sum of this series: $sum_{n=1}^{infty}frac{1}{4n^2-1}$












12












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I am trying to calculate the sum of this infinite series after having read the series chapter of my textbook: $$sum_{n=1}^{infty}frac{1}{4n^2-1}$$



my steps:



$$sum_{n=1}^{infty}frac{1}{4n^2-1}=sum_{n=1}^{infty}frac{2}{4n^2-1}-sum_{n=1}^{infty}frac{1}{4n^2-1}=..help..=sum$$



I am lacking some important properties, I feel I am coming to the right step and cannot spit that out..






sequences-and-series summation







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  • 1




    $begingroup$
    Hint: Write your rational functions as a linear combination of fractions with denominator being a linear function.
    $endgroup$
    – Ofir
    Dec 26 '12 at 10:47








  • 1




    $begingroup$
    see math.stackexchange.com/questions/238728/finding-summation/…
    $endgroup$
    – MathOverview
    Dec 26 '12 at 10:48






  • 1




    $begingroup$
    @doniyor taking $displaystyle lim_{nto infty} left(frac{1}{2} - frac{1}{4n+2}right)$ in the result posted in Elias's link you get $frac{1}{2}$
    $endgroup$
    – Rustyn
    Dec 26 '12 at 10:55


















12












$begingroup$


I am trying to calculate the sum of this infinite series after having read the series chapter of my textbook: $$sum_{n=1}^{infty}frac{1}{4n^2-1}$$



my steps:



$$sum_{n=1}^{infty}frac{1}{4n^2-1}=sum_{n=1}^{infty}frac{2}{4n^2-1}-sum_{n=1}^{infty}frac{1}{4n^2-1}=..help..=sum$$



I am lacking some important properties, I feel I am coming to the right step and cannot spit that out..






sequences-and-series summation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: Write your rational functions as a linear combination of fractions with denominator being a linear function.
    $endgroup$
    – Ofir
    Dec 26 '12 at 10:47








  • 1




    $begingroup$
    see math.stackexchange.com/questions/238728/finding-summation/…
    $endgroup$
    – MathOverview
    Dec 26 '12 at 10:48






  • 1




    $begingroup$
    @doniyor taking $displaystyle lim_{nto infty} left(frac{1}{2} - frac{1}{4n+2}right)$ in the result posted in Elias's link you get $frac{1}{2}$
    $endgroup$
    – Rustyn
    Dec 26 '12 at 10:55
















12












12








12


7



$begingroup$


I am trying to calculate the sum of this infinite series after having read the series chapter of my textbook: $$sum_{n=1}^{infty}frac{1}{4n^2-1}$$



my steps:



$$sum_{n=1}^{infty}frac{1}{4n^2-1}=sum_{n=1}^{infty}frac{2}{4n^2-1}-sum_{n=1}^{infty}frac{1}{4n^2-1}=..help..=sum$$



I am lacking some important properties, I feel I am coming to the right step and cannot spit that out..






sequences-and-series summation







share|cite|improve this question











$endgroup$




I am trying to calculate the sum of this infinite series after having read the series chapter of my textbook: $$sum_{n=1}^{infty}frac{1}{4n^2-1}$$



my steps:



$$sum_{n=1}^{infty}frac{1}{4n^2-1}=sum_{n=1}^{infty}frac{2}{4n^2-1}-sum_{n=1}^{infty}frac{1}{4n^2-1}=..help..=sum$$



I am lacking some important properties, I feel I am coming to the right step and cannot spit that out..






sequences-and-series summation




sequences-and-series summation






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share|cite|improve this question













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share|cite|improve this question








edited Jul 21 '16 at 14:18









Martin Sleziak

45k10122277




45k10122277










asked Dec 26 '12 at 10:43









doniyordoniyor

1,66242348




1,66242348








  • 1




    $begingroup$
    Hint: Write your rational functions as a linear combination of fractions with denominator being a linear function.
    $endgroup$
    – Ofir
    Dec 26 '12 at 10:47








  • 1




    $begingroup$
    see math.stackexchange.com/questions/238728/finding-summation/…
    $endgroup$
    – MathOverview
    Dec 26 '12 at 10:48






  • 1




    $begingroup$
    @doniyor taking $displaystyle lim_{nto infty} left(frac{1}{2} - frac{1}{4n+2}right)$ in the result posted in Elias's link you get $frac{1}{2}$
    $endgroup$
    – Rustyn
    Dec 26 '12 at 10:55
















