Sum of Infinite series with a Geometric series in multiplyThe $n$th derivative of $f(e^x)$Winning Probability / Dice GameFinding formula for the nth partial sumPartial sum of a geometric seriesFormula for partial sum of (10)/(10n+1)Telescoping series order.Proof that a sum is monotonically decreasingSum of infinite series $sum_i=1^infty frac 1 i(i+1)(i+2)…(i+n)$Sequence of Riccati's rootsIs the series $sum_n=1^infty Bigl(1-Bigl(1-frac1n^1+epsilonBigr)^nBigr)$ convergent?Given $S_n = sum dots$ and $a_n = sum dots$ prove that $a_n = S_n + 1over ncdot n!$General formula for a SUM

Plotting level sets of the form f(x,y,c)==0

How did Gollum know Sauron was gathering the Haradrim to make war?

Would there be balance issues if I allowed opportunity attacks against any creature, not just hostile ones?

How can I oppose my advisor granting gift authorship to a collaborator?

What is this red bug infesting some trees in southern Germany?

Divide Numbers by 0

Is torque as fundamental a concept as force?

Why not use futuristic pavise ballistic shields for protection?

Punishment in pacifist society

How to run a command 1 out of N times in Bash

How is total raw calculated for Science Pack 2?

Is mathematics truth?

How do we know if a dialogue sounds unnatural without asking for feedback?

Are there photos of the Apollo LM showing disturbed lunar soil resulting from descent engine exhaust?

Why do we need explainable AI?

In mathematics is there a substitution that is "different" from Vieta's substitution to solve the cubic equation?

Tiny image scraper for xkcd.com

What is the converted mana cost of land cards?

Meaning of "offen balkon machen"?

One hour 10 min layover in Newark; International -> Domestic connection. Enough time to clear customs?

Why is k-means used for non normally distributed data?

How to use multiple criteria for -find

Heuristic argument for the Riemann Hypothesis

Are manifolds admitting a circle foliation covered by manifolds with a (non-trivial) circle action?



Sum of Infinite series with a Geometric series in multiply


The $n$th derivative of $f(e^x)$Winning Probability / Dice GameFinding formula for the nth partial sumPartial sum of a geometric seriesFormula for partial sum of (10)/(10n+1)Telescoping series order.Proof that a sum is monotonically decreasingSum of infinite series $sum_i=1^infty frac 1 i(i+1)(i+2)…(i+n)$Sequence of Riccati's rootsIs the series $sum_n=1^infty Bigl(1-Bigl(1-frac1n^1+epsilonBigr)^nBigr)$ convergent?Given $S_n = sum dots$ and $a_n = sum dots$ prove that $a_n = S_n + 1over ncdot n!$General formula for a SUM






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








4












$begingroup$


I came across these series while solving a probability question.



enter link description here



let |r| < 1 ,



$$S_n=sum_k=1^inftyk^n.r^k$$



For n=0 ,it's a GP. $S_0=frac11-r$



For n=1 ,it's a AGP , $S_1=frac-11-r+frac1(1-r)^2$



for n=2 , This series can be reduced to AGP by substituting $k^2=1.3.5....(2k-1)$ & $S_2=frac-r(1-r)^2+frac2r(1-r)^3$.



Is it possible to find sum further in this series . Is there any pattern.










share|cite|improve this question











$endgroup$




















    4












    $begingroup$


    I came across these series while solving a probability question.



    enter link description here



    let |r| < 1 ,



    $$S_n=sum_k=1^inftyk^n.r^k$$



    For n=0 ,it's a GP. $S_0=frac11-r$



    For n=1 ,it's a AGP , $S_1=frac-11-r+frac1(1-r)^2$



    for n=2 , This series can be reduced to AGP by substituting $k^2=1.3.5....(2k-1)$ & $S_2=frac-r(1-r)^2+frac2r(1-r)^3$.



