Find $\sum_{n=0}^\infty\frac{2^n}{3^{2^{n-1}}+1}$

Start with following infinite product expansion which is valid for $|q| < 1$,

$$\prod_{n=0}^\infty \left(1 + q^{2^n}\right) = \sum_{n=0}^\infty q^n = \frac{1}{1-q}$$ Taking logarithm and apply $q\frac{\partial}{\partial q}$ on both sides, one get

$$\sum_{n=0}^\infty \frac{2^n q^{2^n}}{1 + q^{2^n}} = \frac{q}{1-q}$$

Substitute $q$ by $\frac{1}{\sqrt{3}}$, one obtain

$$\sum_{n=0}^\infty \frac{2^n}{3^{2^{n-1}}+1} = \frac{1}{\sqrt{3}-1} = \frac{\sqrt{3}+1}{2} = 1 + \frac{\sqrt{3}-1}{2} = 1 + \frac{1}{\sqrt{3}+1}$$

the expression you expected.

There's a very simple way to evaluate this sum. Let

$$a_n = \frac{2^n}{3^{2^{n-1}}-1}$$

Then

$$\begin{align}a_n-a_{n+1} &= \frac{2^n}{3^{2^{n-1}}-1} - \frac{2^{n+1}}{3^{2^{n}}-1} \\ &= \frac{2^n}{3^{2^{n-1}}-1} \left [1-\frac{2}{3^{2^{n-1}}+1} \right ]\\ &= \frac{2^n}{3^{2^{n-1}}-1} \frac{3^{2^{n-1}}-1}{3^{2^{n-1}}+1}\\ &= \frac{2^n}{3^{2^{n-1}}+1}\end{align}$$

Thus,

$$\sum_{n=0}^{\infty}\frac{2^n}{3^{2^{n-1}}+1} = a_0-a_{\infty}$$

where

$$a_0 = \frac{1}{\sqrt{3}-1}$$

and

$$a_{\infty} = \lim_{n\to\infty} \frac{2^n}{3^{2^{n-1}}-1} = 0$$

Therefore the sum is $\frac{1}{\sqrt{3}-1}$. Note that

$$1+\frac1{\sqrt{3}+1} = 1+\frac{\sqrt{3}-1}{2} = \frac{\sqrt{3}+1}{2} = \frac1{\sqrt{3}-1}$$

There is no reason for the oddball $\frac{2^0}{3^{2^{0-1}}+1}$ term so I omit it.

$$\begin{array}{ll} \sum_{r=1}^\infty\frac{2^r}{3^{2^{r-1}}+1} & =\sum_{r=1}^\infty2^r\sum_{m=1}^\infty(-1)^{m-1}3^{-2^{r-1}m} \\ & =\sum_{n=1}^\infty3^{-n}\sum_{2^{r-1}\mid n}2^r(-1)^{n/2^r-1} \\ & = \sum_{n=1}^\infty 3^{-n}\cdot\color{Red}{2} & \\ & = 2\frac{1/3}{1-1/3} \\ & =1.\end{array}$$

Note, given $v:=v_2(n)$, $\displaystyle\sum_{2^{r-1}\mid n}2^r(-1)^{n/2^r-1}$ is $\color{Red}{2}$ if $v=0$ (i.e. $n$ is odd) and otherwise

$$=-2\sum_{r=0}^v2^r(-1)^{n/2^r}=-2\left[\sum_{r=0}^{v-1}2^r-2^v\right]=-2\left[\frac{2^v-1}{2-1}-2^v\right]=-2(-1)=\color{Red}{2}$$

if $v>0$ (i.e. $n$ is even).