Show that $sumlimits_{k=1}^infty frac{1}{k(k+1)(k+2)cdots (k+p)}=frac{1}{p!p} $ for every positive integer...












10












$begingroup$


I have to prove that
$$sum_{k=1}^infty frac{1}{k(k+1)(k+2)cdots (k+p)}=dfrac{1}{p!p}$$



How can I do that?






real-analysis sequences-and-series analysis convergence summation







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    what is it? i or k?
    $endgroup$
    – Jasser
    Nov 3 '14 at 12:07










  • $begingroup$
    Sorry, it is k=1 to infinity.
    $endgroup$
    – Juliane
    Nov 3 '14 at 12:08






  • 1




    $begingroup$
    Are you sure that the result is not $frac{1}{(p-1) p!}$ ?
    $endgroup$
    – Claude Leibovici
    Nov 3 '14 at 12:14










  • $begingroup$
    Im sorry again, I forgot a k in the denominator, so probably your solution is right :D
    $endgroup$
    – Juliane
    Nov 3 '14 at 12:16










  • $begingroup$
    Did you mean 'series'?
    $endgroup$
    – 0112
    Nov 3 '14 at 19:00
















10












$begingroup$


I have to prove that
$$sum_{k=1}^infty frac{1}{k(k+1)(k+2)cdots (k+p)}=dfrac{1}{p!p}$$



How can I do that?






real-analysis sequences-and-series analysis convergence summation







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    what is it? i or k?
    $endgroup$
    – Jasser
    Nov 3 '14 at 12:07










  • $begingroup$
    Sorry, it is k=1 to infinity.
    $endgroup$
    – Juliane
    Nov 3 '14 at 12:08






  • 1




    $begingroup$
    Are you sure that the result is not $frac{1}{(p-1) p!}$ ?
    $endgroup$
    – Claude Leibovici
    Nov 3 '14 at 12:14










  • $begingroup$
    Im sorry again, I forgot a k in the denominator, so probably your solution is right :D
    $endgroup$
    – Juliane
    Nov 3 '14 at 12:16










  • $begingroup$
    Did you mean 'series'?
    $endgroup$
    – 0112
    Nov 3 '14 at 19:00














10












10








10


3



$begingroup$


I have to prove that
$$sum_{k=1}^infty frac{1}{k(k+1)(k+2)cdots (k+p)}=dfrac{1}{p!p}$$



How can I do that?






real-analysis sequences-and-series analysis convergence summation







share|cite|improve this question











$endgroup$




I have to prove that
$$sum_{k=1}^infty frac{1}{k(k+1)(k+2)cdots (k+p)}=dfrac{1}{p!p}$$



How can I do that?






real-analysis sequences-and-series analysis convergence summation




real-analysis sequences-and-series analysis convergence summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 18:36









Did

247k23222458




247k23222458










asked Nov 3 '14 at 12:05









JulianeJuliane

186111




186111








  • 2




    $begingroup$
    what is it? i or k?
    $endgroup$
    – Jasser
    Nov 3 '14 at 12:07










  • $begingroup$
    Sorry, it is k=1 to infinity.
    $endgroup$
    – Juliane
    Nov 3 '14 at 12:08






  • 1




    $begingroup$
    Are you sure that the result is not $frac{1}{(p-1) p!}$ ?
    $endgroup$
    – Claude Leibovici
    Nov 3 '14 at 12:14










  • $begingroup$
    Im sorry again, I forgot a k in the denominator, so probably your solution is right :D
    $endgroup$
    – Juliane
    Nov 3 '14 at 12:16










  • $begingroup$
    Did you mean 'series'?
    $endgroup$
    – 0112
    Nov 3 '14 at 19:00














  • 2




    $begingroup$
    what is it? i or k?
    $endgroup$
    – Jasser
    Nov 3 '14 at 12:07










  • $begingroup$
    Sorry, it is k=1 to infinity.
    $endgroup$
    – Juliane
    Nov 3 '14 at 12:08






  • 1




    $begingroup$
    Are you sure that the result is not $frac{1}{(p-1) p!}$ ?
    $endgroup$
    – Claude Leibovici
    Nov 3 '14 at 12:14










  • $begingroup$
    Im sorry again, I forgot a k in the denominator, so probably your solution is right :D
    $endgroup$
    – Juliane
    Nov 3 '14 at 12:16










  • $begingroup$
    Did you mean 'series'?
    $endgroup$
    – 0112
    Nov 3 '14 at 19:00








2




2




$begingroup$
what is it? i or k?
$endgroup$
– Jasser
Nov 3 '14 at 12:07




$begingroup$
what is it? i or k?
$endgroup$
– Jasser
Nov 3 '14 at 12:07












$begingroup$
Sorry, it is k=1 to infinity.
$endgroup$
– Juliane
Nov 3 '14 at 12:08




