Prove the series $\sum_{n=1}^\infty (n(f(\frac{1}{n}) - f(-\frac{1}{n})) - 2f'(0)) $ converges

Question

Prove the series $\sum_{n=1}^\infty (n(f(\frac{1}{n}) - f(-\frac{1}{n})) - 2f'(0)) $ converges where $f$ is defined on $[-1,1]$ and $f''(x)$ is continuous.

I already have a solution for this but I am not quite following it.

It uses the Taylor expansion about $0$ to get: [Note: this is the start of their solution not mine]

$f(x) = f(x) +f'(0)\frac{x}{1} + f''(0)\frac{x^2}{2} + f'''(t)\frac{x^3}{6}$

$f(-x) = f(x) - f'(0)\frac{x}{1} + f''(0)\frac{x^2}{2} - f'''(s)\frac{x^3}{6}$

For some $s,t \in [-1,1]$.

Here is my issue: I am not clear on the $s,t$ used in the third derivative. We know the second derivative is continuous, maybe the statement should be that the third derivative is continuous?

Answer 1

You do not know that $f''$ is differentiable, and so cannot speak of $f'''$; hence the highest degree Taylor expansion is to degree 2:

$f(x) = f(0) +f'(0)x +\frac{f''(s)}{2}x^2$ for some $s$ between $0$ and $x$.

Hence, there are $s_n, t_n \in \left[\frac{-1}{n},\frac{1}{n}\right]$ such that

$n \left(f\left(\frac{1}{n}\right) -f\left(\frac{-1}{n}\right)\right) -2 f'(0)$ $= n \left(f(0) +\frac{f'(0)}{n} +\frac{f''(s_n)}{2n^2} -f(0) +\frac{f'(0)}{n} -\frac{f''(t_n)}{2n^2}\right) -2f'(0)$ $= \frac{f''(s_n)-f''(t_n)}{2n}$

It is the sum of the above terms which one wishes to show convergent. However, I am not sure whether this is always the case, for instance if $f''(x) = \frac{x}{|x|\log |x|}$ and $s_n \approx -t_n \approx \frac{1}{n}$. Unfortunately I do not have enough reputation to post this as a comment instead of an answer.

Answer 2

$f(x)=f(0)+f'(0)x+\frac {f''(t)x^{2}} {2}$ and

$f(-x)=f(0)-f'(0)x+\frac {f''(s)x^{2}} {2}$. So $n(f(\frac 1 n) -f(-\frac 1 n))-2f'(0)$ is bounded by a constant times $\frac 1 {n^{2}}$ by MVT applied to the second derivative provided third derivative exists and is bounded . I think there is a typo in the question and second derivative is not enough.