How to calculate $sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} $











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How to calculate $sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} $
I know that result $frac{331}{910}$ because I checked it in Mathematica but I have troubles with calculate that.
$$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} = ... = $$
$$ sum_{n=1}^{infty}(2^n cdot (frac{2}{11})^n+2cdot(frac{2}{11})^ncdot(-1)^n + frac{1}{11^n})$$ But what should be done after...?






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  • Now you have $3$ geometric series, all convergent, and you can use the fact that for $|p|<1$, $$sum_{p=0}^infty p^n=frac{1}{1-p}$$
    – Jean-Claude Arbaut
    Nov 25 at 21:38












  • Best answer! Repeat that as normal answer and I will mark that as the best :) @Jean-ClaudeArbaut
    – VirtualUser
    Nov 25 at 21:41

















up vote
1
down vote

favorite












How to calculate $sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} $
I know that result $frac{331}{910}$ because I checked it in Mathematica but I have troubles with calculate that.
$$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} = ... = $$
$$ sum_{n=1}^{infty}(2^n cdot (frac{2}{11})^n+2cdot(frac{2}{11})^ncdot(-1)^n + frac{1}{11^n})$$ But what should be done after...?






real-analysis







share|cite|improve this question
























  • Now you have $3$ geometric series, all convergent, and you can use the fact that for $|p|<1$, $$sum_{p=0}^infty p^n=frac{1}{1-p}$$
    – Jean-Claude Arbaut
    Nov 25 at 21:38












  • Best answer! Repeat that as normal answer and I will mark that as the best :) @Jean-ClaudeArbaut
    – VirtualUser
    Nov 25 at 21:41















up vote
1
down vote

favorite









up vote
1
down vote

favorite











How to calculate $sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} $
I know that result $frac{331}{910}$ because I checked it in Mathematica but I have troubles with calculate that.
$$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} = ... = $$
$$ sum_{n=1}^{infty}(2^n cdot (frac{2}{11})^n+2cdot(frac{2}{11})^ncdot(-1)^n + frac{1}{11^n})$$ But what should be done after...?






real-analysis







share|cite|improve this question















How to calculate $sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} $
I know that result $frac{331}{910}$ because I checked it in Mathematica but I have troubles with calculate that.
$$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} = ... = $$
$$ sum_{n=1}^{infty}(2^n cdot (frac{2}{11})^n+2cdot(frac{2}{11})^ncdot(-1)^n + frac{1}{11^n})$$ But what should be done after...?






real-analysis




real-analysis






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 at 21:34

























asked Nov 25 at 21:22









VirtualUser

156




156












  • Now you have $3$ geometric series, all convergent, and you can use the fact that for $|p|<1$, $$sum_{p=0}^infty p^n=frac{1}{1-p}$$
    – Jean-Claude Arbaut
    Nov 25 at 21:38












  • Best answer! Repeat that as normal answer and I will mark that as the best :) @Jean-ClaudeArbaut
    – VirtualUser
    Nov 25 at 21:41




















  • Now you have $3$ geometric series, all convergent, and you can use the fact that for $|p|<1$, $$sum_{p=0}^infty p^n=frac{1}{1-p}$$
    – Jean-Claude Arbaut
    Nov 25 at 21:38












  • Best answer! Repeat that as normal answer and I will mark that as the best :) @Jean-ClaudeArbaut
    – VirtualUser
    Nov 25 at 21:41


















Now you have $3$ geometric series, all convergent, and you can use the fact that for $|p|<1$, $$sum_{p=0}^infty p^n=frac{1}{1-p}$$
– Jean-Claude Arbaut
Nov 25 at 21:38






Now you have $3$ geometric series, all convergent, and you can use the fact that for $|p|<1$, $$sum_{p=0}^infty p^n=frac{1}{1-p}$$
– Jean-Claude Arbaut
Nov 25 at 21:38














Best answer! Repeat that as normal answer and I will mark that as the best :) @Jean-ClaudeArbaut
– VirtualUser
Nov 25 at 21:41






Best answer! Repeat that as normal answer and I will mark that as the best :) @Jean-ClaudeArbaut
– VirtualUser
Nov 25 at 21:41