$begingroup$
Sorry, it is k=1 to infinity.
$endgroup$
– Juliane
Nov 3 '14 at 12:08




1




1




$begingroup$
Are you sure that the result is not $frac{1}{(p-1) p!}$ ?
$endgroup$
– Claude Leibovici
Nov 3 '14 at 12:14




$begingroup$
Are you sure that the result is not $frac{1}{(p-1) p!}$ ?
$endgroup$
– Claude Leibovici
Nov 3 '14 at 12:14












$begingroup$
Im sorry again, I forgot a k in the denominator, so probably your solution is right :D
$endgroup$
– Juliane
Nov 3 '14 at 12:16




$begingroup$
Im sorry again, I forgot a k in the denominator, so probably your solution is right :D
$endgroup$
– Juliane
Nov 3 '14 at 12:16












$begingroup$
Did you mean 'series'?
$endgroup$
– 0112
Nov 3 '14 at 19:00




$begingroup$
Did you mean 'series'?
$endgroup$
– 0112
Nov 3 '14 at 19:00










2 Answers
2






active

oldest

votes


















15












$begingroup$

Note that
$$
frac{1}{k(k!+!1)(k!+!2)cdots (k!+!p)}=frac{1}{p}!left(frac{1}{k(k!+!1)cdots (k!+!p!-!1)}-frac{1}{(k!+!1)(k!+!2)cdots (k!+!p)}right)
$$
Hence
$$
sum_{k=1}^nfrac{1}{k(k+1)(k+2)cdots (k+p)}=\=frac{1}{p}left(frac{1}{p!}-frac{1}{(n+1)(n+2)(n+3)cdots (n+p)}right) to frac{1}{p!p}
$$
as $ntoinfty$.



Similarly,
$$
sum_{k=1}^nfrac{1}{(k+1)(k+2)cdots (k+p)}=\=frac{1}{p-1}left(frac{1}{p!}--frac{1}{(n+2)(n+2)(n+3)cdots (n+p)}right) to frac{1}{p!(p-1)}
$$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Simpler answer: Telescope



    for fixed $pinBbb N$ If we let
    $u_n =frac1{n(n+1)cdots(n+p-1)}$



    then,



    $u_n -u_{n+1} = frac1{n(n+1)cdots(n+p-1)}-frac1{(n+1)(n+2)cdots(n+p)}
    =frac p{n(n+1)cdots(n+p)}$



    Hence By Telescoping sum we get, $$ u_1 = u_1-lim_{nto infty}u_{n+1}=sum_{n=1}^{infty} (u_n -u_{n+1}) = psum_{n=1}^{infty} frac{1}{n(n+1)(n+2)cdots (n+p)}$$



    But $u_1 = frac1{1(1+1)cdots(1+p-1)} = frac{1}{p!}$



    Hence $$ color{red}{sum_{n=1}^{infty} frac{1}{n(n+1)(n+2)cdots (n+p)} = frac{1}{p!p}}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Why a Down vote please?
      $endgroup$
      – Guy Fsone
      Oct 30 '17 at 19:45






    • 2




      $begingroup$
      I did not downvote but your answer is very similar to the first answer, which seems to be quite older then your answer. If the first answer was not correct you should have commented on that instead if giving a new answer.
      $endgroup$
      – MrYouMath
      Oct 30 '17 at 19:55













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    15












    $begingroup$

    Note that
    $$
    frac{1}{k(k!+!1)(k!+!2)cdots (k!+!p)}=frac{1}{p}!left(frac{1}{k(k!+!1)cdots (k!+!p!-!1)}-frac{1}{(k!+!1)(k!+!2)cdots (k!+!p)}right)
    $$
    Hence
    $$
    sum_{k=1}^nfrac{1}{k(k+1)(k+2)cdots (k+p)}=\=frac{1}{p}left(frac{1}{p!}-frac{1}{(n+1)(n+2)(n+3)cdots (n+p)}right) to frac{1}{p!p}
    $$
    as $ntoinfty$.



    Similarly,
    $$
    sum_{k=1}^nfrac{1}{(k+1)(k+2)cdots (k+p)}=\=frac{1}{p-1}left(frac{1}{p!}--frac{1}{(n+2)(n+2)(n+3)cdots (n+p)}right) to frac{1}{p!(p-1)}
    $$






    share|cite|improve this answer











    $endgroup$


















      15












      $begingroup$

      Note that
      $$
      frac{1}{k(k!+!1)(k!+!2)cdots (k!+!p)}=frac{1}{p}!left(frac{1}{k(k!+!1)cdots (k!+!p!-!1)}-frac{1}{(k!+!1)(k!+!2)cdots (k!+!p)}right)
      $$
      Hence
      $$
      sum_{k=1}^nfrac{1}{k(k+1)(k+2)cdots (k+p)}=\=frac{1}{p}left(frac{1}{p!}-frac{1}{(n+1)(n+2)(n+3)cdots (n+p)}right) to frac{1}{p!p}
      $$
      as $ntoinfty$.



