Infinite series problem












5












$begingroup$



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






sequences-and-series







share|cite|improve this question









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  • $begingroup$
    $$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
    $endgroup$
    – lab bhattacharjee
    1 hour ago










  • $begingroup$
    @labbhattacharjee I think he had the right side and wrote the left one...
    $endgroup$
    – DonAntonio
    1 hour ago
















5












$begingroup$



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






sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    $$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
    $endgroup$
    – lab bhattacharjee
    1 hour ago










  • $begingroup$
    @labbhattacharjee I think he had the right side and wrote the left one...
    $endgroup$
    – DonAntonio
    1 hour ago














5












5








5


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






sequences-and-series







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




sequences-and-series






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asked 2 hours ago









DXTDXT

5,6142630




5,6142630












  • $begingroup$
    $$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
    $endgroup$
    – lab bhattacharjee
    1 hour ago










  • $begingroup$
    @labbhattacharjee I think he had the right side and wrote the left one...
    $endgroup$
    – DonAntonio
    1 hour ago


















  • $begingroup$
    $$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
    $endgroup$
    – lab bhattacharjee
    1 hour ago










  • $begingroup$
    @labbhattacharjee I think he had the right side and wrote the left one...
    $endgroup$
    – DonAntonio
    1 hour ago
















$begingroup$
$$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
$endgroup$
– lab bhattacharjee
1 hour ago




$begingroup$
$$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
$endgroup$
– lab bhattacharjee
1 hour ago












$begingroup$
@labbhattacharjee I think he had the right side and wrote the left one...
$endgroup$
– DonAntonio
1 hour ago




$begingroup$
@labbhattacharjee I think he had the right side and wrote the left one...
$endgroup$
– DonAntonio
1 hour ago










2 Answers
2






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oldest

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4












$begingroup$

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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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      4












      $begingroup$

      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









      $endgroup$


















        4












        $begingroup$

        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









        $endgroup$
















          4












          4








          4





          $begingroup$

          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 1 hour ago









          J.G.J.G.

          24.8k22539




          24.8k22539























              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 1 hour ago

























                  answered 1 hour ago









                  SongSong

                  10k627




                  10k627






























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