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









$endgroup$














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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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            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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            Can I use Tabulator js library in my java Spring + Thymeleaf project?

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            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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