  • 1




    $begingroup$
    Hint: Write your rational functions as a linear combination of fractions with denominator being a linear function.
    $endgroup$
    – Ofir
    Dec 26 '12 at 10:47








  • 1




    $begingroup$
    see math.stackexchange.com/questions/238728/finding-summation/…
    $endgroup$
    – MathOverview
    Dec 26 '12 at 10:48






  • 1




    $begingroup$
    @doniyor taking $displaystyle lim_{nto infty} left(frac{1}{2} - frac{1}{4n+2}right)$ in the result posted in Elias's link you get $frac{1}{2}$
    $endgroup$
    – Rustyn
    Dec 26 '12 at 10:55










1




1




$begingroup$
Hint: Write your rational functions as a linear combination of fractions with denominator being a linear function.
$endgroup$
– Ofir
Dec 26 '12 at 10:47






$begingroup$
Hint: Write your rational functions as a linear combination of fractions with denominator being a linear function.
$endgroup$
– Ofir
Dec 26 '12 at 10:47






1




1




$begingroup$
see math.stackexchange.com/questions/238728/finding-summation/…
$endgroup$
– MathOverview
Dec 26 '12 at 10:48




$begingroup$
see math.stackexchange.com/questions/238728/finding-summation/…
$endgroup$
– MathOverview
Dec 26 '12 at 10:48




1




1




$begingroup$
@doniyor taking $displaystyle lim_{nto infty} left(frac{1}{2} - frac{1}{4n+2}right)$ in the result posted in Elias's link you get $frac{1}{2}$
$endgroup$
– Rustyn
Dec 26 '12 at 10:55






$begingroup$
@doniyor taking $displaystyle lim_{nto infty} left(frac{1}{2} - frac{1}{4n+2}right)$ in the result posted in Elias's link you get $frac{1}{2}$
$endgroup$
– Rustyn
Dec 26 '12 at 10:55












5 Answers
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Hint: Partial Fraction decomposition:$$frac{1}{4n^2-1}=frac{1}{(2n-1)(2n+1)}=frac12[frac{1}{2n-1}-frac{1}{2n+1}]$$
You must then compute the closed form of
$$sum_{n=1}^k[frac{1}{2n-1}-frac{1}{2n+1}]$$
Can you do that?
Note that
$$sum_{n=1}^kfrac{1}{2n-1}=frac11+frac13+...+frac1{2k-1}=frac1{2cdot 0+1}+frac1{2cdot 1+1}+...+frac1{2(k-1)+1}=sum_{n=0}^{k-1}frac{1}{2n+1}=sum_{n=1}^{k}frac1{2n+1}+frac{1}{2cdot 0+1}-frac1{2k+1}$$






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  • $begingroup$
    can you pls expand this step? $sum_{n=1}^kfrac{1}{2n-1}=sum_{n=0}^{k-1}frac{1}{2n+1}$
    $endgroup$
    – doniyor
    Dec 26 '12 at 11:29










  • $begingroup$
    @doniyor Sure I can.
    $endgroup$
    – Nameless
    Dec 26 '12 at 11:29



















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Note $frac{1}{4n^2-1}=frac{1}{(2n+1)(2n-1)}={frac{1}{2}}timesfrac{(2n+1)-(2n-1)}{(2n+1)(2n-1)}={frac{1}{2}}times[frac{1}{2n-1}-frac{1}{2n+1}]$ for $ninmathbb N$



Let for $kinmathbb N,$ $S_k=sum_{n=1}^{k}frac{1}{4n^2-1}$ $implies S_k={frac{1}{2}}sum_{n=1}^{k}[frac{1}{2n-1}-frac{1}{2n+1}].$ Thus for $k=1,2,...$



$S_1={frac{1}{2}}sum_{n=1}^{1}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}(1-frac{1}{3})$



$S_2={frac{1}{2}}sum_{n=1}^{2}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}[(1-frac{1}{3})+(frac{1}{3}-frac{1}{5})]=frac{1}{2}(1-frac{1}{5})$



$S_3={frac{1}{2}}sum_{n=1}^{3}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}[(1-frac{1}{3})+(frac{1}{3}-frac{1}{5})+(frac{1}{5}-frac{1}{7})]=frac{1}{2}(1-frac{1}{7})$



...



