Computing a series sumHow to prove $sum_{n=0}^{infty} frac{n^2}{2^n} = 6$?Writing an infinite series as the...

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Computing a series sum


How to prove $sum_{n=0}^{infty} frac{n^2}{2^n} = 6$?Writing an infinite series as the sum of the seriesComputing the sum of an infinite seriescomputing the series $sum_{n=1}^infty frac{1}{n^2 2^n}$Sum of a series to an exact answerConfused computing sum of Fourier seriesClosed form of this series?Prove that the given series is convergent.Sum of reciprocals of the square roots of the first N Natural NumbersSum function for power seriesShowing the sum of a power series is less than P$x$













3












$begingroup$


$$sum_{k=0}^infty frac{k^2}{3^k}$$



I tried with that 2 method but I couldn't get the $n^2$ term.






sequences-and-series power-series







share|cite|improve this question









New contributor



Erinç Emre Çelikten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$








  • 3




    $begingroup$
    You'll need to take two derivatives here and also make use of the first derivative case since taking two derivatives gives you $k(k-1) = k^2 - k$.
    $endgroup$
    – Cameron Williams
    7 hours ago












  • $begingroup$
    math.stackexchange.com/questions/593996/…
    $endgroup$
    – lab bhattacharjee
    4 hours ago










  • $begingroup$
    Possible duplicate of How to prove $sum_{n=0}^{infty} frac{n^2}{2^n} = 6$?
    $endgroup$
    – Hans Lundmark
    40 mins ago
















3












$begingroup$


$$sum_{k=0}^infty frac{k^2}{3^k}$$



I tried with that 2 method but I couldn't get the $n^2$ term.






sequences-and-series power-series







share|cite|improve this question









New contributor



Erinç Emre Çelikten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$








  • 3




    $begingroup$
    You'll need to take two derivatives here and also make use of the first derivative case since taking two derivatives gives you $k(k-1) = k^2 - k$.
    $endgroup$
    – Cameron Williams
    7 hours ago












  • $begingroup$
    math.stackexchange.com/questions/593996/…
    $endgroup$
    – lab bhattacharjee
    4 hours ago










  • $begingroup$
    Possible duplicate of How to prove $sum_{n=0}^{infty} frac{n^2}{2^n} = 6$?
    $endgroup$
    – Hans Lundmark
    40 mins ago














3












3








3





$begingroup$


$$sum_{k=0}^infty frac{k^2}{3^k}$$



I tried with that 2 method but I couldn't get the $n^2$ term.






sequences-and-series power-series







share|cite|improve this question









New contributor



Erinç Emre Çelikten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




$$sum_{k=0}^infty frac{k^2}{3^k}$$



I tried with that 2 method but I couldn't get the $n^2$ term.






sequences-and-series power-series




sequences-and-series power-series






share|cite|improve this question









New contributor



Erinç Emre Çelikten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










share|cite|improve this question









New contributor



Erinç Emre Çelikten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








share|cite|improve this question




share|cite|improve this question








edited 7 hours ago









David G. Stork

12.5k41836




12.5k41836






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Erinç Emre Çelikten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 7 hours ago









Erinç Emre ÇeliktenErinç Emre Çelikten

182




182




New contributor



Erinç Emre Çelikten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




New contributor




Erinç Emre Çelikten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










  • 3




    $begingroup$
    You'll need to take two derivatives here and also make use of the first derivative case since taking two derivatives gives you $k(k-1) = k^2 - k$.
    $endgroup$
    – Cameron Williams
    7 hours ago












  • $begingroup$
    math.stackexchange.com/questions/593996/…
    $endgroup$
    – lab bhattacharjee
    4 hours ago










  • $begingroup$
    Possible duplicate of How to prove $sum_{n=0}^{infty} frac{n^2}{2^n} = 6$?
    $endgroup$
    – Hans Lundmark
    40 mins ago














  • 3




    $begingroup$
    You'll need to take two derivatives here and also make use of the first derivative case since taking two derivatives gives you $k(k-1) = k^2 - k$.
    $endgroup$
    – Cameron Williams
    7 hours ago












  • $begingroup$
    math.stackexchange.com/questions/593996/…
    $endgroup$
    – lab bhattacharjee
    4 hours ago










  • $begingroup$
    Possible duplicate of How to prove $sum_{n=0}^{infty} frac{n^2}{2^n} = 6$?
    $endgroup$
    – Hans Lundmark
    40 mins ago








