A property of the series $\sum_{i=1}^{\infty}\frac{(1-x)x^i}{1+x^i}$ I think this is a hard question:
$$f(x)=\sum_{i=1}^{\infty}\frac{(1-x)x^i}{1+x^i}~~~\text{for}~~~x\in(0,1)$$ 
Prove: $$\lim_{x\to 1^-}f(x)=\ln(2)$$
Find the maximum value of $f(x)$.
 A: We start by expanding $\frac{1}{1+x^i} = \frac{1-x^i}{1-x^{2i}}$ in a geometrical series to get the double sum
$$f(x)=\sum_{i=1}^{\infty}\sum_{n=0}^\infty (1-x)(1-x^i)x^{2n i}$$ 
Since the summands above are non-negative we can, by Tonelli's theorem, interchange the order of summation to get
$$f(x) = \sum_{n=0}^\infty \frac{x^{2n+1}(1-x)^2}{(1-x^{2n+1})(1-x^{2n+2})}$$ 
The function 
$$f_n(x) = \frac{x^{2n+1}(1-x)^2}{(1-x^{2n+1})(1-x^{2n+2})}$$
is monotonely increasing on $[0,1]$ and therefore satisfy
$$f_n(x) \leq \lim_{x\to 1^-} f_n(x) = \frac{1}{(2n+1)(2n+2)}$$
Since $\sum_{n=0}^\infty \frac{1}{(2n+1)(2n+2)}$ converges it follows from Weierstrass M-test that the series $\sum_{n=0}^\infty f_n(x)$ converges uniformly on $[0,1]$ and therefore
$$\lim_{x\to 1^-} f(x) = \lim_{x\to 1^-}\sum_{n=0}^\infty f_n(x) = \sum_{n=0}^\infty \lim_{x\to 1^-} f_n(x) = \sum_{n=0}^\infty \frac{1}{(2n+1)(2n+2)} = \log(2)$$
This is also the maximum value of $f(x)$ on $[0,1]$. The last equality above follows from $\sum_{n=0}^\infty \frac{1}{(2n+1)(2n+2)} =  \sum_{n=0}^\infty \frac{1}{2n+1} - \frac{1}{2n+2} = \sum_{n=0}^\infty \frac{(-1)^n}{n+1} = \log(2)$.