$S_k=frac{1}{2}(1-frac{1}{2k+1})$



$impliessum_{n=1}^{infty}frac{1}{4n^2-1}=lim_{ktoinfty}S_k=frac{1}{2}.$






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  • $begingroup$
    great, thanks Sugata
    $endgroup$
    – doniyor
    Dec 26 '12 at 13:39










  • $begingroup$
    You're most welcome.
    $endgroup$
    – Sugata Adhya
    Dec 26 '12 at 13:41



















5












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Or we want to compute it fastly and use the formula
$$sum_{k=1}^infty frac1{k^2-x^2}=frac1{2x^2}-frac{picot,pi x}{2x}$$
where $x=frac{1}{2}$ because
$$sum_{k=1}^{infty}frac{1}{4k^2-1}=frac{1}{4}sum_{k=1}^{infty}frac{1}{k^2-left(frac{1}{2}right)^2}$$
Here you may find more information about this precious way.






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  • $begingroup$
    @doniyor: you're welcome!
    $endgroup$
    – user 1357113
    Dec 26 '12 at 20:52



















3












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Hint: Work on $S_n=sum_{k=1}^nfrac{1}{4k^2-1}$ and take its limit when $ntoinfty$. Note that $$frac{1}{4n^2-1}=frac{1}{2(2n-1)}-frac{1}{2(2n+1)}$$






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  • 1




    $begingroup$
    Why don't you experts leave such problems for beginners like us ? :(
    $endgroup$
    – Sugata Adhya
    Dec 26 '12 at 13:40










  • $begingroup$
    @SugataAdhya: Dear Sugata, I am not an expert. I am a teacher and like to help others in Maths. That's it. We are here to help eachother in learning Maths well. :-)
    $endgroup$
    – mrs
    Dec 26 '12 at 16:13










  • $begingroup$
    :), yeah, but you guys are doing wonderful job here.. thanks again for all you
    $endgroup$
    – doniyor
    Dec 26 '12 at 20:50










  • $begingroup$
    Yes, as dear doniyor says, you're doing a wonderful job here! +
    $endgroup$
    – Namaste
    Mar 2 '13 at 2:45



















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This is an easy problem by using Fourier's serie of $|sin(x)|$. So,
$|sin(x)|=dfrac2pi-dfrac4pisum_{n=1}^{infty}dfrac{cos(2nx)}{4n^2-1} $. By taking $x=0$, we obtain:



$0=dfrac2pi-dfrac4pisum_{n=1}^{infty}dfrac{1}{4n^2-1} $.



So,



$sum_{n=1}^{infty}dfrac{1}{4n^2-1}=dfrac12$






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    5 Answers
    5






    active

    oldest

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    5 Answers
    5






    active

    oldest

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    active

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    5












    $begingroup$

    Hint: Partial Fraction decomposition:$$frac{1}{4n^2-1}=frac{1}{(2n-1)(2n+1)}=frac12[frac{1}{2n-1}-frac{1}{2n+1}]$$
    You must then compute the closed form of
    $$sum_{n=1}^k[frac{1}{2n-1}-frac{1}{2n+1}]$$
    Can you do that?
    Note that
    $$sum_{n=1}^kfrac{1}{2n-1}=frac11+frac13+...+frac1{2k-1}=frac1{2cdot 0+1}+frac1{2cdot 1+1}+...+frac1{2(k-1)+1}=sum_{n=0}^{k-1}frac{1}{2n+1}=sum_{n=1}^{k}frac1{2n+1}+frac{1}{2cdot 0+1}-frac1{2k+1}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      can you pls expand this step? $sum_{n=1}^kfrac{1}{2n-1}=sum_{n=0}^{k-1}frac{1}{2n+1}$
      $endgroup$
      – doniyor
      Dec 26 '12 at 11:29










    • $begingroup$
      @doniyor Sure I can.
      $endgroup$
      – Nameless
      Dec 26 '12 at 11:29
















    5












    $begingroup$

    Hint: Partial Fraction decomposition:$$frac{1}{4n^2-1}=frac{1}{(2n-1)(2n+1)}=frac12[frac{1}{2n-1}-frac{1}{2n+1}]$$
    You must then compute the closed form of
    $$sum_{n=1}^k[frac{1}{2n-1}-frac{1}{2n+1}]$$
    Can you do that?
    Note that
    $$sum_{n=1}^kfrac{1}{2n-1}=frac11+frac13+...+frac1{2k-1}=frac1{2cdot 0+1}+frac1{2cdot 1+1}+...+frac1{2(k-1)+1}=sum_{n=0}^{k-1}frac{1}{2n+1}=sum_{n=1}^{k}frac1{2n+1}+frac{1}{2cdot 0+1}-frac1{2k+1}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      can you pls expand this step? $sum_{n=1}^kfrac{1}{2n-1}=sum_{n=0}^{k-1}frac{1}{2n+1}$
      $endgroup$
      – doniyor
      Dec 26 '12 at 11:29










