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

8,21711163




8,21711163










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

        8,21711163




        8,21711163






























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            Verónica Boquete

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            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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