Sum of $sum_{n=1}^inftyfrac{ln(n) -ln(n+1)}{n+1}$












0












$begingroup$


The sum series exercise started as: $$sum_{n=1}^inftyfrac{lnleft(frac{n^n}{left(n+1right)^n}right)}{n(n+1)} = sum_{n=1}^infty frac{nlnfrac{n}{n+1}}{n(n+1)} = sum_{n=1}^infty frac{ln(n) - ln(n+1)}{n+1}$$
Looking into calculating it.
A quick hint will help me out!






sequences-and-series closed-form







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please what is the question?
    $endgroup$
    – Olivier Oloa
    Dec 31 '18 at 12:02












  • $begingroup$
    Oh, sorry. I am looking into calculating it.
    $endgroup$
    – Theodossis Papadopoulos
    Dec 31 '18 at 12:05










  • $begingroup$
    A numerical approximation or a closed form?
    $endgroup$
    – Olivier Oloa
    Dec 31 '18 at 12:06






  • 1




    $begingroup$
    Convergence of the series is direct; the value of its sum, though, is not obvious. What makes you think it can be computed?
    $endgroup$
    – Did
    Dec 31 '18 at 12:23








  • 1




    $begingroup$
    Im studying some quick exercises that our Series teacher gave us for Christmas (Im in Kapodistrian University of Athens at Mathematics department). The exercise just says: Calculate: and the sum of the first part my question above.
    $endgroup$
    – Theodossis Papadopoulos
    Dec 31 '18 at 12:44
















0












$begingroup$


The sum series exercise started as: $$sum_{n=1}^inftyfrac{lnleft(frac{n^n}{left(n+1right)^n}right)}{n(n+1)} = sum_{n=1}^infty frac{nlnfrac{n}{n+1}}{n(n+1)} = sum_{n=1}^infty frac{ln(n) - ln(n+1)}{n+1}$$
Looking into calculating it.
A quick hint will help me out!






sequences-and-series closed-form







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please what is the question?
    $endgroup$
    – Olivier Oloa
    Dec 31 '18 at 12:02












  • $begingroup$
    Oh, sorry. I am looking into calculating it.
    $endgroup$
    – Theodossis Papadopoulos
    Dec 31 '18 at 12:05










  • $begingroup$
    A numerical approximation or a closed form?
    $endgroup$
    – Olivier Oloa
    Dec 31 '18 at 12:06






  • 1




    $begingroup$
    Convergence of the series is direct; the value of its sum, though, is not obvious. What makes you think it can be computed?
    $endgroup$
    – Did
    Dec 31 '18 at 12:23








  • 1




    $begingroup$
    Im studying some quick exercises that our Series teacher gave us for Christmas (Im in Kapodistrian University of Athens at Mathematics department). The exercise just says: Calculate: and the sum of the first part my question above.
    $endgroup$
    – Theodossis Papadopoulos
    Dec 31 '18 at 12:44














0












0








0


1



$begingroup$


The sum series exercise started as: $$sum_{n=1}^inftyfrac{lnleft(frac{n^n}{left(n+1right)^n}right)}{n(n+1)} = sum_{n=1}^infty frac{nlnfrac{n}{n+1}}{n(n+1)} = sum_{n=1}^infty frac{ln(n) - ln(n+1)}{n+1}$$
Looking into calculating it.
A quick hint will help me out!






sequences-and-series closed-form







share|cite|improve this question











$endgroup$




The sum series exercise started as: $$sum_{n=1}^inftyfrac{lnleft(frac{n^n}{left(n+1right)^n}right)}{n(n+1)} = sum_{n=1}^infty frac{nlnfrac{n}{n+1}}{n(n+1)} = sum_{n=1}^infty frac{ln(n) - ln(n+1)}{n+1}$$
Looking into calculating it.
A quick hint will help me out!






sequences-and-series closed-form




sequences-and-series closed-form






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 20:01









Zacky

7,92511062




7,92511062










asked Dec 31 '18 at 11:57









Theodossis PapadopoulosTheodossis Papadopoulos

126




126












  • $begingroup$
    Please what is the question?
    $endgroup$
    – Olivier Oloa
    Dec 31 '18 at 12:02












  • $begingroup$
    Oh, sorry. I am looking into calculating it.
    $endgroup$
    – Theodossis Papadopoulos
    Dec 31 '18 at 12:05










  • $begingroup$
    A numerical approximation or a closed form?
    $endgroup$
    – Olivier Oloa
    Dec 31 '18 at 12:06