    • $begingroup$
      @doniyor Sure I can.
      $endgroup$
      – Nameless
      Dec 26 '12 at 11:29














    5












    5








    5





    $begingroup$

    Hint: Partial Fraction decomposition:$$frac{1}{4n^2-1}=frac{1}{(2n-1)(2n+1)}=frac12[frac{1}{2n-1}-frac{1}{2n+1}]$$
    You must then compute the closed form of
    $$sum_{n=1}^k[frac{1}{2n-1}-frac{1}{2n+1}]$$
    Can you do that?
    Note that
    $$sum_{n=1}^kfrac{1}{2n-1}=frac11+frac13+...+frac1{2k-1}=frac1{2cdot 0+1}+frac1{2cdot 1+1}+...+frac1{2(k-1)+1}=sum_{n=0}^{k-1}frac{1}{2n+1}=sum_{n=1}^{k}frac1{2n+1}+frac{1}{2cdot 0+1}-frac1{2k+1}$$






    share|cite|improve this answer











    $endgroup$



    Hint: Partial Fraction decomposition:$$frac{1}{4n^2-1}=frac{1}{(2n-1)(2n+1)}=frac12[frac{1}{2n-1}-frac{1}{2n+1}]$$
    You must then compute the closed form of
    $$sum_{n=1}^k[frac{1}{2n-1}-frac{1}{2n+1}]$$
    Can you do that?
    Note that
    $$sum_{n=1}^kfrac{1}{2n-1}=frac11+frac13+...+frac1{2k-1}=frac1{2cdot 0+1}+frac1{2cdot 1+1}+...+frac1{2(k-1)+1}=sum_{n=0}^{k-1}frac{1}{2n+1}=sum_{n=1}^{k}frac1{2n+1}+frac{1}{2cdot 0+1}-frac1{2k+1}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 26 '12 at 11:31

























    answered Dec 26 '12 at 10:51









    NamelessNameless

    10.5k12255




    10.5k12255












    • $begingroup$
      can you pls expand this step? $sum_{n=1}^kfrac{1}{2n-1}=sum_{n=0}^{k-1}frac{1}{2n+1}$
      $endgroup$
      – doniyor
      Dec 26 '12 at 11:29










    • $begingroup$
      @doniyor Sure I can.
      $endgroup$
      – Nameless
      Dec 26 '12 at 11:29


















    • $begingroup$
      can you pls expand this step? $sum_{n=1}^kfrac{1}{2n-1}=sum_{n=0}^{k-1}frac{1}{2n+1}$
      $endgroup$
      – doniyor
      Dec 26 '12 at 11:29










    • $begingroup$
      @doniyor Sure I can.
      $endgroup$
      – Nameless
      Dec 26 '12 at 11:29
















    $begingroup$
    can you pls expand this step? $sum_{n=1}^kfrac{1}{2n-1}=sum_{n=0}^{k-1}frac{1}{2n+1}$
    $endgroup$
    – doniyor
    Dec 26 '12 at 11:29




    $begingroup$
    can you pls expand this step? $sum_{n=1}^kfrac{1}{2n-1}=sum_{n=0}^{k-1}frac{1}{2n+1}$
    $endgroup$
    – doniyor
    Dec 26 '12 at 11:29












    $begingroup$
    @doniyor Sure I can.
    $endgroup$
    – Nameless
    Dec 26 '12 at 11:29




    $begingroup$
    @doniyor Sure I can.
    $endgroup$
    – Nameless
    Dec 26 '12 at 11:29











    14












    $begingroup$

    Note $frac{1}{4n^2-1}=frac{1}{(2n+1)(2n-1)}={frac{1}{2}}timesfrac{(2n+1)-(2n-1)}{(2n+1)(2n-1)}={frac{1}{2}}times[frac{1}{2n-1}-frac{1}{2n+1}]$ for $ninmathbb N$



    Let for $kinmathbb N,$ $S_k=sum_{n=1}^{k}frac{1}{4n^2-1}$ $implies S_k={frac{1}{2}}sum_{n=1}^{k}[frac{1}{2n-1}-frac{1}{2n+1}].$ Thus for $k=1,2,...$



    $S_1={frac{1}{2}}sum_{n=1}^{1}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}(1-frac{1}{3})$



    $S_2={frac{1}{2}}sum_{n=1}^{2}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}[(1-frac{1}{3})+(frac{1}{3}-frac{1}{5})]=frac{1}{2}(1-frac{1}{5})$



    $S_3={frac{1}{2}}sum_{n=1}^{3}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}[(1-frac{1}{3})+(frac{1}{3}-frac{1}{5})+(frac{1}{5}-frac{1}{7})]=frac{1}{2}(1-frac{1}{7})$



    ...