    Is it possible to find sum further in this series . Is there any pattern.










    share|cite|improve this question











    $endgroup$
















      4












      4








      4


      1



      $begingroup$


      I came across these series while solving a probability question.



      enter link description here



      let |r| < 1 ,



      $$S_n=sum_k=1^inftyk^n.r^k$$



      For n=0 ,it's a GP. $S_0=frac11-r$



      For n=1 ,it's a AGP , $S_1=frac-11-r+frac1(1-r)^2$



      for n=2 , This series can be reduced to AGP by substituting $k^2=1.3.5....(2k-1)$ & $S_2=frac-r(1-r)^2+frac2r(1-r)^3$.



      Is it possible to find sum further in this series . Is there any pattern.










      share|cite|improve this question











      $endgroup$




      I came across these series while solving a probability question.



      enter link description here



      let |r| < 1 ,



      $$S_n=sum_k=1^inftyk^n.r^k$$



      For n=0 ,it's a GP. $S_0=frac11-r$



      For n=1 ,it's a AGP , $S_1=frac-11-r+frac1(1-r)^2$



      for n=2 , This series can be reduced to AGP by substituting $k^2=1.3.5....(2k-1)$ & $S_2=frac-r(1-r)^2+frac2r(1-r)^3$.



      Is it possible to find sum further in this series . Is there any pattern.







      sequences-and-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 11 hours ago







      Rishi

















      asked 12 hours ago









      RishiRishi

      646 bronze badges




      646 bronze badges























          3 Answers
          3






          active

          oldest

          votes


















          6













          $begingroup$

          If you take the (formal) dereivative of $S_n$ with respect to $r$, then
          $$ fracmathrm dmathrm drS_n=sum_k=1^infty k^nfracmathrm dmathrm drr^k=sum_k=1^infty k^ncdot k r^k-1=frac 1rsum_k=1^ infty k^n+1r^k=frac1rS_n+1$$
          so you obtain a recursion formula,
          $$S_n+1=rfracmathrm dmathrm drS_n. $$






          share|cite|improve this answer









          $endgroup$














          • $begingroup$
            fantastic observation! any connection with riemann zeta?
            $endgroup$
            – vidyarthi
            12 hours ago






          • 1




            $begingroup$
            @vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $rto-1$.
            $endgroup$
            – Simply Beautiful Art
            11 hours ago


















          2













          $begingroup$

          Using the $n$th derivative of $f(e^x)$, we have



          $$S_n(e^x)=fracmathrm d^nmathrm dx^nfrac11-e^x=sum_k=0^nnbrace kfrack!e^kx(1-e^x)^k+1$$



          and by extension,



          $$S_n(r)=sum_k=0^nnbrace kfrack!r^k(1-r)^k+1$$



          where $displaystylenbrace k$ are the Stirling numbers of the second kind.






          share|cite|improve this answer









          $endgroup$






















            1













            $begingroup$

            Given $S_n=sum_k=1^inftyk^ncdot r^k$, it must be:
            $$beginalignS_0&=sum_k=1^inftyr^k=fracr1-r=-1+frac11-r; \ S_0'&=frac1(1-r)^2=colorred1(S_0+1)^2;\
            S_1&=sum_k=1^inftykcdot r^k=rS_0'=r(S_0+1)^2; \
            S_1'&=colorred1(S_0+1)^2+colorred2r(S_0+1)^3\
            S_2&=sum_k=1^inftyk^2cdot r^k=rS_1'=r(S_0+1)^2+2r^2(S_0+1)^3; \ S_2'&=colorred1(S_0+1)^2+colorred6r(S_0+1)^3+colorred6r^2(S_0+1)^4\
            S_3&=sum_k=1^inftyk^3cdot r^k=rS_2'; \ S_3'&=colorred1(S_0+1)^2+colorred14r(S_0+1)^3+colorred36r^2(S_0+1)^4+colorred24r^3(S_0+1)^5endalign$$

            where the coefficients in red are the number sequence A019538.






            share|cite|improve this answer









            $endgroup$

















              Your Answer








              StackExchange.ready(function()
              var channelOptions =
              tags: "".split(" "),
              id: "69"
              ;
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function()
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled)
              StackExchange.using("snippets", function()
              createEditor();
              );

              else
              createEditor();

              );

              function createEditor()
              StackExchange.prepareEditor(
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader:
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              ,
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              );



              );













              draft saved

              draft discarded


















              StackExchange.ready(
              function ()
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3341995%2fsum-of-infinite-series-with-a-geometric-series-in-multiply%23new-answer', 'question_page');

