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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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Mh yeah wolfram gives one, mathematica doesn't evaluate it. thats strange – Dominic Michaelis Apr 8 at 11:51
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. – Zev Chonoles Apr 8 at 11:54
@ZevChonoles I get the same with mathematica, btw there is the command NSum which evaluates Sum numerical and is much faster than Sum – Dominic Michaelis Apr 8 at 11:55
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@Dominic: Great, thanks for the tip! :) – Zev Chonoles Apr 8 at 11:56
@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. – robjohn Apr 8 at 14:29
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3 Answers

up vote 32 down vote accepted

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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Great Observation.Would have given you $20\times (+1)'s$ – exploringnet Apr 8 at 12:48
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Math is often an experimental science. Before deciding an attack at a problem it may be a good idea to compute a few special cases. – Jyrki Lahtonen Apr 8 at 13:31
Perfect. Very clear, thank you. – Neves Apr 8 at 13:44
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@exploringnet: he's got those 20 now :-) – robjohn Apr 8 at 14:18

This might be another way of looking at Jyrki's hint, but here is the way I did this: $$ \begin{align} \color{#C0C0C0}{1-\frac1{2-1}+}\frac1{2+1}&=1-\frac2{4-1}\\ 1-\frac{2}{4-1}+\frac{2}{4+1}&=1-\frac{4}{16-1}\\ 1-\frac{4}{16-1}+\frac{4}{16+1}&=1-\frac8{256-1}\\ 1-\frac8{256-1}+\frac8{256+1}&=1-\frac{16}{65536-1}\\ &\vdots \end{align} $$

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+1: Looks like we both want to replace an honest induction with an ellipsis/3dots today. Kids, don't do this at home! – Jyrki Lahtonen Apr 8 at 13:54
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@JyrkiLahtonen: dots the way things are going this morning, but to say we always use an ellipsis would be hyperbole. – robjohn Apr 8 at 14:31
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Going off on a tangent, Rob? – Jyrki Lahtonen Apr 8 at 18:12
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@JyrkiLahtonen: it's a sine of my old age. – robjohn Apr 8 at 18:15

Use the identity $$\frac{1}{t+1} = \frac{1}{t-1} - \frac{2}{t^2 -1}.$$

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