    $S_k=frac{1}{2}(1-frac{1}{2k+1})$



    $impliessum_{n=1}^{infty}frac{1}{4n^2-1}=lim_{ktoinfty}S_k=frac{1}{2}.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      great, thanks Sugata
      $endgroup$
      – doniyor
      Dec 26 '12 at 13:39










    • $begingroup$
      You're most welcome.
      $endgroup$
      – Sugata Adhya
      Dec 26 '12 at 13:41
















    14












    $begingroup$

    Note $frac{1}{4n^2-1}=frac{1}{(2n+1)(2n-1)}={frac{1}{2}}timesfrac{(2n+1)-(2n-1)}{(2n+1)(2n-1)}={frac{1}{2}}times[frac{1}{2n-1}-frac{1}{2n+1}]$ for $ninmathbb N$



    Let for $kinmathbb N,$ $S_k=sum_{n=1}^{k}frac{1}{4n^2-1}$ $implies S_k={frac{1}{2}}sum_{n=1}^{k}[frac{1}{2n-1}-frac{1}{2n+1}].$ Thus for $k=1,2,...$



    $S_1={frac{1}{2}}sum_{n=1}^{1}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}(1-frac{1}{3})$



    $S_2={frac{1}{2}}sum_{n=1}^{2}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}[(1-frac{1}{3})+(frac{1}{3}-frac{1}{5})]=frac{1}{2}(1-frac{1}{5})$



    $S_3={frac{1}{2}}sum_{n=1}^{3}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}[(1-frac{1}{3})+(frac{1}{3}-frac{1}{5})+(frac{1}{5}-frac{1}{7})]=frac{1}{2}(1-frac{1}{7})$



    ...



    $S_k=frac{1}{2}(1-frac{1}{2k+1})$



    $impliessum_{n=1}^{infty}frac{1}{4n^2-1}=lim_{ktoinfty}S_k=frac{1}{2}.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      great, thanks Sugata
      $endgroup$
      – doniyor
      Dec 26 '12 at 13:39










    • $begingroup$
      You're most welcome.
      $endgroup$
      – Sugata Adhya
      Dec 26 '12 at 13:41














    14












    14








    14





    $begingroup$

    Note $frac{1}{4n^2-1}=frac{1}{(2n+1)(2n-1)}={frac{1}{2}}timesfrac{(2n+1)-(2n-1)}{(2n+1)(2n-1)}={frac{1}{2}}times[frac{1}{2n-1}-frac{1}{2n+1}]$ for $ninmathbb N$



    Let for $kinmathbb N,$ $S_k=sum_{n=1}^{k}frac{1}{4n^2-1}$ $implies S_k={frac{1}{2}}sum_{n=1}^{k}[frac{1}{2n-1}-frac{1}{2n+1}].$ Thus for $k=1,2,...$



    $S_1={frac{1}{2}}sum_{n=1}^{1}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}(1-frac{1}{3})$



    $S_2={frac{1}{2}}sum_{n=1}^{2}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}[(1-frac{1}{3})+(frac{1}{3}-frac{1}{5})]=frac{1}{2}(1-frac{1}{5})$



    $S_3={frac{1}{2}}sum_{n=1}^{3}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}[(1-frac{1}{3})+(frac{1}{3}-frac{1}{5})+(frac{1}{5}-frac{1}{7})]=frac{1}{2}(1-frac{1}{7})$



    ...



