Find the infinite sum $\sum_{n\geq0} \frac{2^n}{1+x^{2^n}}$ for $|x|>1$ I tried finding a pattern when expanding the sum to get
$$\frac{x^{n-1}+2x^{n-2}+\cdots+(n-1)x+n}{x^n+x^{n-1}+\cdots+x+1}$$
where nothing seems to cancel out, and the solution to the sum seems to be $\frac{1}{x-1}$, but how would one get to a simplified answer as such?

You are trying to solve $$\sum_{n = 0}^{\infty} \frac{2^n}{1+x^{2^n}}$$
Each term can be be split as $\frac{2^n}{1+x^{2^n}} = \frac{2^n}{-1+x^{2^n}} - \frac{2^{n+1}}{-1+x^{2^{n+1}}}$. Let $a_n = \frac{2^n}{-1+x^{2^n}}$. The sum is then $$\sum_{n = 0}^{\infty} \left( a_n - a_{n+1}\right)$$
which telescopes to $a_0$, or equivalently $$\frac{2^0}{x^{2^0}-1} = \frac{1}{x-1}$$
