$f:[1,+\infty)\to \mathbb{R}_+$ satisfies $\ f''\leq0,\ f(+\infty)=+\infty $. Prove $$\lim_{s\to0^+}\sum_{n=1}^\infty (-1)^n[f(n)]^{-s}=-\frac{1}{2}$$
Some of my friends showed me this question and declared it's difficult. However I've found a quite simple proof. Could you help me examine if it's correct?
Proof: Obviously, $\ f$ is strictly increasing. For $s\in(0,1)$, due to Leibniz's Test, $\sum_{n=1}^\infty (-1)^n[f(n)]^{-s}$ converges and we denote it by $(-1)S(s)$. $$f''\leq0\ \&\ x^{-s}\ \text{is convex}\Rightarrow\ x^{-s}\circ f=[f]^{-s}\ \text{is convex}$$ $$\Rightarrow\ f^{-s}(n)-f^{-s}(n+1)\geq f^{-s}(n+1)-f^{-s}(n+2),\quad n\in\mathbb{N}_+$$
Let $a_n^s$ represents $f^{-s}(n)$, now we have,
$$S(s)=a_1^s-a_2^s+a_3^s-a_4^s\cdots\ .$$ Define $S\tilde(s)$ as $$S\tilde(s)=a_2^s-a_3^s+a_4^s-a_5^s\cdots\ .$$
Notice that $$S(s)\geq S\tilde(s),\ S(s)\leq S\tilde(s)+a_1^s-a_2^s, \ S(s)+S\tilde(s)=a_1^s,$$ which implies $$a_1^s/2\leq S(s)\leq a_1^s-a_2^s/2\ .$$
Let $s\to 0^+$, now we proved $\lim_{s\to 0^+}S(s)=1/2$, that is the conclusion.