Infinite series problem












7












$begingroup$



The sum of $$frac{2}{4-1}+frac{2^2}{4^2-1}+frac{2^4}{4^4-1}+cdots cdots $$




Try: write it as $$S = sum^{infty}_{r=0}frac{2^{2^{r}}}{2^{2^{r+1}}-1}=sum^{infty}_{r=0}frac{2^{2^r}-1+1}{(2^{2^r}-1)(2^{2^r}+1)}$$



d not know how to solve from here, could some help me



to solve it, Thanks










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$endgroup$








  • 1




    $begingroup$
    $$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
    $endgroup$
    – lab bhattacharjee
    Jan 24 at 13:07










  • $begingroup$
    @labbhattacharjee I think he had the right side and wrote the left one...
    $endgroup$
    – DonAntonio
    Jan 24 at 13:09
















7












$begingroup$



The sum of $$frac{2}{4-1}+frac{2^2}{4^2-1}+frac{2^4}{4^4-1}+cdots cdots $$




Try: write it as $$S = sum^{infty}_{r=0}frac{2^{2^{r}}}{2^{2^{r+1}}-1}=sum^{infty}_{r=0}frac{2^{2^r}-1+1}{(2^{2^r}-1)(2^{2^r}+1)}$$



d not know how to solve from here, could some help me



to solve it, Thanks










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
    $endgroup$
    – lab bhattacharjee
    Jan 24 at 13:07










  • $begingroup$
    @labbhattacharjee I think he had the right side and wrote the left one...
    $endgroup$
    – DonAntonio
    Jan 24 at 13:09














7












7








7


1



$begingroup$



The sum of $$frac{2}{4-1}+frac{2^2}{4^2-1}+frac{2^4}{4^4-1}+cdots cdots $$




Try: write it as $$S = sum^{infty}_{r=0}frac{2^{2^{r}}}{2^{2^{r+1}}-1}=sum^{infty}_{r=0}frac{2^{2^r}-1+1}{(2^{2^r}-1)(2^{2^r}+1)}$$



d not know how to solve from here, could some help me



to solve it, Thanks










share|cite|improve this question









$endgroup$





The sum of $$frac{2}{4-1}+frac{2^2}{4^2-1}+frac{2^4}{4^4-1}+cdots cdots $$




Try: write it as $$S = sum^{infty}_{r=0}frac{2^{2^{r}}}{2^{2^{r+1}}-1}=sum^{infty}_{r=0}frac{2^{2^r}-1+1}{(2^{2^r}-1)(2^{2^r}+1)}$$



d not know how to solve from here, could some help me



to solve it, Thanks







sequences-and-series






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share|cite|improve this question











share|cite|improve this question




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asked Jan 24 at 13:00









DXTDXT

6,0612732




6,0612732








  • 1




    $begingroup$
    $$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
    $endgroup$
    – lab bhattacharjee
    Jan 24 at 13:07










  • $begingroup$
    @labbhattacharjee I think he had the right side and wrote the left one...
    $endgroup$
    – DonAntonio
    Jan 24 at 13:09














  • 1




    $begingroup$
    $$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
    $endgroup$
    – lab bhattacharjee
    Jan 24 at 13:07










  • $begingroup$
    @labbhattacharjee I think he had the right side and wrote the left one...
    $endgroup$
    – DonAntonio
    Jan 24 at 13:09








1




1




$begingroup$
$$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
$endgroup$
– lab bhattacharjee
Jan 24 at 13:07




$begingroup$
$$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
$endgroup$
– lab bhattacharjee
Jan 24 at 13:07












$begingroup$
@labbhattacharjee I think he had the right side and wrote the left one...
$endgroup$
– DonAntonio
Jan 24 at 13:09




$begingroup$
@labbhattacharjee I think he had the right side and wrote the left one...
$endgroup$
– DonAntonio
Jan 24 at 13:09










3 Answers
3






active

oldest

votes


















3












$begingroup$

You may continue as:



$$frac{2^n}{(2^n-1)(2^n+1)} =frac{(2^n+1)-1}{(2^n-1)(2^n+1)} =frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}$$
Where $n = 2^r$. Now write the sum as:



$$left(frac 11 - frac 13right) + left(frac 13 - frac 1{15}right) + ... + left(frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}right)$$



After cancelling the terms you are left with:



$$1 - frac 1{(2^n-1)(2^n+1)}$$



When $n$ tends to infinity, the expression becomes $1$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Welcome to Math.SE. Note that mathematical notation can be properly typeset in posts, which makes them easier to read.
    $endgroup$
    – hardmath
    Jan 24 at 16:40






