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

Solution 1:

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 $1-S(n)$?

Solution 2:

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

Solution 3:

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