      Similarly,
      $$
      sum_{k=1}^nfrac{1}{(k+1)(k+2)cdots (k+p)}=\=frac{1}{p-1}left(frac{1}{p!}--frac{1}{(n+2)(n+2)(n+3)cdots (n+p)}right) to frac{1}{p!(p-1)}
      $$






      share|cite|improve this answer











      $endgroup$
















        15












        15








        15





        $begingroup$

        Note that
        $$
        frac{1}{k(k!+!1)(k!+!2)cdots (k!+!p)}=frac{1}{p}!left(frac{1}{k(k!+!1)cdots (k!+!p!-!1)}-frac{1}{(k!+!1)(k!+!2)cdots (k!+!p)}right)
        $$
        Hence
        $$
        sum_{k=1}^nfrac{1}{k(k+1)(k+2)cdots (k+p)}=\=frac{1}{p}left(frac{1}{p!}-frac{1}{(n+1)(n+2)(n+3)cdots (n+p)}right) to frac{1}{p!p}
        $$
        as $ntoinfty$.



        Similarly,
        $$
        sum_{k=1}^nfrac{1}{(k+1)(k+2)cdots (k+p)}=\=frac{1}{p-1}left(frac{1}{p!}--frac{1}{(n+2)(n+2)(n+3)cdots (n+p)}right) to frac{1}{p!(p-1)}
        $$






        share|cite|improve this answer











        $endgroup$



        Note that
        $$
        frac{1}{k(k!+!1)(k!+!2)cdots (k!+!p)}=frac{1}{p}!left(frac{1}{k(k!+!1)cdots (k!+!p!-!1)}-frac{1}{(k!+!1)(k!+!2)cdots (k!+!p)}right)
        $$
        Hence
        $$
        sum_{k=1}^nfrac{1}{k(k+1)(k+2)cdots (k+p)}=\=frac{1}{p}left(frac{1}{p!}-frac{1}{(n+1)(n+2)(n+3)cdots (n+p)}right) to frac{1}{p!p}
        $$
        as $ntoinfty$.



        Similarly,
        $$
        sum_{k=1}^nfrac{1}{(k+1)(k+2)cdots (k+p)}=\=frac{1}{p-1}left(frac{1}{p!}--frac{1}{(n+2)(n+2)(n+3)cdots (n+p)}right) to frac{1}{p!(p-1)}
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 1 at 20:30

























        answered Nov 3 '14 at 12:15









        Yiorgos S. SmyrlisYiorgos S. Smyrlis

        63k1384163




        63k1384163























            2












            $begingroup$

            Simpler answer: Telescope



            for fixed $pinBbb N$ If we let
            $u_n =frac1{n(n+1)cdots(n+p-1)}$



            then,



            $u_n -u_{n+1} = frac1{n(n+1)cdots(n+p-1)}-frac1{(n+1)(n+2)cdots(n+p)}
            =frac p{n(n+1)cdots(n+p)}$



            Hence By Telescoping sum we get, $$ u_1 = u_1-lim_{nto infty}u_{n+1}=sum_{n=1}^{infty} (u_n -u_{n+1}) = psum_{n=1}^{infty} frac{1}{n(n+1)(n+2)cdots (n+p)}$$



            But $u_1 = frac1{1(1+1)cdots(1+p-1)} = frac{1}{p!}$



            Hence $$ color{red}{sum_{n=1}^{infty} frac{1}{n(n+1)(n+2)cdots (n+p)} = frac{1}{p!p}}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Why a Down vote please?
              $endgroup$
              – Guy Fsone
              Oct 30 '17 at 19:45






            • 2




              $begingroup$
              I did not downvote but your answer is very similar to the first answer, which seems to be quite older then your answer. If the first answer was not correct you should have commented on that instead if giving a new answer.
              $endgroup$
              – MrYouMath
              Oct 30 '17 at 19:55


















            2












            $begingroup$

            Simpler answer: Telescope



            for fixed $pinBbb N$ If we let
            $u_n =frac1{n(n+1)cdots(n+p-1)}$



            then,



            $u_n -u_{n+1} = frac1{n(n+1)cdots(n+p-1)}-frac1{(n+1)(n+2)cdots(n+p)}
            =frac p{n(n+1)cdots(n+p)}$