              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              6













              $begingroup$

              If you take the (formal) dereivative of $S_n$ with respect to $r$, then
              $$ fracmathrm dmathrm drS_n=sum_k=1^infty k^nfracmathrm dmathrm drr^k=sum_k=1^infty k^ncdot k r^k-1=frac 1rsum_k=1^ infty k^n+1r^k=frac1rS_n+1$$
              so you obtain a recursion formula,
              $$S_n+1=rfracmathrm dmathrm drS_n. $$






              share|cite|improve this answer









              $endgroup$














              • $begingroup$
                fantastic observation! any connection with riemann zeta?
                $endgroup$
                – vidyarthi
                12 hours ago






              • 1




                $begingroup$
                @vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $rto-1$.
                $endgroup$
                – Simply Beautiful Art
                11 hours ago















              6













              $begingroup$

              If you take the (formal) dereivative of $S_n$ with respect to $r$, then
              $$ fracmathrm dmathrm drS_n=sum_k=1^infty k^nfracmathrm dmathrm drr^k=sum_k=1^infty k^ncdot k r^k-1=frac 1rsum_k=1^ infty k^n+1r^k=frac1rS_n+1$$
              so you obtain a recursion formula,
              $$S_n+1=rfracmathrm dmathrm drS_n. $$






              share|cite|improve this answer









              $endgroup$














              • $begingroup$
                fantastic observation! any connection with riemann zeta?
                $endgroup$
                – vidyarthi
                12 hours ago






              • 1




                $begingroup$
                @vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $rto-1$.
                $endgroup$
                – Simply Beautiful Art
                11 hours ago













              6














              6










              6







              $begingroup$

              If you take the (formal) dereivative of $S_n$ with respect to $r$, then
              $$ fracmathrm dmathrm drS_n=sum_k=1^infty k^nfracmathrm dmathrm drr^k=sum_k=1^infty k^ncdot k r^k-1=frac 1rsum_k=1^ infty k^n+1r^k=frac1rS_n+1$$
              so you obtain a recursion formula,
              $$S_n+1=rfracmathrm dmathrm drS_n. $$






              share|cite|improve this answer









              $endgroup$



              If you take the (formal) dereivative of $S_n$ with respect to $r$, then
              $$ fracmathrm dmathrm drS_n=sum_k=1^infty k^nfracmathrm dmathrm drr^k=sum_k=1^infty k^ncdot k r^k-1=frac 1rsum_k=1^ infty k^n+1r^k=frac1rS_n+1$$
              so you obtain a recursion formula,
              $$S_n+1=rfracmathrm dmathrm drS_n. $$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 12 hours ago









              Hagen von EitzenHagen von Eitzen

              296k24 gold badges285 silver badges522 bronze badges




              296k24 gold badges285 silver badges522 bronze badges














              • $begingroup$
                fantastic observation! any connection with riemann zeta?
                $endgroup$
                – vidyarthi
                12 hours ago






              • 1




                $begingroup$
                @vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $rto-1$.
                $endgroup$
                – Simply Beautiful Art
                11 hours ago
















              • $begingroup$
                fantastic observation! any connection with riemann zeta?
                $endgroup$
                – vidyarthi
                12 hours ago






              • 1




                $begingroup$
                @vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $rto-1$.
                $endgroup$
                – Simply Beautiful Art
                11 hours ago















              $begingroup$
              fantastic observation! any connection with riemann zeta?
              $endgroup$
              – vidyarthi
              12 hours ago




              $begingroup$
              fantastic observation! any connection with riemann zeta?
              $endgroup$
              – vidyarthi
              12 hours ago




              1




              1




              $begingroup$
              @vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $rto-1$.
              $endgroup$
              – Simply Beautiful Art
              11 hours ago




              $begingroup$
              @vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $rto-1$.
              $endgroup$
              – Simply Beautiful Art
              11 hours ago













              2













              $begingroup$

              Using the $n$th derivative of $f(e^x)$, we have



              $$S_n(e^x)=fracmathrm d^nmathrm dx^nfrac11-e^x=sum_k=0^nnbrace kfrack!e^kx(1-e^x)^k+1$$



              and by extension,



              $$S_n(r)=sum_k=0^nnbrace kfrack!r^k(1-r)^k+1$$



              where $displaystylenbrace k$ are the Stirling numbers of the second kind.






