What is the value of $1+frac{1}{2^2}+ frac{1}{4^2}+ frac{1}{8^2}+… $?












1












$begingroup$


I need to find the value of the following series:



$$ 1+frac{1}{2^2}+ frac{1}{4^2}+ frac{1}{8^2}+cdots $$



That is
$$
1+ frac{1}{2^2}+ frac{1}{(2^2)^2} +frac {1}{(2^3)^2}+ cdots$$



It is the summation of a geometric progression where all the terms are squared. I am unable to go further with it.






calculus sequences-and-series summation







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  • $begingroup$
    That's a geometric series...
    $endgroup$
    – Yanko
    Feb 1 at 16:12












  • $begingroup$
    What is a "G.P."?
    $endgroup$
    – user587192
    Feb 1 at 16:18










  • $begingroup$
    @user587192 Simply Geometric Progression
    $endgroup$
    – user638473
    Feb 1 at 16:19










  • $begingroup$
    If you didn't spot the identity $1+x+x^2+cdots=frac{1}{1-x}$ for $|x|<1$ implies $1+y^2+y^4+cdots=frac{1}{1-y^2}$ from $x=y^2$, another option would be $(1+y)(1+y^2+y^4+cdots)=1+y+y^2+cdots=frac{1}{1-y}$.
    $endgroup$
    – J.G.
    Feb 1 at 16:55
















1












$begingroup$


I need to find the value of the following series:



$$ 1+frac{1}{2^2}+ frac{1}{4^2}+ frac{1}{8^2}+cdots $$



That is
$$
1+ frac{1}{2^2}+ frac{1}{(2^2)^2} +frac {1}{(2^3)^2}+ cdots$$



It is the summation of a geometric progression where all the terms are squared. I am unable to go further with it.






calculus sequences-and-series summation







share|cite|improve this question











$endgroup$












  • $begingroup$
    That's a geometric series...
    $endgroup$
    – Yanko
    Feb 1 at 16:12












  • $begingroup$
    What is a "G.P."?
    $endgroup$
    – user587192
    Feb 1 at 16:18










  • $begingroup$
    @user587192 Simply Geometric Progression
    $endgroup$
    – user638473
    Feb 1 at 16:19










  • $begingroup$
    If you didn't spot the identity $1+x+x^2+cdots=frac{1}{1-x}$ for $|x|<1$ implies $1+y^2+y^4+cdots=frac{1}{1-y^2}$ from $x=y^2$, another option would be $(1+y)(1+y^2+y^4+cdots)=1+y+y^2+cdots=frac{1}{1-y}$.
    $endgroup$
    – J.G.
    Feb 1 at 16:55














1












1








1


0



$begingroup$


I need to find the value of the following series:



$$ 1+frac{1}{2^2}+ frac{1}{4^2}+ frac{1}{8^2}+cdots $$



That is
$$
1+ frac{1}{2^2}+ frac{1}{(2^2)^2} +frac {1}{(2^3)^2}+ cdots$$



It is the summation of a geometric progression where all the terms are squared. I am unable to go further with it.






calculus sequences-and-series summation







share|cite|improve this question











$endgroup$




I need to find the value of the following series:



$$ 1+frac{1}{2^2}+ frac{1}{4^2}+ frac{1}{8^2}+cdots $$



That is
$$
1+ frac{1}{2^2}+ frac{1}{(2^2)^2} +frac {1}{(2^3)^2}+ cdots$$



It is the summation of a geometric progression where all the terms are squared. I am unable to go further with it.






calculus sequences-and-series summation




calculus sequences-and-series summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 1 at 16:20







user587192

















asked Feb 1 at 16:09









user638473user638473

505




505












  • $begingroup$
    That's a geometric series...
    $endgroup$
    – Yanko
    Feb 1 at 16:12












  • $begingroup$
    What is a "G.P."?
    $endgroup$
    – user587192
    Feb 1 at 16:18










  • $begingroup$
    @user587192 Simply Geometric Progression
    $endgroup$
    – user638473
    Feb 1 at 16:19










  • $begingroup$
    If you didn't spot the identity $1+x+x^2+cdots=frac{1}{1-x}$ for $|x|<1$ implies $1+y^2+y^4+cdots=frac{1}{1-y^2}$ from $x=y^2$, another option would be $(1+y)(1+y^2+y^4+cdots)=1+y+y^2+cdots=frac{1}{1-y}$.
    $endgroup$
    – J.G.
    Feb 1 at 16:55


















