,dt.$$
    Your sum is $f(1/2)=(1-ln2)/2$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Let $f(x)=displaystylesum_{n=1}^{infty}left(nlnfrac{n+x}{n-x}-2xright)$ for $xin(-1,1)$. Then
      $$f'(x)=displaystylesum_{n=1}^{infty}frac{2x^2}{n^2-x^2}=1-pi xcotpi x$$
      (termwise differentiation is admissible because of uniform convergence of the latter series in $[-a,a]$ for any $0<a<1$; the second equality is known). Thus,
      $$f(x)=x-frac{1}{pi}int_{0}^{pi x}tcot t,dt=x(1-lnsinpi x)+frac{1}{pi}int_{0}^{pi x}lnsin t,dt.$$
      Your sum is $f(1/2)=(1-ln2)/2$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Let $f(x)=displaystylesum_{n=1}^{infty}left(nlnfrac{n+x}{n-x}-2xright)$ for $xin(-1,1)$. Then
        $$f'(x)=displaystylesum_{n=1}^{infty}frac{2x^2}{n^2-x^2}=1-pi xcotpi x$$
        (termwise differentiation is admissible because of uniform convergence of the latter series in $[-a,a]$ for any $0<a<1$; the second equality is known). Thus,
        $$f(x)=x-frac{1}{pi}int_{0}^{pi x}tcot t,dt=x(1-lnsinpi x)+frac{1}{pi}int_{0}^{pi x}lnsin t,dt.$$
        Your sum is $f(1/2)=(1-ln2)/2$.






        share|cite|improve this answer









        $endgroup$



        Let $f(x)=displaystylesum_{n=1}^{infty}left(nlnfrac{n+x}{n-x}-2xright)$ for $xin(-1,1)$. Then
        $$f'(x)=displaystylesum_{n=1}^{infty}frac{2x^2}{n^2-x^2}=1-pi xcotpi x$$
        (termwise differentiation is admissible because of uniform convergence of the latter series in $[-a,a]$ for any $0<a<1$; the second equality is known). Thus,
        $$f(x)=x-frac{1}{pi}int_{0}^{pi x}tcot t,dt=x(1-lnsinpi x)+frac{1}{pi}int_{0}^{pi x}lnsin t,dt.$$
        Your sum is $f(1/2)=(1-ln2)/2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 19:53









        metamorphymetamorphy

        3,6821621




        3,6821621























            1












            $begingroup$

            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$




            With $ds{N in mathbb{N}_{geq 1}}$:




            begin{align}
            &bbox[#ffd,10px]{sum_{n = 1}^{N}bracks{nlnpars{2n + 1 over 2n - 1} - 1}}
            \[5mm] = &
            sum_{n = 1}^{N}nlnpars{2n + 1} - sum_{n = 1}^{N}nlnpars{2n - 1} - N
            \[5mm] = &
            -N + sum_{n = 0}^{N}nlnpars{2n + 1} -
            sum_{n = 0}^{N - 1}pars{n + 1}lnpars{2n + 1}
            \[5mm] = &
            -N + Nlnpars{2N + 1} -Nlnpars{2} -
            sum_{n = 0}^{N - 1}lnpars{n + {1 over 2}}
            \[5mm] = &
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{prod_{n = 0}^{N - 1}bracks{n + {1 over 2}}}
            \[5mm] = &
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{bracks{N - 1/2}! over Gammapars{1/2}}
            \[5mm] stackrel{mrm{as} N to infty}{sim}&
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{root{2pi}bracks{N - 1/2}^{N}expo{-N + 1/2} over root{pi}}
            \[5mm] = &
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{2^{1/2}N^{N}bracks{1 - {1/2 over N}}^{N}
            expo{-N + 1/2}}
            \[5mm] stackrel{mrm{as} N to infty}{sim} &
            -N + Nlnpars{N + {1 over 2}} -
            bracks{{1 over 2},lnpars{2} + Nlnpars{N} - N}
            \[5mm] = &
            underbrace{Nlnpars{1 + {1 over 2N}}}
            _{ds{stackrel{mrm{as} N to infty}{to} {1 over 2}}}
            - {1 over 2},lnpars{2}label{1}tag{1}
            \[5mm] stackrel{mrm{as} N to infty}{to} &
            bbx{1 - lnpars{2} over 2} approx 0.1534
            end{align}






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              How have you moved from one before last to last line?
              $endgroup$
              – mvxxx
              Nov 29 '18 at 21:16