              share|cite|improve this answer









              $endgroup$



















                2













                $begingroup$

                Using the $n$th derivative of $f(e^x)$, we have



                $$S_n(e^x)=fracmathrm d^nmathrm dx^nfrac11-e^x=sum_k=0^nnbrace kfrack!e^kx(1-e^x)^k+1$$



                and by extension,



                $$S_n(r)=sum_k=0^nnbrace kfrack!r^k(1-r)^k+1$$



                where $displaystylenbrace k$ are the Stirling numbers of the second kind.






                share|cite|improve this answer









                $endgroup$

















                  2














                  2










                  2







                  $begingroup$

                  Using the $n$th derivative of $f(e^x)$, we have



                  $$S_n(e^x)=fracmathrm d^nmathrm dx^nfrac11-e^x=sum_k=0^nnbrace kfrack!e^kx(1-e^x)^k+1$$



                  and by extension,



                  $$S_n(r)=sum_k=0^nnbrace kfrack!r^k(1-r)^k+1$$



                  where $displaystylenbrace k$ are the Stirling numbers of the second kind.






                  share|cite|improve this answer









                  $endgroup$



                  Using the $n$th derivative of $f(e^x)$, we have



                  $$S_n(e^x)=fracmathrm d^nmathrm dx^nfrac11-e^x=sum_k=0^nnbrace kfrack!e^kx(1-e^x)^k+1$$



                  and by extension,



                  $$S_n(r)=sum_k=0^nnbrace kfrack!r^k(1-r)^k+1$$



                  where $displaystylenbrace k$ are the Stirling numbers of the second kind.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 11 hours ago









                  Simply Beautiful ArtSimply Beautiful Art

                  53.7k6 gold badges85 silver badges195 bronze badges




                  53.7k6 gold badges85 silver badges195 bronze badges
























                      1













                      $begingroup$

                      Given $S_n=sum_k=1^inftyk^ncdot r^k$, it must be:
                      $$beginalignS_0&=sum_k=1^inftyr^k=fracr1-r=-1+frac11-r; \ S_0'&=frac1(1-r)^2=colorred1(S_0+1)^2;\
                      S_1&=sum_k=1^inftykcdot r^k=rS_0'=r(S_0+1)^2; \
                      S_1'&=colorred1(S_0+1)^2+colorred2r(S_0+1)^3\
                      S_2&=sum_k=1^inftyk^2cdot r^k=rS_1'=r(S_0+1)^2+2r^2(S_0+1)^3; \ S_2'&=colorred1(S_0+1)^2+colorred6r(S_0+1)^3+colorred6r^2(S_0+1)^4\
                      S_3&=sum_k=1^inftyk^3cdot r^k=rS_2'; \ S_3'&=colorred1(S_0+1)^2+colorred14r(S_0+1)^3+colorred36r^2(S_0+1)^4+colorred24r^3(S_0+1)^5endalign$$

                      where the coefficients in red are the number sequence A019538.






                      share|cite|improve this answer









                      $endgroup$



















                        1













                        $begingroup$

                        Given $S_n=sum_k=1^inftyk^ncdot r^k$, it must be:
                        $$beginalignS_0&=sum_k=1^inftyr^k=fracr1-r=-1+frac11-r; \ S_0'&=frac1(1-r)^2=colorred1(S_0+1)^2;\
                        S_1&=sum_k=1^inftykcdot r^k=rS_0'=r(S_0+1)^2; \
                        S_1'&=colorred1(S_0+1)^2+colorred2r(S_0+1)^3\
                        S_2&=sum_k=1^inftyk^2cdot r^k=rS_1'=r(S_0+1)^2+2r^2(S_0+1)^3; \ S_2'&=colorred1(S_0+1)^2+colorred6r(S_0+1)^3+colorred6r^2(S_0+1)^4\
                        S_3&=sum_k=1^inftyk^3cdot r^k=rS_2'; \ S_3'&=colorred1(S_0+1)^2+colorred14r(S_0+1)^3+colorred36r^2(S_0+1)^4+colorred24r^3(S_0+1)^5endalign$$

                        where the coefficients in red are the number sequence A019538.