  • 1




    $begingroup$
    @PaulSinclair In the solution , $n=2^j$ , so the pattern does indeed follow , as $$frac{1}{2^2^{j+1}-1}=frac{1}{(2^2^j+1)(2^2^j-1)}$$ . While the solution should have mentioned this , I hope this clears it up .
    $endgroup$
    – Sinπ
    Jan 24 at 19:05












  • $begingroup$
    @Rahuboy For future reference you can't do double exponentials without braces, e.g. $2^{2^{j+1}}$ is 2^{2^{j+1}}, not 2^2^{j+1}. That's why your comment malfunctioned.
    $endgroup$
    – J.G.
    Jan 24 at 22:29










  • $begingroup$
    @J.G. Oh I see.... thanks for the help !
    $endgroup$
    – Sinπ
    Jan 25 at 0:02










  • $begingroup$
    That is embarrassing. My apologies for failing to realize this.
    $endgroup$
    – Paul Sinclair
    Jan 25 at 0:14



















7












$begingroup$

You may prove by induction that $$sum_{r=0}^nfrac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{n+1}}-1}.$$(If yoiu didn't spot this conjecture at first, you will after calculating the first few partial sums.) Indeed the claim is correct if $n=0$, and if it holds for $n=k$ then $$sum_{r=0}^{k+1}frac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{k+1}}-1}+frac{2^{2^{k+1}}}{2^{2^{k+2}}-1}=1-frac{1}{2^{2^{k+2}}-1}$$as required, where the final calculation is the $a=2^{2^{k+1}}$ special case of $$1-frac{1}{a-1}+frac{a}{a^2-1}=1-frac{1}{a^2-1}.$$You may wish to rewrite this argument as the computation of a telescoping series.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Note first that
    $$
    frac{2^n}{2^{2n}-1}=2^{-n}frac{1}{1-2^{-2n}}=sum_{k:text{odd},kinBbb N} 2^{-nk}.
    $$
    If we sum over $n=2^j$, $jge 0$, we have
    $$
    S=sum_{j=0}^infty frac{2^{2^j}}{2^{2^{j+1}}-1}=sum_{j=0}^inftysum_{k:text{odd},kinBbb N} 2^{-2^jcdot k}=sum_{l=1}^infty 2^{-l}=1
    $$
    since every $lge 1$ has a unique representation $l=2^jcdot k$ for some $jge 0$ and odd $kin Bbb N$.






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

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      active

      oldest

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      active

      oldest

      votes









      3












      $begingroup$

      You may continue as:



      $$frac{2^n}{(2^n-1)(2^n+1)} =frac{(2^n+1)-1}{(2^n-1)(2^n+1)} =frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}$$
      Where $n = 2^r$. Now write the sum as:



      $$left(frac 11 - frac 13right) + left(frac 13 - frac 1{15}right) + ... + left(frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}right)$$



      After cancelling the terms you are left with:



      $$1 - frac 1{(2^n-1)(2^n+1)}$$



      When $n$ tends to infinity, the expression becomes $1$.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        Welcome to Math.SE. Note that mathematical notation can be properly typeset in posts, which makes them easier to read.
        $endgroup$
        – hardmath
        Jan 24 at 16:40






      • 1




        $begingroup$
        @PaulSinclair In the solution , $n=2^j$ , so the pattern does indeed follow , as $$frac{1}{2^2^{j+1}-1}=frac{1}{(2^2^j+1)(2^2^j-1)}$$ . While the solution should have mentioned this , I hope this clears it up .
        $endgroup$
        – Sinπ
        Jan 24 at 19:05












      • $begingroup$
        @Rahuboy For future reference you can't do double exponentials without braces, e.g. $2^{2^{j+1}}$ is 2^{2^{j+1}}, not 2^2^{j+1}. That's why your comment malfunctioned.
        $endgroup$
        – J.G.
        Jan 24 at 22:29










      • $begingroup$
        @J.G. Oh I see.... thanks for the help !
        $endgroup$
        – Sinπ
        Jan 25 at 0:02










      • $begingroup$
        That is embarrassing. My apologies for failing to realize this.
        $endgroup$
        – Paul Sinclair
        Jan 25 at 0:14
















      3












      $begingroup$

      You may continue as:



      $$frac{2^n}{(2^n-1)(2^n+1)} =frac{(2^n+1)-1}{(2^n-1)(2^n+1)} =frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}$$
      Where $n = 2^r$. Now write the sum as:



      $$left(frac 11 - frac 13right) + left(frac 13 - frac 1{15}right) + ... + left(frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}right)$$



