Computing a series sumHow to prove $sum_n=0^infty fracn^22^n = 6$?Writing an infinite series as the sum of the seriesComputing the sum of an infinite seriescomputing the series $sum_n=1^infty frac1n^2 2^n$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$x$

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Computing a series sum


How to prove $sum_n=0^infty fracn^22^n = 6$?Writing an infinite series as the sum of the seriesComputing the sum of an infinite seriescomputing the series $sum_n=1^infty frac1n^2 2^n$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












$begingroup$


$$sum_k=0^infty frack^23^k$$



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 $sum_n=0^infty fracn^22^n = 6$?
    $endgroup$
    – Hans Lundmark
    46 mins ago















3












$begingroup$


$$sum_k=0^infty frack^23^k$$



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 $sum_n=0^infty fracn^22^n = 6$?
    $endgroup$
    – Hans Lundmark
    46 mins ago













3












3








3





$begingroup$


$$sum_k=0^infty frack^23^k$$



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.






$endgroup$




$$sum_k=0^infty frack^23^k$$



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 $sum_n=0^infty fracn^22^n = 6$?
    $endgroup$
    – Hans Lundmark
    46 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 $sum_n=0^infty fracn^22^n = 6$?
    $endgroup$
    – Hans Lundmark
    46 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 $sum_n=0^infty fracn^22^n = 6$?
$endgroup$
– Hans Lundmark
46 mins ago




$begingroup$
Possible duplicate of How to prove $sum_n=0^infty fracn^22^n = 6$?
$endgroup$
– Hans Lundmark
46 mins ago










3 Answers
3






active

oldest

votes


















5












$begingroup$

$$frac11-x=sum_k=0^inftyx^k$$
$$frac1(1-x)^2=sum_k=1^inftykx^k-1$$
Multiply by $x$
$$fracx(1-x)^2=sum_k=1^inftykx^k$$
$$left(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k-1$$
Multiply by $x$
$$xleft(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k$$
then let $x=frac13$






share|cite|improve this answer











$endgroup$




















    1












    $begingroup$

    Hint try to show
    begineqnarray*
    sum_k=0^infty k^2 x^k =fracx(1+4x+x^2)(1-x)^3.
    endeqnarray*






    share|cite|improve this answer









    $endgroup$




















      0












      $begingroup$

      There is another way to deal with this problem.



      Denote $S_n=sum_k=0^n frack^23^k$ then$frac13S_n=sum_k=1^n+1 frac(k-1)^23^k$ so $frac23S_n=sum_k=1^nfrac2k-13^k-fracn^23^n+1$.



      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












        $begingroup$

        $$frac11-x=sum_k=0^inftyx^k$$
        $$frac1(1-x)^2=sum_k=1^inftykx^k-1$$
        Multiply by $x$
        $$fracx(1-x)^2=sum_k=1^inftykx^k$$
        $$left(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k-1$$
        Multiply by $x$
        $$xleft(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k$$
        then let $x=frac13$






        share|cite|improve this answer











        $endgroup$

















          5












          $begingroup$

          $$frac11-x=sum_k=0^inftyx^k$$
          $$frac1(1-x)^2=sum_k=1^inftykx^k-1$$
          Multiply by $x$
          $$fracx(1-x)^2=sum_k=1^inftykx^k$$
          $$left(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k-1$$
          Multiply by $x$
          $$xleft(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k$$
          then let $x=frac13$






          share|cite|improve this answer











          $endgroup$















            5












            5








            5





            $begingroup$

            $$frac11-x=sum_k=0^inftyx^k$$
            $$frac1(1-x)^2=sum_k=1^inftykx^k-1$$
            Multiply by $x$
            $$fracx(1-x)^2=sum_k=1^inftykx^k$$
            $$left(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k-1$$
            Multiply by $x$
            $$xleft(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k$$
            then let $x=frac13$






            share|cite|improve this answer











            $endgroup$



            $$frac11-x=sum_k=0^inftyx^k$$
            $$frac1(1-x)^2=sum_k=1^inftykx^k-1$$
            Multiply by $x$
            $$fracx(1-x)^2=sum_k=1^inftykx^k$$
            $$left(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k-1$$
            Multiply by $x$
            $$xleft(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k$$
            then let $x=frac13$







            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
                begineqnarray*
                sum_k=0^infty k^2 x^k =fracx(1+4x+x^2)(1-x)^3.
                endeqnarray*






                share|cite|improve this answer









                $endgroup$

















                  1












                  $begingroup$

                  Hint try to show
                  begineqnarray*
                  sum_k=0^infty k^2 x^k =fracx(1+4x+x^2)(1-x)^3.
                  endeqnarray*






                  share|cite|improve this answer









                  $endgroup$















                    1












                    1








                    1





                    $begingroup$

                    Hint try to show
                    begineqnarray*
                    sum_k=0^infty k^2 x^k =fracx(1+4x+x^2)(1-x)^3.
                    endeqnarray*






                    share|cite|improve this answer









                    $endgroup$



                    Hint try to show
                    begineqnarray*
                    sum_k=0^infty k^2 x^k =fracx(1+4x+x^2)(1-x)^3.
                    endeqnarray*







                    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 $S_n=sum_k=0^n frack^23^k$ then$frac13S_n=sum_k=1^n+1 frac(k-1)^23^k$ so $frac23S_n=sum_k=1^nfrac2k-13^k-fracn^23^n+1$.



                        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 $S_n=sum_k=0^n frack^23^k$ then$frac13S_n=sum_k=1^n+1 frac(k-1)^23^k$ so $frac23S_n=sum_k=1^nfrac2k-13^k-fracn^23^n+1$.



                          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 $S_n=sum_k=0^n frack^23^k$ then$frac13S_n=sum_k=1^n+1 frac(k-1)^23^k$ so $frac23S_n=sum_k=1^nfrac2k-13^k-fracn^23^n+1$.



                            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 $S_n=sum_k=0^n frack^23^k$ then$frac13S_n=sum_k=1^n+1 frac(k-1)^23^k$ so $frac23S_n=sum_k=1^nfrac2k-13^k-fracn^23^n+1$.



                            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




















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









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