            • $begingroup$
              @mvxxx I add one more line ( line (1) ) where you can see that the first $displaystyle -N$ cancels with the $displaystyle -left(-Nright)$ at the far right and $displaystyle Nlnleft(N + {1 over 2}right) - Nlnleft(Nright) = Nlnleft(1 + {1 over 2N}right) to {large{1 over 2}}$
              $endgroup$
              – Felix Marin
              Nov 29 '18 at 21:34


















            1












            $begingroup$

            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$




            With $ds{N in mathbb{N}_{geq 1}}$:




            begin{align}
            &bbox[#ffd,10px]{sum_{n = 1}^{N}bracks{nlnpars{2n + 1 over 2n - 1} - 1}}
            \[5mm] = &
            sum_{n = 1}^{N}nlnpars{2n + 1} - sum_{n = 1}^{N}nlnpars{2n - 1} - N
            \[5mm] = &
            -N + sum_{n = 0}^{N}nlnpars{2n + 1} -
            sum_{n = 0}^{N - 1}pars{n + 1}lnpars{2n + 1}
            \[5mm] = &
            -N + Nlnpars{2N + 1} -Nlnpars{2} -
            sum_{n = 0}^{N - 1}lnpars{n + {1 over 2}}
            \[5mm] = &
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{prod_{n = 0}^{N - 1}bracks{n + {1 over 2}}}
            \[5mm] = &
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{bracks{N - 1/2}! over Gammapars{1/2}}
            \[5mm] stackrel{mrm{as} N to infty}{sim}&
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{root{2pi}bracks{N - 1/2}^{N}expo{-N + 1/2} over root{pi}}
            \[5mm] = &
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{2^{1/2}N^{N}bracks{1 - {1/2 over N}}^{N}
            expo{-N + 1/2}}
            \[5mm] stackrel{mrm{as} N to infty}{sim} &
            -N + Nlnpars{N + {1 over 2}} -
            bracks{{1 over 2},lnpars{2} + Nlnpars{N} - N}
            \[5mm] = &
            underbrace{Nlnpars{1 + {1 over 2N}}}
            _{ds{stackrel{mrm{as} N to infty}{to} {1 over 2}}}
            - {1 over 2},lnpars{2}label{1}tag{1}
            \[5mm] stackrel{mrm{as} N to infty}{to} &
            bbx{1 - lnpars{2} over 2} approx 0.1534
            end{align}






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              How have you moved from one before last to last line?
              $endgroup$
              – mvxxx
              Nov 29 '18 at 21:16










            • $begingroup$
              @mvxxx I add one more line ( line (1) ) where you can see that the first $displaystyle -N$ cancels with the $displaystyle -left(-Nright)$ at the far right and $displaystyle Nlnleft(N + {1 over 2}right) - Nlnleft(Nright) = Nlnleft(1 + {1 over 2N}right) to {large{1 over 2}}$
              $endgroup$
              – Felix Marin
              Nov 29 '18 at 21:34
















            1












            1








            1





            $begingroup$

            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$




            With $ds{N in mathbb{N}_{geq 1}}$:




            begin{align}
            &bbox[#ffd,10px]{sum_{n = 1}^{N}bracks{nlnpars{2n + 1 over 2n - 1} - 1}}
            \[5mm] = &
            sum_{n = 1}^{N}nlnpars{2n + 1} - sum_{n = 1}^{N}nlnpars{2n - 1} - N
            \[5mm] = &
            -N + sum_{n = 0}^{N}nlnpars{2n + 1} -
            sum_{n = 0}^{N - 1}pars{n + 1}lnpars{2n + 1}
            \[5mm] = &
            -N + Nlnpars{2N + 1} -Nlnpars{2} -
            sum_{n = 0}^{N - 1}lnpars{n + {1 over 2}}
            \[5mm] = &
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{prod_{n = 0}^{N - 1}bracks{n + {1 over 2}}}
            \[5mm] = &
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{bracks{N - 1/2}! over Gammapars{1/2}}
            \[5mm] stackrel{mrm{as} N to infty}{sim}&
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{root{2pi}bracks{N - 1/2}^{N}expo{-N + 1/2} over root{pi}}
            \[5mm] = &
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{2^{1/2}N^{N}bracks{1 - {1/2 over N}}^{N}
            expo{-N + 1/2}}
            \[5mm] stackrel{mrm{as} N to infty}{sim} &
            -N + Nlnpars{N + {1 over 2}} -
            bracks{{1 over 2},lnpars{2} + Nlnpars{N} - N}
            \[5mm] = &
            underbrace{Nlnpars{1 + {1 over 2N}}}
            _{ds{stackrel{mrm{as} N to infty}{to} {1 over 2}}}
            - {1 over 2},lnpars{2}label{1}tag{1}
            \[5mm] stackrel{mrm{as} N to infty}{to} &
            bbx{1 - lnpars{2} over 2} approx 0.1534
            end{align}