      After cancelling the terms you are left with:



      $$1 - frac 1{(2^n-1)(2^n+1)}$$



      When $n$ tends to infinity, the expression becomes $1$.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        Welcome to Math.SE. Note that mathematical notation can be properly typeset in posts, which makes them easier to read.
        $endgroup$
        – hardmath
        Jan 24 at 16:40






      • 1




        $begingroup$
        @PaulSinclair In the solution , $n=2^j$ , so the pattern does indeed follow , as $$frac{1}{2^2^{j+1}-1}=frac{1}{(2^2^j+1)(2^2^j-1)}$$ . While the solution should have mentioned this , I hope this clears it up .
        $endgroup$
        – Sinπ
        Jan 24 at 19:05












      • $begingroup$
        @Rahuboy For future reference you can't do double exponentials without braces, e.g. $2^{2^{j+1}}$ is 2^{2^{j+1}}, not 2^2^{j+1}. That's why your comment malfunctioned.
        $endgroup$
        – J.G.
        Jan 24 at 22:29










      • $begingroup$
        @J.G. Oh I see.... thanks for the help !
        $endgroup$
        – Sinπ
        Jan 25 at 0:02










      • $begingroup$
        That is embarrassing. My apologies for failing to realize this.
        $endgroup$
        – Paul Sinclair
        Jan 25 at 0:14














      3












      3








      3





      $begingroup$

      You may continue as:



      $$frac{2^n}{(2^n-1)(2^n+1)} =frac{(2^n+1)-1}{(2^n-1)(2^n+1)} =frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}$$
      Where $n = 2^r$. Now write the sum as:



      $$left(frac 11 - frac 13right) + left(frac 13 - frac 1{15}right) + ... + left(frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}right)$$



      After cancelling the terms you are left with:



      $$1 - frac 1{(2^n-1)(2^n+1)}$$



      When $n$ tends to infinity, the expression becomes $1$.






      share|cite|improve this answer











      $endgroup$



      You may continue as:



      $$frac{2^n}{(2^n-1)(2^n+1)} =frac{(2^n+1)-1}{(2^n-1)(2^n+1)} =frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}$$
      Where $n = 2^r$. Now write the sum as:



      $$left(frac 11 - frac 13right) + left(frac 13 - frac 1{15}right) + ... + left(frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}right)$$



      After cancelling the terms you are left with:



      $$1 - frac 1{(2^n-1)(2^n+1)}$$



      When $n$ tends to infinity, the expression becomes $1$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 25 at 0:21









      Paul Sinclair

      20.5k21543




      20.5k21543










      answered Jan 24 at 16:02









      AspiringEngineerAspiringEngineer

      514




      514








      • 1




        $begingroup$
        Welcome to Math.SE. Note that mathematical notation can be properly typeset in posts, which makes them easier to read.
        $endgroup$
        – hardmath
        Jan 24 at 16:40






      • 1




        $begingroup$
        @PaulSinclair In the solution , $n=2^j$ , so the pattern does indeed follow , as $$frac{1}{2^2^{j+1}-1}=frac{1}{(2^2^j+1)(2^2^j-1)}$$ . While the solution should have mentioned this , I hope this clears it up .
        $endgroup$
        – Sinπ
        Jan 24 at 19:05












      • $begingroup$
        @Rahuboy For future reference you can't do double exponentials without braces, e.g. $2^{2^{j+1}}$ is 2^{2^{j+1}}, not 2^2^{j+1}. That's why your comment malfunctioned.
        $endgroup$
        – J.G.
        Jan 24 at 22:29










      • $begingroup$
        @J.G. Oh I see.... thanks for the help !
        $endgroup$
        – Sinπ
        Jan 25 at 0:02










      • $begingroup$
        That is embarrassing. My apologies for failing to realize this.
        $endgroup$
        – Paul Sinclair
        Jan 25 at 0:14














      • 1




        $begingroup$
        Welcome to Math.SE. Note that mathematical notation can be properly typeset in posts, which makes them easier to read.
        $endgroup$
        – hardmath
        Jan 24 at 16:40






      • 1




        $begingroup$
        @PaulSinclair In the solution , $n=2^j$ , so the pattern does indeed follow , as $$frac{1}{2^2^{j+1}-1}=frac{1}{(2^2^j+1)(2^2^j-1)}$$ . While the solution should have mentioned this , I hope this clears it up .
        $endgroup$
        – Sinπ
        Jan 24 at 19:05