    $S_k=frac{1}{2}(1-frac{1}{2k+1})$



    $impliessum_{n=1}^{infty}frac{1}{4n^2-1}=lim_{ktoinfty}S_k=frac{1}{2}.$






    share|cite|improve this answer









    $endgroup$



    Note $frac{1}{4n^2-1}=frac{1}{(2n+1)(2n-1)}={frac{1}{2}}timesfrac{(2n+1)-(2n-1)}{(2n+1)(2n-1)}={frac{1}{2}}times[frac{1}{2n-1}-frac{1}{2n+1}]$ for $ninmathbb N$



    Let for $kinmathbb N,$ $S_k=sum_{n=1}^{k}frac{1}{4n^2-1}$ $implies S_k={frac{1}{2}}sum_{n=1}^{k}[frac{1}{2n-1}-frac{1}{2n+1}].$ Thus for $k=1,2,...$



    $S_1={frac{1}{2}}sum_{n=1}^{1}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}(1-frac{1}{3})$



    $S_2={frac{1}{2}}sum_{n=1}^{2}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}[(1-frac{1}{3})+(frac{1}{3}-frac{1}{5})]=frac{1}{2}(1-frac{1}{5})$



    $S_3={frac{1}{2}}sum_{n=1}^{3}[frac{1}{2n-1}-frac{1}{2n+1}]=frac{1}{2}[(1-frac{1}{3})+(frac{1}{3}-frac{1}{5})+(frac{1}{5}-frac{1}{7})]=frac{1}{2}(1-frac{1}{7})$



    ...



    $S_k=frac{1}{2}(1-frac{1}{2k+1})$



    $impliessum_{n=1}^{infty}frac{1}{4n^2-1}=lim_{ktoinfty}S_k=frac{1}{2}.$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 26 '12 at 13:30









    Sugata AdhyaSugata Adhya

    3,0221224




    3,0221224












    • $begingroup$
      great, thanks Sugata
      $endgroup$
      – doniyor
      Dec 26 '12 at 13:39










    • $begingroup$
      You're most welcome.
      $endgroup$
      – Sugata Adhya
      Dec 26 '12 at 13:41


















    • $begingroup$
      great, thanks Sugata
      $endgroup$
      – doniyor
      Dec 26 '12 at 13:39










    • $begingroup$
      You're most welcome.
      $endgroup$
      – Sugata Adhya
      Dec 26 '12 at 13:41
















    $begingroup$
    great, thanks Sugata
    $endgroup$
    – doniyor
    Dec 26 '12 at 13:39




    $begingroup$
    great, thanks Sugata
    $endgroup$
    – doniyor
    Dec 26 '12 at 13:39












    $begingroup$
    You're most welcome.
    $endgroup$
    – Sugata Adhya
    Dec 26 '12 at 13:41




    $begingroup$
    You're most welcome.
    $endgroup$
    – Sugata Adhya
    Dec 26 '12 at 13:41











    5












    $begingroup$

    Or we want to compute it fastly and use the formula
    $$sum_{k=1}^infty frac1{k^2-x^2}=frac1{2x^2}-frac{picot,pi x}{2x}$$
    where $x=frac{1}{2}$ because
    $$sum_{k=1}^{infty}frac{1}{4k^2-1}=frac{1}{4}sum_{k=1}^{infty}frac{1}{k^2-left(frac{1}{2}right)^2}$$
    Here you may find more information about this precious way.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @doniyor: you're welcome!
      $endgroup$
      – user 1357113
      Dec 26 '12 at 20:52
















    5












    $begingroup$

    Or we want to compute it fastly and use the formula
    $$sum_{k=1}^infty frac1{k^2-x^2}=frac1{2x^2}-frac{picot,pi x}{2x}$$
    where $x=frac{1}{2}$ because
    $$sum_{k=1}^{infty}frac{1}{4k^2-1}=frac{1}{4}sum_{k=1}^{infty}frac{1}{k^2-left(frac{1}{2}right)^2}$$
    Here you may find more information about this precious way.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @doniyor: you're welcome!
      $endgroup$
      – user 1357113
      Dec 26 '12 at 20:52














    5












    5








    5





    $begingroup$

    Or we want to compute it fastly and use the formula
    $$sum_{k=1}^infty frac1{k^2-x^2}=frac1{2x^2}-frac{picot,pi x}{2x}$$
    where $x=frac{1}{2}$ because
    $$sum_{k=1}^{infty}frac{1}{4k^2-1}=frac{1}{4}sum_{k=1}^{infty}frac{1}{k^2-left(frac{1}{2}right)^2}$$
    Here you may find more information about this precious way.






    share|cite|improve this answer











    $endgroup$



    Or we want to compute it fastly and use the formula
    $$sum_{k=1}^infty frac1{k^2-x^2}=frac1{2x^2}-frac{picot,pi x}{2x}$$
    where $x=frac{1}{2}$ because
    $$sum_{k=1}^{infty}frac{1}{4k^2-1}=frac{1}{4}sum_{k=1}^{infty}frac{1}{k^2-left(frac{1}{2}right)^2}$$
    Here you may find more information about this precious way.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 13 '17 at 12:21









