Does $\sum_{n=1}^\infty \left[n\left(f\left(\frac{1}{n}\right)-f\left(-\frac{1}{n}\right)\right)-2f'(0)\right]$ converge?

Question

Let $f\in C^3([-1,1])$

Is the series $\sum\limits_{n=1}^\infty \left[n\left(f\left(\frac{1}{n}\right)-f\left(-\frac{1}{n}\right)\right)-2f'(0)\right]$ convergent?

I'm trying to use Taylor's polynomial and remainder to prove that it is, but so far had no success. Any help would be greatly appreciated.

Answer 1

Yes.

Consider $f({1\over n})=f(0)+f'(0)/n+f''(0)/2n^2+R_3(n^{-1})/6n^3$ with $|R_3|\le M$ for some constant $M$.

Similarly $f(-{1\over n})=f(0)-f'(0)/n+f''(0)/2n^2+R_3(-n^{-1})/6n^3$

So we get

$$\sum_{n=1}^\infty{R_3(n^{-1})-R_3(-n^{-1})\over n^3}$$ which is absolutely convergent since $R_3$ is bounded.

Answer 2

You can use some Taylor expansion of $f$ at $0$ to prove that $$n\left( f\left(\frac{1}{n}\right)-f\left(-\frac{1}{n}\right)\right)-2f'(0) = O\left(\frac{1}{n^2}\right)$$

This yields the absolute convergence, hence convergence of the series.

The continuity of $f'''$ over $[-1,1]$ is not necessary. We just need to get a bound on $f'''$.