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 $f(x)$.

  • $\begingroup$ @Winther Could you provide a full work through as an answer? I'm interested in seeing the result myself. $\endgroup$ – mattos Jan 6 '16 at 11:00
  • $\begingroup$ @Mattos I get time later (it's a busy day) and no one else steps up in the meantime I will have a go at doing it properly. $\endgroup$ – Winther Jan 6 '16 at 11:02
  • $\begingroup$ @Winther Thanks mate, that would be great. No worries if you can't though. $\endgroup$ – mattos Jan 6 '16 at 11:09

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 $\sum_{n=0}^\infty \frac{1}{(2n+1)(2n+2)} = \sum_{n=0}^\infty \frac{1}{2n+1} - \frac{1}{2n+2} = \sum_{n=0}^\infty \frac{(-1)^n}{n+1} = \log(2)$.

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