Jun 8, 2021 - 22:04

Problem

pic.twitter.com/N32grPeT1c

— 級数bot (@infseriesbot) June 8, 2021

Solution

$$ \begin{aligned} \sum_{n=0}^\infty \frac{2^n x^{2^n}}{1 + x^{2^n}} &= \frac{x}{1 + x} + \frac{2x^2}{1 + x^2} + \frac{4x^4}{1 + x^4} \dots \\
&= \frac{x}{1-x} - \frac{x}{1-x} + \frac{x}{1 + x} + \frac{2x^2}{1 + x^2} + \frac{4x^4}{1 + x^4} + \dots \\
& = \frac{x}{1-x} - \frac{2x^2}{1-x^2} + \frac{2x^2}{1 + x^2} + \frac{4x^4}{1 + x^4} + \dots \\
& = \frac{x}{1-x} - \frac{4x^4}{1-x^4} + \frac{4x^4}{1 + x^4} + \dots \\
& = \frac{x}{1-x} - \frac{8x^8}{1-x^8} + \dots \\
& = \frac{x}{1-x} \end{aligned} $$

For the second part, we need to differentiate both sides of the above wrt $x$.

Derivative of $\frac{2^{n} x^{2^n}}{1 + x^{2^n}}$ is

$$ \frac{(1 + x^{2^n})2^{2n}x^{2^n-1} - 2^{2n} x^{2^{n+1}-1}}{(1 + x^{2^n})^2} = \frac{2^{2n} x^{2^{n}-1}}{(1 + x^{2^n})^2} $$

Derivative of $\frac{x}{1-x}$ is

$$ \frac{1}{(1-x)^2} $$