      • $begingroup$
        @Rahuboy For future reference you can't do double exponentials without braces, e.g. $2^{2^{j+1}}$ is 2^{2^{j+1}}, not 2^2^{j+1}. That's why your comment malfunctioned.
        $endgroup$
        – J.G.
        Jan 24 at 22:29










      • $begingroup$
        @J.G. Oh I see.... thanks for the help !
        $endgroup$
        – Sinπ
        Jan 25 at 0:02










      • $begingroup$
        That is embarrassing. My apologies for failing to realize this.
        $endgroup$
        – Paul Sinclair
        Jan 25 at 0:14








      1




      1




      $begingroup$
      Welcome to Math.SE. Note that mathematical notation can be properly typeset in posts, which makes them easier to read.
      $endgroup$
      – hardmath
      Jan 24 at 16:40




      $begingroup$
      Welcome to Math.SE. Note that mathematical notation can be properly typeset in posts, which makes them easier to read.
      $endgroup$
      – hardmath
      Jan 24 at 16:40




      1




      1




      $begingroup$
      @PaulSinclair In the solution , $n=2^j$ , so the pattern does indeed follow , as $$frac{1}{2^2^{j+1}-1}=frac{1}{(2^2^j+1)(2^2^j-1)}$$ . While the solution should have mentioned this , I hope this clears it up .
      $endgroup$
      – Sinπ
      Jan 24 at 19:05






      $begingroup$
      @PaulSinclair In the solution , $n=2^j$ , so the pattern does indeed follow , as $$frac{1}{2^2^{j+1}-1}=frac{1}{(2^2^j+1)(2^2^j-1)}$$ . While the solution should have mentioned this , I hope this clears it up .
      $endgroup$
      – Sinπ
      Jan 24 at 19:05














      $begingroup$
      @Rahuboy For future reference you can't do double exponentials without braces, e.g. $2^{2^{j+1}}$ is 2^{2^{j+1}}, not 2^2^{j+1}. That's why your comment malfunctioned.
      $endgroup$
      – J.G.
      Jan 24 at 22:29




      $begingroup$
      @Rahuboy For future reference you can't do double exponentials without braces, e.g. $2^{2^{j+1}}$ is 2^{2^{j+1}}, not 2^2^{j+1}. That's why your comment malfunctioned.
      $endgroup$
      – J.G.
      Jan 24 at 22:29












      $begingroup$
      @J.G. Oh I see.... thanks for the help !
      $endgroup$
      – Sinπ
      Jan 25 at 0:02




      $begingroup$
      @J.G. Oh I see.... thanks for the help !
      $endgroup$
      – Sinπ
      Jan 25 at 0:02












      $begingroup$
      That is embarrassing. My apologies for failing to realize this.
      $endgroup$
      – Paul Sinclair
      Jan 25 at 0:14




      $begingroup$
      That is embarrassing. My apologies for failing to realize this.
      $endgroup$
      – Paul Sinclair
      Jan 25 at 0:14











      7












      $begingroup$

      You may prove by induction that $$sum_{r=0}^nfrac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{n+1}}-1}.$$(If yoiu didn't spot this conjecture at first, you will after calculating the first few partial sums.) Indeed the claim is correct if $n=0$, and if it holds for $n=k$ then $$sum_{r=0}^{k+1}frac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{k+1}}-1}+frac{2^{2^{k+1}}}{2^{2^{k+2}}-1}=1-frac{1}{2^{2^{k+2}}-1}$$as required, where the final calculation is the $a=2^{2^{k+1}}$ special case of $$1-frac{1}{a-1}+frac{a}{a^2-1}=1-frac{1}{a^2-1}.$$You may wish to rewrite this argument as the computation of a telescoping series.






      share|cite|improve this answer









      $endgroup$


















        7












        $begingroup$

        You may prove by induction that $$sum_{r=0}^nfrac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{n+1}}-1}.$$(If yoiu didn't spot this conjecture at first, you will after calculating the first few partial sums.) Indeed the claim is correct if $n=0$, and if it holds for $n=k$ then $$sum_{r=0}^{k+1}frac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{k+1}}-1}+frac{2^{2^{k+1}}}{2^{2^{k+2}}-1}=1-frac{1}{2^{2^{k+2}}-1}$$as required, where the final calculation is the $a=2^{2^{k+1}}$ special case of $$1-frac{1}{a-1}+frac{a}{a^2-1}=1-frac{1}{a^2-1}.$$You may wish to rewrite this argument as the computation of a telescoping series.