  • $begingroup$
    That's a geometric series...
    $endgroup$
    – Yanko
    Feb 1 at 16:12












  • $begingroup$
    What is a "G.P."?
    $endgroup$
    – user587192
    Feb 1 at 16:18










  • $begingroup$
    @user587192 Simply Geometric Progression
    $endgroup$
    – user638473
    Feb 1 at 16:19










  • $begingroup$
    If you didn't spot the identity $1+x+x^2+cdots=frac{1}{1-x}$ for $|x|<1$ implies $1+y^2+y^4+cdots=frac{1}{1-y^2}$ from $x=y^2$, another option would be $(1+y)(1+y^2+y^4+cdots)=1+y+y^2+cdots=frac{1}{1-y}$.
    $endgroup$
    – J.G.
    Feb 1 at 16:55
















$begingroup$
That's a geometric series...
$endgroup$
– Yanko
Feb 1 at 16:12






$begingroup$
That's a geometric series...
$endgroup$
– Yanko
Feb 1 at 16:12














$begingroup$
What is a "G.P."?
$endgroup$
– user587192
Feb 1 at 16:18




$begingroup$
What is a "G.P."?
$endgroup$
– user587192
Feb 1 at 16:18












$begingroup$
@user587192 Simply Geometric Progression
$endgroup$
– user638473
Feb 1 at 16:19




$begingroup$
@user587192 Simply Geometric Progression
$endgroup$
– user638473
Feb 1 at 16:19












$begingroup$
If you didn't spot the identity $1+x+x^2+cdots=frac{1}{1-x}$ for $|x|<1$ implies $1+y^2+y^4+cdots=frac{1}{1-y^2}$ from $x=y^2$, another option would be $(1+y)(1+y^2+y^4+cdots)=1+y+y^2+cdots=frac{1}{1-y}$.
$endgroup$
– J.G.
Feb 1 at 16:55




$begingroup$
If you didn't spot the identity $1+x+x^2+cdots=frac{1}{1-x}$ for $|x|<1$ implies $1+y^2+y^4+cdots=frac{1}{1-y^2}$ from $x=y^2$, another option would be $(1+y)(1+y^2+y^4+cdots)=1+y+y^2+cdots=frac{1}{1-y}$.
$endgroup$
– J.G.
Feb 1 at 16:55










3 Answers
3






active

oldest

votes


















5












$begingroup$

You're right. You're summing:



$$sum_{n=0}^infty{bigg[color{green}1cdotcolor{red}{bigg(frac14bigg)}^nbigg]}=frac{color{green}1}{1-color{red}{frac14}}=frac43$$






share|cite|improve this answer











$endgroup$





















    8












    $begingroup$

    Why is it that the squares give you trouble? Your series is simply$$1+frac1{2^2}+frac1{(2^2)^2}+frac1{(2^2)^3}+cdots$$and therefore its sum is$$frac1{1-frac1{2^2}}=frac43.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Oh my! Of course. I totally missed out on that
      $endgroup$
      – user638473
      Feb 1 at 16:13



















    4












    $begingroup$

    $$
    begin{align}
    1+frac{1}{2^2}+ frac{1}{4^2}+ frac{1}{8^2}+cdots
    &=frac{1}{(2^0)^2}+frac{1}{(2^1)^2}+ frac{1}{(2^2)^2}+ frac{1}{(2^3)^2}+cdots\
    &=frac{1}{(2^2)^0}+frac{1}{(2^2)^1}+ frac{1}{(2^2)^2}+ frac{1}{(2^2)^3}+cdots\
    &=sumlimits_{n=0}^{infty}frac{1}{left(2^2right)^n}\
    &=sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n.
    end{align}
    $$



    $sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n$ is a geometric series because it's an expression of the form $sumlimits_{n=0}^{infty}ar^n$ where $a=1$ and $r=frac{1}{4}$. We also know that if $|r|<1$, a geometric series converges (otherwise it diverges which means that the sum does not approach a finite number, in other words, it blows up to either positive or negative infinity) and we can even find what it converges to using the formula $sumlimits_{n=0}^{infty}ar^n=frac{a}{1-r}$. In our case, $left|frac{1}{4}right|<1$. So, we know that our series converges and we can find what it converges to:



    $$
    sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n=frac{1}{1-frac{1}{4}}=frac{4}{3}=1frac{1}{3}.
    $$






