How to calculate the limit of zeta function

Question

Suppose $f(x)>0$,$f''(x)\leqslant0$,and $\lim\limits_{x\to+\infty}f(x)=+\infty$ on$[0,+\infty)$.prove that $$\lim\limits_{s\to0^+}\sum\limits_{n=0}^{\infty}\dfrac{(-1)^n}{f^s(n)}=\frac{1}{2}.$$ I tried to do it ,first of all we have $$\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)}=\sum\limits_{n=0}^{\infty} \left( \frac{1}{f^s(2n)}-\frac{1}{f^s(2n+1)} \right).$$ Use MVT we have $$ \frac{1}{f^s(2n)}-\frac{1}{f^s(2n+1)}=-\frac{sf'(\xi_n)}{f^{s+1}(\xi_n)}(\xi_n\in(2n,2n+1)).$$ But next, I don't know how to deal with it. I want to ask that this problem can be solved by the property that the function is concave.

Answer 1

We can generalize @zhw's solution from Prove that $\lim\limits_{x\to 0^+}\sum\limits_{n=1}^\infty\frac{(-1)^n}{n^x}=-\frac12$.

Starting with your application of the mean-value theorem (note that you have a sign error), $$ \frac{1}{f^s(2n)}-\frac{1}{f^s(2n+1)}=\frac{sf'(\xi_n)}{f^{s+1}(\xi_n)} \le \frac{sf'(2n)}{f^{s+1}(2n)} $$ since $f$ is increasing and $f'$ is decreasing. It follows that $$ \sum_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)} \le \sum_{n=0}^{\infty} \frac{sf'(2n)}{f^{s+1}(2n)} \le \frac{sf'(0)}{f^{s+1}(0)} + s\int_0^\infty \frac{f'(2t)}{f^{s+1}(2t)} \, dt = \frac{sf'(0)}{f^{s+1}(0)} + \frac{1}{2f^s(0)} \, . $$ Similarly, $$ \sum_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)} \ge \sum_{n=0}^{\infty} \frac{sf'(2n+1)}{f^{s+1}(2n+1)} \ge s\int_1^\infty \frac{f'(2t)}{f^{s+1}(2t)} \, dt = \frac{1}{2f^s(1)} \, . $$ The result now follows by squeezing these estimates for $s \to 0$.