Closed form for $\sum_{n=1}^\infty\frac{1}{2^n\left(1+\sqrt[2^n]{2}\right)}$ Here is another infinite sum I need you help with:
$$\sum_{n=1}^\infty\frac{1}{2^n\left(1+\sqrt[2^n]{2}\right)}.$$
I was told it could be represented in terms of elementary functions and integers.
 A: Note that 
$$\frac{2^{-n}}{2^{2^{-n}}-1}-\frac{2^{-(n-1)}}{2^{2^{-(n-1)}}-1}  = \frac{2^{-n}}{2^{2^{-n}}+1} $$
Thus we have a telescoping sum.  However, note that
$$\lim_{n \to \infty} \frac{2^{-n}}{2^{2^{-n}}-1} = \frac{1}{\log{2}}$$
Therefore the sum is
$$a_1-a_0 + a_2-a_1 + a_3-a_2 + \ldots + \frac{1}{\log{2}} = \frac{1}{\log{2}}- a_0$$
where
$$a_n = \frac{1}{2^n \left ( 2^{2^{-n}}-1\right)}$$
or
$$\sum_{n=1}^{\infty} \frac{1}{2^n \left ( 1+ \sqrt[2^n]{2}\right)}= \frac{1}{\log{2}}-1$$
A: Notice that
$$
\frac1{2^n(\sqrt[2^n]{2}-1)}
-\frac1{2^n(\sqrt[2^n]{2}+1)}
=\frac1{2^{n-1}(\sqrt[2^{n-1}]{2}-1)}
$$
We can rearrange this to
$$
\left(\frac1{2^n(\sqrt[2^n]{2}-1)}-1\right)
=\frac1{2^n(\sqrt[2^n]{2}+1)}
+\left(\frac1{2^{n-1}(\sqrt[2^{n-1}]{2}-1)}-1\right)
$$
and for $n=1$,
$$
\frac1{2^{n-1}(\sqrt[2^{n-1}]{2}-1)}-1=0
$$
therefore, the partial sum is
$$
\sum_{n=1}^m\frac1{2^n(\sqrt[2^n]{2}+1)}
=\frac1{2^m(\sqrt[2^m]{2}-1)}-1
$$
Taking the limit as $m\to\infty$, we get
$$
\sum_{n=1}^\infty\frac1{2^n(\sqrt[2^n]{2}+1)}
=\frac1{\log(2)}-1
$$
