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.

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

$$\sum_{n=1}^{\infty} \frac{1}{2^n \left ( 1+ \sqrt[2^n]{2}\right)}= \frac{1}{\log{2}}-1$$

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 $$ \sum_{n=1}^\infty\frac1{2^n(\sqrt[2^n]{2}+1)} =\frac1{\log(2)}-1 $$