# Finding the value of $\sum\limits_{k=0}^{\infty}\frac{2^{k}}{2^{2^{k}}+1}$

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.

• Mh yeah wolfram gives one, mathematica doesn't evaluate it. thats strange Commented Apr 8, 2013 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. Commented Apr 8, 2013 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 Commented Apr 8, 2013 at 11:55
• @Dominic: Great, thanks for the tip! :) Commented Apr 8, 2013 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
Commented Apr 8, 2013 at 14:29

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 $1-S(n)$?

• Great Observation.Would have given you $20\times (+1)'s$
– ABC
Commented Apr 8, 2013 at 12:48
• 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. Commented Apr 8, 2013 at 13:31
• Perfect. Very clear, thank you. Commented Apr 8, 2013 at 13:44
• @exploringnet: he's got those 20 now :-)
– robjohn
Commented Apr 8, 2013 at 14:18
• Great answer, Sir! Commented Jun 17, 2015 at 0:21

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 $n\to\infty$, $$\sum_{k=0}^\infty\frac{2^k}{2^{2^k}+1}=1$$

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

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