        share|cite|improve this answer









        $endgroup$
















          7












          7








          7





          $begingroup$

          You may prove by induction that $$sum_{r=0}^nfrac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{n+1}}-1}.$$(If yoiu didn't spot this conjecture at first, you will after calculating the first few partial sums.) Indeed the claim is correct if $n=0$, and if it holds for $n=k$ then $$sum_{r=0}^{k+1}frac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{k+1}}-1}+frac{2^{2^{k+1}}}{2^{2^{k+2}}-1}=1-frac{1}{2^{2^{k+2}}-1}$$as required, where the final calculation is the $a=2^{2^{k+1}}$ special case of $$1-frac{1}{a-1}+frac{a}{a^2-1}=1-frac{1}{a^2-1}.$$You may wish to rewrite this argument as the computation of a telescoping series.






          share|cite|improve this answer









          $endgroup$



          You may prove by induction that $$sum_{r=0}^nfrac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{n+1}}-1}.$$(If yoiu didn't spot this conjecture at first, you will after calculating the first few partial sums.) Indeed the claim is correct if $n=0$, and if it holds for $n=k$ then $$sum_{r=0}^{k+1}frac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{k+1}}-1}+frac{2^{2^{k+1}}}{2^{2^{k+2}}-1}=1-frac{1}{2^{2^{k+2}}-1}$$as required, where the final calculation is the $a=2^{2^{k+1}}$ special case of $$1-frac{1}{a-1}+frac{a}{a^2-1}=1-frac{1}{a^2-1}.$$You may wish to rewrite this argument as the computation of a telescoping series.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 24 at 13:13









          J.G.J.G.

          30.3k23148




          30.3k23148























              2












              $begingroup$

              Note first that
              $$
              frac{2^n}{2^{2n}-1}=2^{-n}frac{1}{1-2^{-2n}}=sum_{k:text{odd},kinBbb N} 2^{-nk}.
              $$
              If we sum over $n=2^j$, $jge 0$, we have
              $$
              S=sum_{j=0}^infty frac{2^{2^j}}{2^{2^{j+1}}-1}=sum_{j=0}^inftysum_{k:text{odd},kinBbb N} 2^{-2^jcdot k}=sum_{l=1}^infty 2^{-l}=1
              $$
              since every $lge 1$ has a unique representation $l=2^jcdot k$ for some $jge 0$ and odd $kin Bbb N$.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Note first that
                $$
                frac{2^n}{2^{2n}-1}=2^{-n}frac{1}{1-2^{-2n}}=sum_{k:text{odd},kinBbb N} 2^{-nk}.
                $$
                If we sum over $n=2^j$, $jge 0$, we have
                $$
                S=sum_{j=0}^infty frac{2^{2^j}}{2^{2^{j+1}}-1}=sum_{j=0}^inftysum_{k:text{odd},kinBbb N} 2^{-2^jcdot k}=sum_{l=1}^infty 2^{-l}=1
                $$
                since every $lge 1$ has a unique representation $l=2^jcdot k$ for some $jge 0$ and odd $kin Bbb N$.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Note first that
                  $$
                  frac{2^n}{2^{2n}-1}=2^{-n}frac{1}{1-2^{-2n}}=sum_{k:text{odd},kinBbb N} 2^{-nk}.
                  $$
                  If we sum over $n=2^j$, $jge 0$, we have
                  $$
                  S=sum_{j=0}^infty frac{2^{2^j}}{2^{2^{j+1}}-1}=sum_{j=0}^inftysum_{k:text{odd},kinBbb N} 2^{-2^jcdot k}=sum_{l=1}^infty 2^{-l}=1
                  $$
                  since every $lge 1$ has a unique representation $l=2^jcdot k$ for some $jge 0$ and odd $kin Bbb N$.






                  share|cite|improve this answer











                  $endgroup$



                  Note first that
                  $$
                  frac{2^n}{2^{2n}-1}=2^{-n}frac{1}{1-2^{-2n}}=sum_{k:text{odd},kinBbb N} 2^{-nk}.
                  $$
                  If we sum over $n=2^j$, $jge 0$, we have
                  $$
                  S=sum_{j=0}^infty frac{2^{2^j}}{2^{2^{j+1}}-1}=sum_{j=0}^inftysum_{k:text{odd},kinBbb N} 2^{-2^jcdot k}=sum_{l=1}^infty 2^{-l}=1
                  $$
                  since every $lge 1$ has a unique representation $l=2^jcdot k$ for some $jge 0$ and odd $kin Bbb N$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 24 at 13:19

























                  answered Jan 24 at 13:11









                  SongSong

                  18k21449




                  18k21449






























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