  • 1




    $begingroup$
    Convergence of the series is direct; the value of its sum, though, is not obvious. What makes you think it can be computed?
    $endgroup$
    – Did
    Dec 31 '18 at 12:23








  • 1




    $begingroup$
    Im studying some quick exercises that our Series teacher gave us for Christmas (Im in Kapodistrian University of Athens at Mathematics department). The exercise just says: Calculate: and the sum of the first part my question above.
    $endgroup$
    – Theodossis Papadopoulos
    Dec 31 '18 at 12:44


















  • $begingroup$
    Please what is the question?
    $endgroup$
    – Olivier Oloa
    Dec 31 '18 at 12:02












  • $begingroup$
    Oh, sorry. I am looking into calculating it.
    $endgroup$
    – Theodossis Papadopoulos
    Dec 31 '18 at 12:05










  • $begingroup$
    A numerical approximation or a closed form?
    $endgroup$
    – Olivier Oloa
    Dec 31 '18 at 12:06






  • 1




    $begingroup$
    Convergence of the series is direct; the value of its sum, though, is not obvious. What makes you think it can be computed?
    $endgroup$
    – Did
    Dec 31 '18 at 12:23








  • 1




    $begingroup$
    Im studying some quick exercises that our Series teacher gave us for Christmas (Im in Kapodistrian University of Athens at Mathematics department). The exercise just says: Calculate: and the sum of the first part my question above.
    $endgroup$
    – Theodossis Papadopoulos
    Dec 31 '18 at 12:44
















$begingroup$
Please what is the question?
$endgroup$
– Olivier Oloa
Dec 31 '18 at 12:02






$begingroup$
Please what is the question?
$endgroup$
– Olivier Oloa
Dec 31 '18 at 12:02














$begingroup$
Oh, sorry. I am looking into calculating it.
$endgroup$
– Theodossis Papadopoulos
Dec 31 '18 at 12:05




$begingroup$
Oh, sorry. I am looking into calculating it.
$endgroup$
– Theodossis Papadopoulos
Dec 31 '18 at 12:05












$begingroup$
A numerical approximation or a closed form?
$endgroup$
– Olivier Oloa
Dec 31 '18 at 12:06




$begingroup$
A numerical approximation or a closed form?
$endgroup$
– Olivier Oloa
Dec 31 '18 at 12:06




1




1




$begingroup$
Convergence of the series is direct; the value of its sum, though, is not obvious. What makes you think it can be computed?
$endgroup$
– Did
Dec 31 '18 at 12:23






$begingroup$
Convergence of the series is direct; the value of its sum, though, is not obvious. What makes you think it can be computed?
$endgroup$
– Did
Dec 31 '18 at 12:23






1




1




$begingroup$
Im studying some quick exercises that our Series teacher gave us for Christmas (Im in Kapodistrian University of Athens at Mathematics department). The exercise just says: Calculate: and the sum of the first part my question above.
$endgroup$
– Theodossis Papadopoulos
Dec 31 '18 at 12:44




$begingroup$
Im studying some quick exercises that our Series teacher gave us for Christmas (Im in Kapodistrian University of Athens at Mathematics department). The exercise just says: Calculate: and the sum of the first part my question above.
$endgroup$
– Theodossis Papadopoulos
Dec 31 '18 at 12:44










2 Answers
2






active

oldest

votes


















1












$begingroup$

As $n to infty$, the general term of the series satisfies
$$
frac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-frac{1}{n+1}ln{small{left(1+frac1nright)}}sim -frac1{n^2}
$$
giving the convergence of the series.



From
$$
sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=sum_{n=1}^inftyfrac{ln n-ln(n+1)}{n+1}
$$ one may use Theorem 2 (16) to get a closed form in terms of poly-Stieltjes constants:




$$
sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=gamma_1(0,1)-gamma_1 tag1
$$




where
$$
gamma_1(a,b) = lim_{Nto+infty}left(sum_{n=1}^N frac{log (n+a)}{n+b}-frac{log^2 !N}2right)
$$
and $gamma_1=gamma_1(1,1)$ is an ordinary Stieltjes constant.