2 Answers
2






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up vote
2
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For $|p|<1$, $sum_{n=1}^{infty}p^n=frac{p}{1-p}$, so



$$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n}=sum_{n=1}^{infty}frac{4^n}{11^n}+2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}\
= frac{frac{4}{11}}{1-frac{4}{11}}-2frac{frac{2}{11}}{1+frac{2}{11}}+frac{frac{1}{11}}{1-frac{1}{11}}\=frac{4}{7}-frac{4}{13}+frac{1}{10}=frac{331}{910}$$






share|cite|improve this answer




























    up vote
    0
    down vote













    Once you've expanded $(2^n-(-1)^n)^2=4^n-2(-2)^n+1$, split up the summation as



    $$sum_{n=1}^{infty} frac{4^n}{11^n}-2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}.$$



    These are just geometric series, so you can apply the formula



    $$sum_{n=1}^{infty} x^n=frac{x}{1-x}.$$






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      2
      down vote



      accepted










      For $|p|<1$, $sum_{n=1}^{infty}p^n=frac{p}{1-p}$, so



      $$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n}=sum_{n=1}^{infty}frac{4^n}{11^n}+2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}\
      = frac{frac{4}{11}}{1-frac{4}{11}}-2frac{frac{2}{11}}{1+frac{2}{11}}+frac{frac{1}{11}}{1-frac{1}{11}}\=frac{4}{7}-frac{4}{13}+frac{1}{10}=frac{331}{910}$$






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted










        For $|p|<1$, $sum_{n=1}^{infty}p^n=frac{p}{1-p}$, so



        $$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n}=sum_{n=1}^{infty}frac{4^n}{11^n}+2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}\
        = frac{frac{4}{11}}{1-frac{4}{11}}-2frac{frac{2}{11}}{1+frac{2}{11}}+frac{frac{1}{11}}{1-frac{1}{11}}\=frac{4}{7}-frac{4}{13}+frac{1}{10}=frac{331}{910}$$






        share|cite|improve this answer























          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          For $|p|<1$, $sum_{n=1}^{infty}p^n=frac{p}{1-p}$, so



          $$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n}=sum_{n=1}^{infty}frac{4^n}{11^n}+2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}\
          = frac{frac{4}{11}}{1-frac{4}{11}}-2frac{frac{2}{11}}{1+frac{2}{11}}+frac{frac{1}{11}}{1-frac{1}{11}}\=frac{4}{7}-frac{4}{13}+frac{1}{10}=frac{331}{910}$$






          share|cite|improve this answer












          For $|p|<1$, $sum_{n=1}^{infty}p^n=frac{p}{1-p}$, so



          $$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n}=sum_{n=1}^{infty}frac{4^n}{11^n}+2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}\
          = frac{frac{4}{11}}{1-frac{4}{11}}-2frac{frac{2}{11}}{1+frac{2}{11}}+frac{frac{1}{11}}{1-frac{1}{11}}\=frac{4}{7}-frac{4}{13}+frac{1}{10}=frac{331}{910}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 at 21:44









          Jean-Claude Arbaut

          14.7k63363




          14.7k63363






















              up vote
              0
              down vote













              Once you've expanded $(2^n-(-1)^n)^2=4^n-2(-2)^n+1$, split up the summation as



              $$sum_{n=1}^{infty} frac{4^n}{11^n}-2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}.$$



              These are just geometric series, so you can apply the formula



              $$sum_{n=1}^{infty} x^n=frac{x}{1-x}.$$






              share|cite|improve this answer

























                up vote
                0
                down vote













                Once you've expanded $(2^n-(-1)^n)^2=4^n-2(-2)^n+1$, split up the summation as



                $$sum_{n=1}^{infty} frac{4^n}{11^n}-2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}.$$



                These are just geometric series, so you can apply the formula



                $$sum_{n=1}^{infty} x^n=frac{x}{1-x}.$$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Once you've expanded $(2^n-(-1)^n)^2=4^n-2(-2)^n+1$, split up the summation as



                  $$sum_{n=1}^{infty} frac{4^n}{11^n}-2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}.$$



                  These are just geometric series, so you can apply the formula



                  $$sum_{n=1}^{infty} x^n=frac{x}{1-x}.$$






                  share|cite|improve this answer












                  Once you've expanded $(2^n-(-1)^n)^2=4^n-2(-2)^n+1$, split up the summation as



                  $$sum_{n=1}^{infty} frac{4^n}{11^n}-2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}.$$



                  These are just geometric series, so you can apply the formula



                  $$sum_{n=1}^{infty} x^n=frac{x}{1-x}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 25 at 21:45









                  Carl Schildkraut

                  10.9k11439




                  10.9k11439






























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