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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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2
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