    Community♦

    1




    1










    answered Dec 26 '12 at 19:57









    user 1357113user 1357113

    22.5k878227




    22.5k878227












    • $begingroup$
      @doniyor: you're welcome!
      $endgroup$
      – user 1357113
      Dec 26 '12 at 20:52


















    • $begingroup$
      @doniyor: you're welcome!
      $endgroup$
      – user 1357113
      Dec 26 '12 at 20:52
















    $begingroup$
    @doniyor: you're welcome!
    $endgroup$
    – user 1357113
    Dec 26 '12 at 20:52




    $begingroup$
    @doniyor: you're welcome!
    $endgroup$
    – user 1357113
    Dec 26 '12 at 20:52











    3












    $begingroup$

    Hint: Work on $S_n=sum_{k=1}^nfrac{1}{4k^2-1}$ and take its limit when $ntoinfty$. Note that $$frac{1}{4n^2-1}=frac{1}{2(2n-1)}-frac{1}{2(2n+1)}$$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Why don't you experts leave such problems for beginners like us ? :(
      $endgroup$
      – Sugata Adhya
      Dec 26 '12 at 13:40










    • $begingroup$
      @SugataAdhya: Dear Sugata, I am not an expert. I am a teacher and like to help others in Maths. That's it. We are here to help eachother in learning Maths well. :-)
      $endgroup$
      – mrs
      Dec 26 '12 at 16:13










    • $begingroup$
      :), yeah, but you guys are doing wonderful job here.. thanks again for all you
      $endgroup$
      – doniyor
      Dec 26 '12 at 20:50










    • $begingroup$
      Yes, as dear doniyor says, you're doing a wonderful job here! +
      $endgroup$
      – Namaste
      Mar 2 '13 at 2:45
















    3












    $begingroup$

    Hint: Work on $S_n=sum_{k=1}^nfrac{1}{4k^2-1}$ and take its limit when $ntoinfty$. Note that $$frac{1}{4n^2-1}=frac{1}{2(2n-1)}-frac{1}{2(2n+1)}$$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Why don't you experts leave such problems for beginners like us ? :(
      $endgroup$
      – Sugata Adhya
      Dec 26 '12 at 13:40










    • $begingroup$
      @SugataAdhya: Dear Sugata, I am not an expert. I am a teacher and like to help others in Maths. That's it. We are here to help eachother in learning Maths well. :-)
      $endgroup$
      – mrs
      Dec 26 '12 at 16:13










    • $begingroup$
      :), yeah, but you guys are doing wonderful job here.. thanks again for all you
      $endgroup$
      – doniyor
      Dec 26 '12 at 20:50










    • $begingroup$
      Yes, as dear doniyor says, you're doing a wonderful job here! +
      $endgroup$
      – Namaste
      Mar 2 '13 at 2:45














    3












    3








    3





    $begingroup$

    Hint: Work on $S_n=sum_{k=1}^nfrac{1}{4k^2-1}$ and take its limit when $ntoinfty$. Note that $$frac{1}{4n^2-1}=frac{1}{2(2n-1)}-frac{1}{2(2n+1)}$$






    share|cite|improve this answer









    $endgroup$



    Hint: Work on $S_n=sum_{k=1}^nfrac{1}{4k^2-1}$ and take its limit when $ntoinfty$. Note that $$frac{1}{4n^2-1}=frac{1}{2(2n-1)}-frac{1}{2(2n+1)}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 26 '12 at 10:50









    mrsmrs

    1




    1








    • 1




      $begingroup$
      Why don't you experts leave such problems for beginners like us ? :(
      $endgroup$
      – Sugata Adhya
      Dec 26 '12 at 13:40










    • $begingroup$
      @SugataAdhya: Dear Sugata, I am not an expert. I am a teacher and like to help others in Maths. That's it. We are here to help eachother in learning Maths well. :-)
      $endgroup$
      – mrs
      Dec 26 '12 at 16:13










    • $begingroup$
      :), yeah, but you guys are doing wonderful job here.. thanks again for all you
      $endgroup$
      – doniyor
      Dec 26 '12 at 20:50










    • $begingroup$
      Yes, as dear doniyor says, you're doing a wonderful job here! +
      $endgroup$
      – Namaste
      Mar 2 '13 at 2:45