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      You're right. You're summing:



      $$sum_{n=0}^infty{bigg[color{green}1cdotcolor{red}{bigg(frac14bigg)}^nbigg]}=frac{color{green}1}{1-color{red}{frac14}}=frac43$$






      share|cite|improve this answer











      $endgroup$


















        5












        $begingroup$

        You're right. You're summing:



        $$sum_{n=0}^infty{bigg[color{green}1cdotcolor{red}{bigg(frac14bigg)}^nbigg]}=frac{color{green}1}{1-color{red}{frac14}}=frac43$$






        share|cite|improve this answer











        $endgroup$
















          5












          5








          5





          $begingroup$

          You're right. You're summing:



          $$sum_{n=0}^infty{bigg[color{green}1cdotcolor{red}{bigg(frac14bigg)}^nbigg]}=frac{color{green}1}{1-color{red}{frac14}}=frac43$$






          share|cite|improve this answer











          $endgroup$



          You're right. You're summing:



          $$sum_{n=0}^infty{bigg[color{green}1cdotcolor{red}{bigg(frac14bigg)}^nbigg]}=frac{color{green}1}{1-color{red}{frac14}}=frac43$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 1 at 16:47

























          answered Feb 1 at 16:14









          Rhys HughesRhys Hughes

          7,0501630




          7,0501630























              8












              $begingroup$

              Why is it that the squares give you trouble? Your series is simply$$1+frac1{2^2}+frac1{(2^2)^2}+frac1{(2^2)^3}+cdots$$and therefore its sum is$$frac1{1-frac1{2^2}}=frac43.$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Oh my! Of course. I totally missed out on that
                $endgroup$
                – user638473
                Feb 1 at 16:13
















              8












              $begingroup$

              Why is it that the squares give you trouble? Your series is simply$$1+frac1{2^2}+frac1{(2^2)^2}+frac1{(2^2)^3}+cdots$$and therefore its sum is$$frac1{1-frac1{2^2}}=frac43.$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Oh my! Of course. I totally missed out on that
                $endgroup$
                – user638473
                Feb 1 at 16:13














              8












              8








              8





              $begingroup$

              Why is it that the squares give you trouble? Your series is simply$$1+frac1{2^2}+frac1{(2^2)^2}+frac1{(2^2)^3}+cdots$$and therefore its sum is$$frac1{1-frac1{2^2}}=frac43.$$






              share|cite|improve this answer









              $endgroup$



              Why is it that the squares give you trouble? Your series is simply$$1+frac1{2^2}+frac1{(2^2)^2}+frac1{(2^2)^3}+cdots$$and therefore its sum is$$frac1{1-frac1{2^2}}=frac43.$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Feb 1 at 16:12









              José Carlos SantosJosé Carlos Santos

              174k23133242




              174k23133242












              • $begingroup$
                Oh my! Of course. I totally missed out on that
                $endgroup$
                – user638473
                Feb 1 at 16:13


















              • $begingroup$
                Oh my! Of course. I totally missed out on that
                $endgroup$
                – user638473
                Feb 1 at 16:13
















              $begingroup$
              Oh my! Of course. I totally missed out on that
              $endgroup$
              – user638473
              Feb 1 at 16:13




              $begingroup$
              Oh my! Of course. I totally missed out on that
              $endgroup$
              – user638473
              Feb 1 at 16:13











              4












              $begingroup$

              $$
              begin{align}
              1+frac{1}{2^2}+ frac{1}{4^2}+ frac{1}{8^2}+cdots
              &=frac{1}{(2^0)^2}+frac{1}{(2^1)^2}+ frac{1}{(2^2)^2}+ frac{1}{(2^3)^2}+cdots\
              &=frac{1}{(2^2)^0}+frac{1}{(2^2)^1}+ frac{1}{(2^2)^2}+ frac{1}{(2^2)^3}+cdots\
              &=sumlimits_{n=0}^{infty}frac{1}{left(2^2right)^n}\
              &=sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n.
              end{align}
              $$