                        share|cite|improve this answer









                        $endgroup$

















                          1














                          1










                          1







                          $begingroup$

                          Given $S_n=sum_k=1^inftyk^ncdot r^k$, it must be:
                          $$beginalignS_0&=sum_k=1^inftyr^k=fracr1-r=-1+frac11-r; \ S_0'&=frac1(1-r)^2=colorred1(S_0+1)^2;\
                          S_1&=sum_k=1^inftykcdot r^k=rS_0'=r(S_0+1)^2; \
                          S_1'&=colorred1(S_0+1)^2+colorred2r(S_0+1)^3\
                          S_2&=sum_k=1^inftyk^2cdot r^k=rS_1'=r(S_0+1)^2+2r^2(S_0+1)^3; \ S_2'&=colorred1(S_0+1)^2+colorred6r(S_0+1)^3+colorred6r^2(S_0+1)^4\
                          S_3&=sum_k=1^inftyk^3cdot r^k=rS_2'; \ S_3'&=colorred1(S_0+1)^2+colorred14r(S_0+1)^3+colorred36r^2(S_0+1)^4+colorred24r^3(S_0+1)^5endalign$$

                          where the coefficients in red are the number sequence A019538.






                          share|cite|improve this answer









                          $endgroup$



                          Given $S_n=sum_k=1^inftyk^ncdot r^k$, it must be:
                          $$beginalignS_0&=sum_k=1^inftyr^k=fracr1-r=-1+frac11-r; \ S_0'&=frac1(1-r)^2=colorred1(S_0+1)^2;\
                          S_1&=sum_k=1^inftykcdot r^k=rS_0'=r(S_0+1)^2; \
                          S_1'&=colorred1(S_0+1)^2+colorred2r(S_0+1)^3\
                          S_2&=sum_k=1^inftyk^2cdot r^k=rS_1'=r(S_0+1)^2+2r^2(S_0+1)^3; \ S_2'&=colorred1(S_0+1)^2+colorred6r(S_0+1)^3+colorred6r^2(S_0+1)^4\
                          S_3&=sum_k=1^inftyk^3cdot r^k=rS_2'; \ S_3'&=colorred1(S_0+1)^2+colorred14r(S_0+1)^3+colorred36r^2(S_0+1)^4+colorred24r^3(S_0+1)^5endalign$$

                          where the coefficients in red are the number sequence A019538.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 6 hours ago









                          farruhotafarruhota

                          25.2k2 gold badges10 silver badges46 bronze badges




                          25.2k2 gold badges10 silver badges46 bronze badges






























                              draft saved

                              draft discarded
















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid


                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.

                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function ()
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3341995%2fsum-of-infinite-series-with-a-geometric-series-in-multiply%23new-answer', 'question_page');

                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              Canceling a color specificationRandomly assigning color to Graphics3D objects?Default color for Filling in Mathematica 9Coloring specific elements of sets with a prime modified order in an array plotHow to pick a color differing significantly from the colors already in a given color list?Detection of the text colorColor numbers based on their valueCan color schemes for use with ColorData include opacity specification?My dynamic color schemes

                              Invision Community Contents History See also References External links Navigation menuProprietaryinvisioncommunity.comIPS Community ForumsIPS Community Forumsthis blog entry"License Changes, IP.Board 3.4, and the Future""Interview -- Matt Mecham of Ibforums""CEO Invision Power Board, Matt Mecham Is a Liar, Thief!"IPB License Explanation 1.3, 1.3.1, 2.0, and 2.1ArchivedSecurity Fixes, Updates And Enhancements For IPB 1.3.1Archived"New Demo Accounts - Invision Power Services"the original"New Default Skin"the original"Invision Power Board 3.0.0 and Applications Released"the original"Archived copy"the original"Perpetual licenses being done away with""Release Notes - Invision Power Services""Introducing: IPS Community Suite 4!"Invision Community Release Notes

                              199年 目錄 大件事 到箇年出世嗰人 到箇年死嗰人 節慶、風俗習慣 導覽選單