3




3




$begingroup$
You'll need to take two derivatives here and also make use of the first derivative case since taking two derivatives gives you $k(k-1) = k^2 - k$.
$endgroup$
– Cameron Williams
7 hours ago






$begingroup$
You'll need to take two derivatives here and also make use of the first derivative case since taking two derivatives gives you $k(k-1) = k^2 - k$.
$endgroup$
– Cameron Williams
7 hours ago














$begingroup$
math.stackexchange.com/questions/593996/…
$endgroup$
– lab bhattacharjee
4 hours ago




$begingroup$
math.stackexchange.com/questions/593996/…
$endgroup$
– lab bhattacharjee
4 hours ago












$begingroup$
Possible duplicate of How to prove $sum_{n=0}^{infty} frac{n^2}{2^n} = 6$?
$endgroup$
– Hans Lundmark
40 mins ago




$begingroup$
Possible duplicate of How to prove $sum_{n=0}^{infty} frac{n^2}{2^n} = 6$?
$endgroup$
– Hans Lundmark
40 mins ago










3 Answers
3






active

oldest

votes


















5












$begingroup$

$$frac{1}{1-x}=sum_{k=0}^{infty}x^k$$
$$frac{1}{(1-x)^2}=sum_{k=1}^{infty}kx^{k-1}$$
Multiply by $x$
$$frac{x}{(1-x)^2}=sum_{k=1}^{infty}kx^{k}$$
$$left(frac{x}{(1-x)^2}right)'=sum_{k=1}^{infty}k^2x^{k-1}$$
Multiply by $x$
$$xleft(frac{x}{(1-x)^2}right)'=sum_{k=1}^{infty}k^2x^{k}$$
then let $x=frac{1}{3}$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Hint try to show
    begin{eqnarray*}
    sum_{k=0}^{infty} k^2 x^k =frac{x(1+4x+x^2)}{(1-x)^3}.
    end{eqnarray*}






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      There is another way to deal with this problem.



      Denote $S_n=sum_{k=0}^n frac{k^2}{3^k}$ then$frac{1}{3}S_n=sum_{k=1}^{n+1} frac{(k-1)^2}{3^k}$ so $frac{2}{3}S_n=sum_{k=1}^nfrac{2k-1}{3^k}-frac{n^2}{3^{n+1}}$.



      In this way we change the numerator from twice power of $k$ to the lower power. Using the same operation you can change the numerator to numbers without the appearance of $k$ then you can use the summation formula for geometric series.






      share|cite|improve this answer









      $endgroup$














        Your Answer








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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        $$frac{1}{1-x}=sum_{k=0}^{infty}x^k$$
        $$frac{1}{(1-x)^2}=sum_{k=1}^{infty}kx^{k-1}$$
        Multiply by $x$
        $$frac{x}{(1-x)^2}=sum_{k=1}^{infty}kx^{k}$$
        $$left(frac{x}{(1-x)^2}right)'=sum_{k=1}^{infty}k^2x^{k-1}$$
        Multiply by $x$
        $$xleft(frac{x}{(1-x)^2}right)'=sum_{k=1}^{infty}k^2x^{k}$$
        then let $x=frac{1}{3}$






        share|cite|improve this answer











        $endgroup$


















          5












          $begingroup$

          $$frac{1}{1-x}=sum_{k=0}^{infty}x^k$$
          $$frac{1}{(1-x)^2}=sum_{k=1}^{infty}kx^{k-1}$$
          Multiply by $x$
          $$frac{x}{(1-x)^2}=sum_{k=1}^{infty}kx^{k}$$
          $$left(frac{x}{(1-x)^2}right)'=sum_{k=1}^{infty}k^2x^{k-1}$$
          Multiply by $x$
          $$xleft(frac{x}{(1-x)^2}right)'=sum_{k=1}^{infty}k^2x^{k}$$
          then let $x=frac{1}{3}$






          share|cite|improve this answer











          $endgroup$
















            5












            5








            5





            $begingroup$

            $$frac{1}{1-x}=sum_{k=0}^{infty}x^k$$
            $$frac{1}{(1-x)^2}=sum_{k=1}^{infty}kx^{k-1}$$
            Multiply by $x$
            $$frac{x}{(1-x)^2}=sum_{k=1}^{infty}kx^{k}$$
            $$left(frac{x}{(1-x)^2}right)'=sum_{k=1}^{infty}k^2x^{k-1}$$
            Multiply by $x$
            $$xleft(frac{x}{(1-x)^2}right)'=sum_{k=1}^{infty}k^2x^{k}$$
            then let $x=frac{1}{3}$