              $sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n$ is a geometric series because it's an expression of the form $sumlimits_{n=0}^{infty}ar^n$ where $a=1$ and $r=frac{1}{4}$. We also know that if $|r|<1$, a geometric series converges (otherwise it diverges which means that the sum does not approach a finite number, in other words, it blows up to either positive or negative infinity) and we can even find what it converges to using the formula $sumlimits_{n=0}^{infty}ar^n=frac{a}{1-r}$. In our case, $left|frac{1}{4}right|<1$. So, we know that our series converges and we can find what it converges to:



              $$
              sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n=frac{1}{1-frac{1}{4}}=frac{4}{3}=1frac{1}{3}.
              $$






              share|cite|improve this answer











              $endgroup$


















                4












                $begingroup$

                $$
                begin{align}
                1+frac{1}{2^2}+ frac{1}{4^2}+ frac{1}{8^2}+cdots
                &=frac{1}{(2^0)^2}+frac{1}{(2^1)^2}+ frac{1}{(2^2)^2}+ frac{1}{(2^3)^2}+cdots\
                &=frac{1}{(2^2)^0}+frac{1}{(2^2)^1}+ frac{1}{(2^2)^2}+ frac{1}{(2^2)^3}+cdots\
                &=sumlimits_{n=0}^{infty}frac{1}{left(2^2right)^n}\
                &=sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n.
                end{align}
                $$



                $sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n$ is a geometric series because it's an expression of the form $sumlimits_{n=0}^{infty}ar^n$ where $a=1$ and $r=frac{1}{4}$. We also know that if $|r|<1$, a geometric series converges (otherwise it diverges which means that the sum does not approach a finite number, in other words, it blows up to either positive or negative infinity) and we can even find what it converges to using the formula $sumlimits_{n=0}^{infty}ar^n=frac{a}{1-r}$. In our case, $left|frac{1}{4}right|<1$. So, we know that our series converges and we can find what it converges to:



                $$
                sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n=frac{1}{1-frac{1}{4}}=frac{4}{3}=1frac{1}{3}.
                $$






                share|cite|improve this answer











                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  $$
                  begin{align}
                  1+frac{1}{2^2}+ frac{1}{4^2}+ frac{1}{8^2}+cdots
                  &=frac{1}{(2^0)^2}+frac{1}{(2^1)^2}+ frac{1}{(2^2)^2}+ frac{1}{(2^3)^2}+cdots\
                  &=frac{1}{(2^2)^0}+frac{1}{(2^2)^1}+ frac{1}{(2^2)^2}+ frac{1}{(2^2)^3}+cdots\
                  &=sumlimits_{n=0}^{infty}frac{1}{left(2^2right)^n}\
                  &=sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n.
                  end{align}
                  $$



                  $sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n$ is a geometric series because it's an expression of the form $sumlimits_{n=0}^{infty}ar^n$ where $a=1$ and $r=frac{1}{4}$. We also know that if $|r|<1$, a geometric series converges (otherwise it diverges which means that the sum does not approach a finite number, in other words, it blows up to either positive or negative infinity) and we can even find what it converges to using the formula $sumlimits_{n=0}^{infty}ar^n=frac{a}{1-r}$. In our case, $left|frac{1}{4}right|<1$. So, we know that our series converges and we can find what it converges to:



                  $$
                  sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n=frac{1}{1-frac{1}{4}}=frac{4}{3}=1frac{1}{3}.
                  $$






                  share|cite|improve this answer











                  $endgroup$



                  $$
                  begin{align}
                  1+frac{1}{2^2}+ frac{1}{4^2}+ frac{1}{8^2}+cdots
                  &=frac{1}{(2^0)^2}+frac{1}{(2^1)^2}+ frac{1}{(2^2)^2}+ frac{1}{(2^3)^2}+cdots\
                  &=frac{1}{(2^2)^0}+frac{1}{(2^2)^1}+ frac{1}{(2^2)^2}+ frac{1}{(2^2)^3}+cdots\
                  &=sumlimits_{n=0}^{infty}frac{1}{left(2^2right)^n}\
                  &=sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n.
                  end{align}
                  $$



                  $sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n$ is a geometric series because it's an expression of the form $sumlimits_{n=0}^{infty}ar^n$ where $a=1$ and $r=frac{1}{4}$. We also know that if $|r|<1$, a geometric series converges (otherwise it diverges which means that the sum does not approach a finite number, in other words, it blows up to either positive or negative infinity) and we can even find what it converges to using the formula $sumlimits_{n=0}^{infty}ar^n=frac{a}{1-r}$. In our case, $left|frac{1}{4}right|<1$. So, we know that our series converges and we can find what it converges to:



                  $$
                  sumlimits_{n=0}^{infty}left(frac{1}{4}right)^n=frac{1}{1-frac{1}{4}}=frac{4}{3}=1frac{1}{3}.
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Feb 1 at 17:22

























                  answered Feb 1 at 16:35









                  Michael RybkinMichael Rybkin

                  4,264422




                  4,264422






























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