    • 1




      $begingroup$
      Why don't you experts leave such problems for beginners like us ? :(
      $endgroup$
      – Sugata Adhya
      Dec 26 '12 at 13:40










    • $begingroup$
      @SugataAdhya: Dear Sugata, I am not an expert. I am a teacher and like to help others in Maths. That's it. We are here to help eachother in learning Maths well. :-)
      $endgroup$
      – mrs
      Dec 26 '12 at 16:13










    • $begingroup$
      :), yeah, but you guys are doing wonderful job here.. thanks again for all you
      $endgroup$
      – doniyor
      Dec 26 '12 at 20:50










    • $begingroup$
      Yes, as dear doniyor says, you're doing a wonderful job here! +
      $endgroup$
      – Namaste
      Mar 2 '13 at 2:45








    1




    1




    $begingroup$
    Why don't you experts leave such problems for beginners like us ? :(
    $endgroup$
    – Sugata Adhya
    Dec 26 '12 at 13:40




    $begingroup$
    Why don't you experts leave such problems for beginners like us ? :(
    $endgroup$
    – Sugata Adhya
    Dec 26 '12 at 13:40












    $begingroup$
    @SugataAdhya: Dear Sugata, I am not an expert. I am a teacher and like to help others in Maths. That's it. We are here to help eachother in learning Maths well. :-)
    $endgroup$
    – mrs
    Dec 26 '12 at 16:13




    $begingroup$
    @SugataAdhya: Dear Sugata, I am not an expert. I am a teacher and like to help others in Maths. That's it. We are here to help eachother in learning Maths well. :-)
    $endgroup$
    – mrs
    Dec 26 '12 at 16:13












    $begingroup$
    :), yeah, but you guys are doing wonderful job here.. thanks again for all you
    $endgroup$
    – doniyor
    Dec 26 '12 at 20:50




    $begingroup$
    :), yeah, but you guys are doing wonderful job here.. thanks again for all you
    $endgroup$
    – doniyor
    Dec 26 '12 at 20:50












    $begingroup$
    Yes, as dear doniyor says, you're doing a wonderful job here! +
    $endgroup$
    – Namaste
    Mar 2 '13 at 2:45




    $begingroup$
    Yes, as dear doniyor says, you're doing a wonderful job here! +
    $endgroup$
    – Namaste
    Mar 2 '13 at 2:45











    0












    $begingroup$

    This is an easy problem by using Fourier's serie of $|sin(x)|$. So,
    $|sin(x)|=dfrac2pi-dfrac4pisum_{n=1}^{infty}dfrac{cos(2nx)}{4n^2-1} $. By taking $x=0$, we obtain:



    $0=dfrac2pi-dfrac4pisum_{n=1}^{infty}dfrac{1}{4n^2-1} $.



    So,



    $sum_{n=1}^{infty}dfrac{1}{4n^2-1}=dfrac12$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      This is an easy problem by using Fourier's serie of $|sin(x)|$. So,
      $|sin(x)|=dfrac2pi-dfrac4pisum_{n=1}^{infty}dfrac{cos(2nx)}{4n^2-1} $. By taking $x=0$, we obtain:



      $0=dfrac2pi-dfrac4pisum_{n=1}^{infty}dfrac{1}{4n^2-1} $.



      So,



      $sum_{n=1}^{infty}dfrac{1}{4n^2-1}=dfrac12$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        This is an easy problem by using Fourier's serie of $|sin(x)|$. So,
        $|sin(x)|=dfrac2pi-dfrac4pisum_{n=1}^{infty}dfrac{cos(2nx)}{4n^2-1} $. By taking $x=0$, we obtain:



        $0=dfrac2pi-dfrac4pisum_{n=1}^{infty}dfrac{1}{4n^2-1} $.



        So,



        $sum_{n=1}^{infty}dfrac{1}{4n^2-1}=dfrac12$






        share|cite|improve this answer









        $endgroup$



        This is an easy problem by using Fourier's serie of $|sin(x)|$. So,
        $|sin(x)|=dfrac2pi-dfrac4pisum_{n=1}^{infty}dfrac{cos(2nx)}{4n^2-1} $. By taking $x=0$, we obtain:



        $0=dfrac2pi-dfrac4pisum_{n=1}^{infty}dfrac{1}{4n^2-1} $.



        So,



        $sum_{n=1}^{infty}dfrac{1}{4n^2-1}=dfrac12$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 18:13









        Toni Van Hul MirallesToni Van Hul Miralles

        1




        1






























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