            share|cite|improve this answer











            $endgroup$



            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$




            With $ds{N in mathbb{N}_{geq 1}}$:




            begin{align}
            &bbox[#ffd,10px]{sum_{n = 1}^{N}bracks{nlnpars{2n + 1 over 2n - 1} - 1}}
            \[5mm] = &
            sum_{n = 1}^{N}nlnpars{2n + 1} - sum_{n = 1}^{N}nlnpars{2n - 1} - N
            \[5mm] = &
            -N + sum_{n = 0}^{N}nlnpars{2n + 1} -
            sum_{n = 0}^{N - 1}pars{n + 1}lnpars{2n + 1}
            \[5mm] = &
            -N + Nlnpars{2N + 1} -Nlnpars{2} -
            sum_{n = 0}^{N - 1}lnpars{n + {1 over 2}}
            \[5mm] = &
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{prod_{n = 0}^{N - 1}bracks{n + {1 over 2}}}
            \[5mm] = &
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{bracks{N - 1/2}! over Gammapars{1/2}}
            \[5mm] stackrel{mrm{as} N to infty}{sim}&
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{root{2pi}bracks{N - 1/2}^{N}expo{-N + 1/2} over root{pi}}
            \[5mm] = &
            -N + Nlnpars{N + {1 over 2}} -
            lnpars{2^{1/2}N^{N}bracks{1 - {1/2 over N}}^{N}
            expo{-N + 1/2}}
            \[5mm] stackrel{mrm{as} N to infty}{sim} &
            -N + Nlnpars{N + {1 over 2}} -
            bracks{{1 over 2},lnpars{2} + Nlnpars{N} - N}
            \[5mm] = &
            underbrace{Nlnpars{1 + {1 over 2N}}}
            _{ds{stackrel{mrm{as} N to infty}{to} {1 over 2}}}
            - {1 over 2},lnpars{2}label{1}tag{1}
            \[5mm] stackrel{mrm{as} N to infty}{to} &
            bbx{1 - lnpars{2} over 2} approx 0.1534
            end{align}







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 29 '18 at 21:30

























            answered Nov 29 '18 at 21:00









            Felix MarinFelix Marin

            67.8k7107142




            67.8k7107142












            • $begingroup$
              How have you moved from one before last to last line?
              $endgroup$
              – mvxxx
              Nov 29 '18 at 21:16










            • $begingroup$
              @mvxxx I add one more line ( line (1) ) where you can see that the first $displaystyle -N$ cancels with the $displaystyle -left(-Nright)$ at the far right and $displaystyle Nlnleft(N + {1 over 2}right) - Nlnleft(Nright) = Nlnleft(1 + {1 over 2N}right) to {large{1 over 2}}$
              $endgroup$
              – Felix Marin
              Nov 29 '18 at 21:34




















            • $begingroup$
              How have you moved from one before last to last line?
              $endgroup$
              – mvxxx
              Nov 29 '18 at 21:16










            • $begingroup$
              @mvxxx I add one more line ( line (1) ) where you can see that the first $displaystyle -N$ cancels with the $displaystyle -left(-Nright)$ at the far right and $displaystyle Nlnleft(N + {1 over 2}right) - Nlnleft(Nright) = Nlnleft(1 + {1 over 2N}right) to {large{1 over 2}}$
              $endgroup$
              – Felix Marin
              Nov 29 '18 at 21:34


















            $begingroup$
            How have you moved from one before last to last line?
            $endgroup$
            – mvxxx
            Nov 29 '18 at 21:16




            $begingroup$
            How have you moved from one before last to last line?
            $endgroup$
            – mvxxx
            Nov 29 '18 at 21:16












            $begingroup$
            @mvxxx I add one more line ( line (1) ) where you can see that the first $displaystyle -N$ cancels with the $displaystyle -left(-Nright)$ at the far right and $displaystyle Nlnleft(N + {1 over 2}right) - Nlnleft(Nright) = Nlnleft(1 + {1 over 2N}right) to {large{1 over 2}}$
            $endgroup$
            – Felix Marin
            Nov 29 '18 at 21:34






            $begingroup$
            @mvxxx I add one more line ( line (1) ) where you can see that the first $displaystyle -N$ cancels with the $displaystyle -left(-Nright)$ at the far right and $displaystyle Nlnleft(N + {1 over 2}right) - Nlnleft(Nright) = Nlnleft(1 + {1 over 2N}right) to {large{1 over 2}}$
            $endgroup$
            – Felix Marin
            Nov 29 '18 at 21:34




















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