            share|cite|improve this answer











            $endgroup$



            $$frac{1}{1-x}=sum_{k=0}^{infty}x^k$$
            $$frac{1}{(1-x)^2}=sum_{k=1}^{infty}kx^{k-1}$$
            Multiply by $x$
            $$frac{x}{(1-x)^2}=sum_{k=1}^{infty}kx^{k}$$
            $$left(frac{x}{(1-x)^2}right)'=sum_{k=1}^{infty}k^2x^{k-1}$$
            Multiply by $x$
            $$xleft(frac{x}{(1-x)^2}right)'=sum_{k=1}^{infty}k^2x^{k}$$
            then let $x=frac{1}{3}$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 7 hours ago









            clathratus

            5,6861441




            5,6861441










            answered 7 hours ago









            E.H.EE.H.E

            17.5k11969




            17.5k11969























                1












                $begingroup$

                Hint try to show
                begin{eqnarray*}
                sum_{k=0}^{infty} k^2 x^k =frac{x(1+4x+x^2)}{(1-x)^3}.
                end{eqnarray*}






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Hint try to show
                  begin{eqnarray*}
                  sum_{k=0}^{infty} k^2 x^k =frac{x(1+4x+x^2)}{(1-x)^3}.
                  end{eqnarray*}






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Hint try to show
                    begin{eqnarray*}
                    sum_{k=0}^{infty} k^2 x^k =frac{x(1+4x+x^2)}{(1-x)^3}.
                    end{eqnarray*}






                    share|cite|improve this answer









                    $endgroup$



                    Hint try to show
                    begin{eqnarray*}
                    sum_{k=0}^{infty} k^2 x^k =frac{x(1+4x+x^2)}{(1-x)^3}.
                    end{eqnarray*}







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 7 hours ago









                    Donald SplutterwitDonald Splutterwit

                    23.4k21448




                    23.4k21448























                        0












                        $begingroup$

                        There is another way to deal with this problem.



                        Denote $S_n=sum_{k=0}^n frac{k^2}{3^k}$ then$frac{1}{3}S_n=sum_{k=1}^{n+1} frac{(k-1)^2}{3^k}$ so $frac{2}{3}S_n=sum_{k=1}^nfrac{2k-1}{3^k}-frac{n^2}{3^{n+1}}$.



                        In this way we change the numerator from twice power of $k$ to the lower power. Using the same operation you can change the numerator to numbers without the appearance of $k$ then you can use the summation formula for geometric series.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          There is another way to deal with this problem.



                          Denote $S_n=sum_{k=0}^n frac{k^2}{3^k}$ then$frac{1}{3}S_n=sum_{k=1}^{n+1} frac{(k-1)^2}{3^k}$ so $frac{2}{3}S_n=sum_{k=1}^nfrac{2k-1}{3^k}-frac{n^2}{3^{n+1}}$.



                          In this way we change the numerator from twice power of $k$ to the lower power. Using the same operation you can change the numerator to numbers without the appearance of $k$ then you can use the summation formula for geometric series.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            There is another way to deal with this problem.



                            Denote $S_n=sum_{k=0}^n frac{k^2}{3^k}$ then$frac{1}{3}S_n=sum_{k=1}^{n+1} frac{(k-1)^2}{3^k}$ so $frac{2}{3}S_n=sum_{k=1}^nfrac{2k-1}{3^k}-frac{n^2}{3^{n+1}}$.



                            In this way we change the numerator from twice power of $k$ to the lower power. Using the same operation you can change the numerator to numbers without the appearance of $k$ then you can use the summation formula for geometric series.






                            share|cite|improve this answer









                            $endgroup$



                            There is another way to deal with this problem.



                            Denote $S_n=sum_{k=0}^n frac{k^2}{3^k}$ then$frac{1}{3}S_n=sum_{k=1}^{n+1} frac{(k-1)^2}{3^k}$ so $frac{2}{3}S_n=sum_{k=1}^nfrac{2k-1}{3^k}-frac{n^2}{3^{n+1}}$.



                            In this way we change the numerator from twice power of $k$ to the lower power. Using the same operation you can change the numerator to numbers without the appearance of $k$ then you can use the summation formula for geometric series.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 7 hours ago









                            Feng ShaoFeng Shao

                            18010




                            18010






















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                                Erinç Emre Çelikten is a new contributor. Be nice, and check out our Code of Conduct.
















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