By using Mathematica, one gets




$$
sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-0.7885205660cdots. tag2
$$







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Awesome! Thanks
    $endgroup$
    – Theodossis Papadopoulos
    Dec 31 '18 at 12:44










  • $begingroup$
    Why can't we use telescoping series method for this sum?
    $endgroup$
    – harshit54
    Dec 31 '18 at 13:00










  • $begingroup$
    Please, I'm not sure I get your point. Do you mean you can obtain a closed form with a telescoping series?
    $endgroup$
    – Olivier Oloa
    Dec 31 '18 at 14:24





















1












$begingroup$

We can also express the series into a more fancy way:$$sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-sum_{n=1}^infty frac{lnleft(frac{n+1}{n}right)}{n+1}=-sum_{n=2}^infty frac{lnleft(frac{n}{n-1}right)}{n}=-1-ln(C)$$
Where $C$ is Alladi-Grinstead Constant.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    As $n to infty$, the general term of the series satisfies
    $$
    frac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-frac{1}{n+1}ln{small{left(1+frac1nright)}}sim -frac1{n^2}
    $$
    giving the convergence of the series.



    From
    $$
    sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=sum_{n=1}^inftyfrac{ln n-ln(n+1)}{n+1}
    $$     one may use Theorem 2 (16) to get a closed form in terms of poly-Stieltjes constants:




    $$
    sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=gamma_1(0,1)-gamma_1 tag1
    $$




    where
    $$
    gamma_1(a,b) = lim_{Nto+infty}left(sum_{n=1}^N frac{log (n+a)}{n+b}-frac{log^2 !N}2right)
    $$
    and $gamma_1=gamma_1(1,1)$ is an ordinary Stieltjes constant.



    By using Mathematica, one gets




    $$
    sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-0.7885205660cdots. tag2
    $$







    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Awesome! Thanks
      $endgroup$
      – Theodossis Papadopoulos
      Dec 31 '18 at 12:44










    • $begingroup$
      Why can't we use telescoping series method for this sum?
      $endgroup$
      – harshit54
      Dec 31 '18 at 13:00










    • $begingroup$
      Please, I'm not sure I get your point. Do you mean you can obtain a closed form with a telescoping series?
      $endgroup$
      – Olivier Oloa
      Dec 31 '18 at 14:24


















    1












    $begingroup$

    As $n to infty$, the general term of the series satisfies
    $$
    frac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-frac{1}{n+1}ln{small{left(1+frac1nright)}}sim -frac1{n^2}
    $$
    giving the convergence of the series.



    From
    $$
    sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=sum_{n=1}^inftyfrac{ln n-ln(n+1)}{n+1}
    $$     one may use Theorem 2 (16) to get a closed form in terms of poly-Stieltjes constants:




    $$
    sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=gamma_1(0,1)-gamma_1 tag1
    $$




    where
    $$
    gamma_1(a,b) = lim_{Nto+infty}left(sum_{n=1}^N frac{log (n+a)}{n+b}-frac{log^2 !N}2right)
    $$
    and $gamma_1=gamma_1(1,1)$ is an ordinary Stieltjes constant.



    By using Mathematica, one gets




    $$
    sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-0.7885205660cdots. tag2
    $$







    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Awesome! Thanks
      $endgroup$
      – Theodossis Papadopoulos
      Dec 31 '18 at 12:44










    • $begingroup$
      Why can't we use telescoping series method for this sum?
      $endgroup$
      – harshit54
      Dec 31 '18 at 13:00










    • $begingroup$
      Please, I'm not sure I get your point. Do you mean you can obtain a closed form with a telescoping series?
      $endgroup$
      – Olivier Oloa
      Dec 31 '18 at 14:24
















    1












    1








    1





    $begingroup$

    As $n to infty$, the general term of the series satisfies
    $$
    frac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-frac{1}{n+1}ln{small{left(1+frac1nright)}}sim -frac1{n^2}
    $$
    giving the convergence of the series.



    From
    $$
    sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=sum_{n=1}^inftyfrac{ln n-ln(n+1)}{n+1}
    $$     one may use Theorem 2 (16) to get a closed form in terms of poly-Stieltjes constants:




    $$
    sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=gamma_1(0,1)-gamma_1 tag1
    $$




    where
    $$
    gamma_1(a,b) = lim_{Nto+infty}left(sum_{n=1}^N frac{log (n+a)}{n+b}-frac{log^2 !N}2right)
    $$
    and $gamma_1=gamma_1(1,1)$ is an ordinary Stieltjes constant.



    By using Mathematica, one gets




    $$
    sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-0.7885205660cdots. tag2
    $$







    share|cite|improve this answer









    $endgroup$



    As $n to infty$, the general term of the series satisfies
    $$
    frac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-frac{1}{n+1}ln{small{left(1+frac1nright)}}sim -frac1{n^2}
    $$
    giving the convergence of the series.



