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Does this weighted sum of reciprocals of Fermat numbers,

$$ F=\sum_{k=0}^{\infty}\dfrac{2^{k}}{2^{2^{k}}+1} $$

have a nice closed form? Wolfram says it's $1$.

Thanks.

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  • $\begingroup$ Mh yeah wolfram gives one, mathematica doesn't evaluate it. thats strange $\endgroup$ Commented Apr 8, 2013 at 11:51
  • $\begingroup$ I think Wolfram Alpha is adding the numerical approximation automatically, whereas Mathematica doesn't try to guess what you mean. Here's what I get in Mathematica. $\endgroup$ Commented Apr 8, 2013 at 11:54
  • $\begingroup$ @ZevChonoles I get the same with mathematica, btw there is the command NSum which evaluates Sum numerical and is much faster than Sum $\endgroup$ Commented Apr 8, 2013 at 11:55
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    $\begingroup$ @Dominic: Great, thanks for the tip! :) $\endgroup$ Commented Apr 8, 2013 at 11:56
  • $\begingroup$ @DominicMichaelis: NSum only does numerical sums. If Sum cannot find a closed form, it calls NSum to compute numerically. Therefore, if you are only looking for a numerical approximation, then NSum will always be faster. $\endgroup$
    – robjohn
    Commented Apr 8, 2013 at 14:29

3 Answers 3

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Hint:

Try to guess and prove a formula for partial sums $$ S(n)=\sum_{k=0}^n\frac{2^k}{2^{2^k}+1}. $$ Here $$ S(0)=\frac13,\ S(1)=\frac{11}{15},\ S(2)=\frac{247}{255},\ldots $$ See a pattern for $1-S(n)$?

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  • $\begingroup$ Great Observation.Would have given you $20\times (+1)'s$ $\endgroup$
    – ABC
    Commented Apr 8, 2013 at 12:48
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    $\begingroup$ Math is often an experimental science. Before deciding on an attack at a problem it may be a good idea to compute a few special cases. $\endgroup$ Commented Apr 8, 2013 at 13:31
  • $\begingroup$ Perfect. Very clear, thank you. $\endgroup$
    – Neves
    Commented Apr 8, 2013 at 13:44
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    $\begingroup$ @exploringnet: he's got those 20 now :-) $\endgroup$
    – robjohn
    Commented Apr 8, 2013 at 14:18
  • $\begingroup$ Great answer, Sir! $\endgroup$ Commented Jun 17, 2015 at 0:21
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This might be another way of looking at Jyrki's hint, but here is the way I did this: $$ \begin{align} \color{#C0C0C0}{1-\frac1{2-1}+}\frac1{2+1}&=1-\frac2{4-1}\\ 1-\frac{2}{4-1}+\frac{2}{4+1}&=1-\frac{4}{16-1}\\ 1-\frac{4}{16-1}+\frac{4}{16+1}&=1-\frac8{256-1}\\ 1-\frac8{256-1}+\frac8{256+1}&=1-\frac{16}{65536-1}\\ &\vdots \end{align} $$


Motivation behind this approach

One trick to try is conjugates. Noting that $$ \frac{2^k}{2^{2^k}-1}-\frac{2^k}{2^{2^k}+1}=\frac{2^{k+1}}{2^{2^{k+1}}-1} $$ we see, using Telescoping Series, that $$ \begin{align} \sum_{k=0}^{n-1}\frac{2^k}{2^{2^k}+1} &=\sum_{k=0}^{n-1}\left(\frac{2^k}{2^{2^k}-1}-\frac{2^{k+1}}{2^{2^{k+1}}-1}\right)\\ &=1-\frac{2^n}{2^{2^n}-1} \end{align} $$ Therefore, letting $n\to\infty$, $$ \sum_{k=0}^\infty\frac{2^k}{2^{2^k}+1}=1 $$

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    $\begingroup$ +1: Looks like we both want to replace an honest induction with an ellipsis/3dots today. Kids, don't do this at home! $\endgroup$ Commented Apr 8, 2013 at 13:54
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    $\begingroup$ @JyrkiLahtonen: dots the way things are going this morning, but to say we always use an ellipsis would be hyperbole. $\endgroup$
    – robjohn
    Commented Apr 8, 2013 at 14:31
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    $\begingroup$ Going off on a tangent, Rob? $\endgroup$ Commented Apr 8, 2013 at 18:12
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    $\begingroup$ @JyrkiLahtonen: it's a sine of my old age. $\endgroup$
    – robjohn
    Commented Apr 8, 2013 at 18:15
  • $\begingroup$ Very good use of colors. $\endgroup$ Commented Dec 9, 2016 at 22:38
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Use the identity $$\frac{1}{t+1} = \frac{1}{t-1} - \frac{2}{t^2 -1}.$$

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