            Hence By Telescoping sum we get, $$ u_1 = u_1-lim_{nto infty}u_{n+1}=sum_{n=1}^{infty} (u_n -u_{n+1}) = psum_{n=1}^{infty} frac{1}{n(n+1)(n+2)cdots (n+p)}$$



            But $u_1 = frac1{1(1+1)cdots(1+p-1)} = frac{1}{p!}$



            Hence $$ color{red}{sum_{n=1}^{infty} frac{1}{n(n+1)(n+2)cdots (n+p)} = frac{1}{p!p}}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Why a Down vote please?
              $endgroup$
              – Guy Fsone
              Oct 30 '17 at 19:45






            • 2




              $begingroup$
              I did not downvote but your answer is very similar to the first answer, which seems to be quite older then your answer. If the first answer was not correct you should have commented on that instead if giving a new answer.
              $endgroup$
              – MrYouMath
              Oct 30 '17 at 19:55
















            2












            2








            2





            $begingroup$

            Simpler answer: Telescope



            for fixed $pinBbb N$ If we let
            $u_n =frac1{n(n+1)cdots(n+p-1)}$



            then,



            $u_n -u_{n+1} = frac1{n(n+1)cdots(n+p-1)}-frac1{(n+1)(n+2)cdots(n+p)}
            =frac p{n(n+1)cdots(n+p)}$



            Hence By Telescoping sum we get, $$ u_1 = u_1-lim_{nto infty}u_{n+1}=sum_{n=1}^{infty} (u_n -u_{n+1}) = psum_{n=1}^{infty} frac{1}{n(n+1)(n+2)cdots (n+p)}$$



            But $u_1 = frac1{1(1+1)cdots(1+p-1)} = frac{1}{p!}$



            Hence $$ color{red}{sum_{n=1}^{infty} frac{1}{n(n+1)(n+2)cdots (n+p)} = frac{1}{p!p}}$$






            share|cite|improve this answer











            $endgroup$



            Simpler answer: Telescope



            for fixed $pinBbb N$ If we let
            $u_n =frac1{n(n+1)cdots(n+p-1)}$



            then,



            $u_n -u_{n+1} = frac1{n(n+1)cdots(n+p-1)}-frac1{(n+1)(n+2)cdots(n+p)}
            =frac p{n(n+1)cdots(n+p)}$



            Hence By Telescoping sum we get, $$ u_1 = u_1-lim_{nto infty}u_{n+1}=sum_{n=1}^{infty} (u_n -u_{n+1}) = psum_{n=1}^{infty} frac{1}{n(n+1)(n+2)cdots (n+p)}$$



            But $u_1 = frac1{1(1+1)cdots(1+p-1)} = frac{1}{p!}$



            Hence $$ color{red}{sum_{n=1}^{infty} frac{1}{n(n+1)(n+2)cdots (n+p)} = frac{1}{p!p}}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Oct 30 '17 at 19:45

























            answered Oct 30 '17 at 19:37









            Guy FsoneGuy Fsone

            17.2k42873




            17.2k42873












            • $begingroup$
              Why a Down vote please?
              $endgroup$
              – Guy Fsone
              Oct 30 '17 at 19:45






            • 2




              $begingroup$
              I did not downvote but your answer is very similar to the first answer, which seems to be quite older then your answer. If the first answer was not correct you should have commented on that instead if giving a new answer.
              $endgroup$
              – MrYouMath
              Oct 30 '17 at 19:55




















            • $begingroup$
              Why a Down vote please?
              $endgroup$
              – Guy Fsone
              Oct 30 '17 at 19:45






            • 2




              $begingroup$
              I did not downvote but your answer is very similar to the first answer, which seems to be quite older then your answer. If the first answer was not correct you should have commented on that instead if giving a new answer.
              $endgroup$
              – MrYouMath
              Oct 30 '17 at 19:55


















            $begingroup$
            Why a Down vote please?
            $endgroup$
            – Guy Fsone
            Oct 30 '17 at 19:45




            $begingroup$
            Why a Down vote please?
            $endgroup$
            – Guy Fsone
            Oct 30 '17 at 19:45




            2




            2




            $begingroup$
            I did not downvote but your answer is very similar to the first answer, which seems to be quite older then your answer. If the first answer was not correct you should have commented on that instead if giving a new answer.
            $endgroup$
            – MrYouMath
            Oct 30 '17 at 19:55






            $begingroup$
            I did not downvote but your answer is very similar to the first answer, which seems to be quite older then your answer. If the first answer was not correct you should have commented on that instead if giving a new answer.
            $endgroup$
            – MrYouMath
            Oct 30 '17 at 19:55




















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            File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
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