    From
    $$
    sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=sum_{n=1}^inftyfrac{ln n-ln(n+1)}{n+1}
    $$     one may use Theorem 2 (16) to get a closed form in terms of poly-Stieltjes constants:




    $$
    sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=gamma_1(0,1)-gamma_1 tag1
    $$




    where
    $$
    gamma_1(a,b) = lim_{Nto+infty}left(sum_{n=1}^N frac{log (n+a)}{n+b}-frac{log^2 !N}2right)
    $$
    and $gamma_1=gamma_1(1,1)$ is an ordinary Stieltjes constant.



    By using Mathematica, one gets




    $$
    sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-0.7885205660cdots. tag2
    $$








    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 31 '18 at 12:38









    Olivier OloaOlivier Oloa

    109k17178294




    109k17178294








    • 1




      $begingroup$
      Awesome! Thanks
      $endgroup$
      – Theodossis Papadopoulos
      Dec 31 '18 at 12:44










    • $begingroup$
      Why can't we use telescoping series method for this sum?
      $endgroup$
      – harshit54
      Dec 31 '18 at 13:00










    • $begingroup$
      Please, I'm not sure I get your point. Do you mean you can obtain a closed form with a telescoping series?
      $endgroup$
      – Olivier Oloa
      Dec 31 '18 at 14:24
















    • 1




      $begingroup$
      Awesome! Thanks
      $endgroup$
      – Theodossis Papadopoulos
      Dec 31 '18 at 12:44










    • $begingroup$
      Why can't we use telescoping series method for this sum?
      $endgroup$
      – harshit54
      Dec 31 '18 at 13:00










    • $begingroup$
      Please, I'm not sure I get your point. Do you mean you can obtain a closed form with a telescoping series?
      $endgroup$
      – Olivier Oloa
      Dec 31 '18 at 14:24










    1




    1




    $begingroup$
    Awesome! Thanks
    $endgroup$
    – Theodossis Papadopoulos
    Dec 31 '18 at 12:44




    $begingroup$
    Awesome! Thanks
    $endgroup$
    – Theodossis Papadopoulos
    Dec 31 '18 at 12:44












    $begingroup$
    Why can't we use telescoping series method for this sum?
    $endgroup$
    – harshit54
    Dec 31 '18 at 13:00




    $begingroup$
    Why can't we use telescoping series method for this sum?
    $endgroup$
    – harshit54
    Dec 31 '18 at 13:00












    $begingroup$
    Please, I'm not sure I get your point. Do you mean you can obtain a closed form with a telescoping series?
    $endgroup$
    – Olivier Oloa
    Dec 31 '18 at 14:24






    $begingroup$
    Please, I'm not sure I get your point. Do you mean you can obtain a closed form with a telescoping series?
    $endgroup$
    – Olivier Oloa
    Dec 31 '18 at 14:24













    1












    $begingroup$

    We can also express the series into a more fancy way:$$sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-sum_{n=1}^infty frac{lnleft(frac{n+1}{n}right)}{n+1}=-sum_{n=2}^infty frac{lnleft(frac{n}{n-1}right)}{n}=-1-ln(C)$$
    Where $C$ is Alladi-Grinstead Constant.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      We can also express the series into a more fancy way:$$sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-sum_{n=1}^infty frac{lnleft(frac{n+1}{n}right)}{n+1}=-sum_{n=2}^infty frac{lnleft(frac{n}{n-1}right)}{n}=-1-ln(C)$$
      Where $C$ is Alladi-Grinstead Constant.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        We can also express the series into a more fancy way:$$sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-sum_{n=1}^infty frac{lnleft(frac{n+1}{n}right)}{n+1}=-sum_{n=2}^infty frac{lnleft(frac{n}{n-1}right)}{n}=-1-ln(C)$$
        Where $C$ is Alladi-Grinstead Constant.






        share|cite|improve this answer









        $endgroup$



        We can also express the series into a more fancy way:$$sum_{n=1}^inftyfrac{lnleft(frac{n^n}{(n+1)^n}right)}{n(n+1)}=-sum_{n=1}^infty frac{lnleft(frac{n+1}{n}right)}{n+1}=-sum_{n=2}^infty frac{lnleft(frac{n}{n-1}right)}{n}=-1-ln(C)$$
        Where $C$ is Alladi-Grinstead Constant.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 19:48









        ZackyZacky

        7,92511062




        7,92511062






























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            ComboBox Display Member on multiple fields

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            2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
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            Is it possible to collect Nectar points via Trainline?

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            2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
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