<s>[INST] Find $\sum_{n=0}^\infty\frac{2^n}{3^{2^{n-1}}+1}$ Find $\sum_{n=0}^\infty\frac{2^n}{3^{2^{n-1}}+1}$. 
I already proved that it converges but I can't find the sum, Although I think it should be $1+\frac{1}{\sqrt3 +1}$, from calculations.
 [/INST] 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
$$1+\frac1{\sqrt{3}+1} = 1+\frac{\sqrt{3}-1}{2} = \frac{\sqrt{3}+1}{2} = \frac1{\sqrt{3}-1}$$
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