The general term of a sequence is given by
$$a_n = {{2^n}\over{1+x^{2^n}}} $$
Find the sum of this series using differentials.
I thought of taking log but couldn't solve it. Although I can sum up this series by adding and subtracting $1\over1-x$ but don't know how to use differentials.
From the elementary identity $$(1+x)(1+x^2)\ldots(1+x^{2^{n-1}})=\frac{1-x^{2^n}}{1-x},$$ we obtain $$\sum^{n-1}_{k=0}\log(1+x^{2^k})=\log(1-x^{2^n})-\log(1-x),$$ and by differentiating $$\sum^{n-1}_{k=0}\frac{2^kx^{2^k-1}}{1+x^{2^k}}=\frac1{1-x}-\frac{2^nx^{2^n-1}}{1-x^{2^n}}.$$ Multiplying by $x$ and replacing $x\to1/x$, we arrive at $$\sum^{n-1}_{k=0}\frac{2^k}{1+x^{2^k}}=\frac1{x-1}-\frac{2^n}{x^{2^n}-1}.$$ The limit as $n\to\infty$ clearly exists only for $|x|>1$, and then, $$\sum^\infty_{k=0}\frac{2^k}{1+x^{2^k}}=\frac1{x-1}.$$