Show that $\sum_{n=1}^\infty \left\{n \left[f \left(\frac{1}{n}\right)- f \left(-\frac{1}{n}\right)\right]-2 f'(0)\right\}$ converges 
For $f \in C^3 (\mathbb{R})$ show that
$$\sum_{n=1}^\infty \left\{n \left[f \left(\frac{1}{n}\right)- f \left(-\frac{1}{n}\right)\right]-2 f'(0)\right\}$$
converges.

I am stuck on this problem. What I tried is that
$$n \left[f \left(\frac{1}{n}\right)- f \left(-\frac{1}{n}\right)\right] = \frac{f \left(\frac{1}{n}\right) - f(0) + f(0) - f \left(-\frac{1}{n}\right)}{\frac{1}{n}} = \frac{f \left(\frac{1}{n}\right)-f(0)}{\frac{1}{n}} + \frac{f \left(-\frac{1}{n}\right)-f(0)}{-\frac{1}{n}} \to 2f'(0)$$
But it doesn't say that the given series is convergent. Also, I'm not sure how to use the assumption that $f$ is $C^3$ function.
 A: hint
By Taylor-Lagrange formula,
$$f \in C^3(\Bbb R) \implies $$
$$(\forall n>0) \;\exists c_1\in(-\frac 1n,0) \;\exists c_2\in(0,\frac 1n)\;:$$
$$f(\frac 1n)=f(0)+\frac 1nf'(0)+\frac{1}{2n^2}f''(0)+\frac{1}{6n^3}f'''(c_2)$$
$$f(-\frac 1n)=f(0)-\frac 1nf'(0)+\frac{1}{2n^2}f''(0)-\frac{1}{6n^3}f'''(c_1)$$
$ f^{(3)} $ is continuous at $ [-1,1] $, so it is bounded by $ M $ and
$$|f'''(c_1)+f'''(c_2)|\le 2M$$
Thus
$$|u_n|\le \frac{M}{3n^2}$$
The series is absolutely convergent.
A: We have by Taylor expanding $f$ at $0$
$$n \left[f \left(\frac{1}{n}\right)- f \left(-\frac{1}{n}\right)\right]-2 f'(0)$$
$$=n\left[ \left(f(0)+\frac{f'(0)}{n}+\frac{f''(0)}{2!n^2}+\frac{f'''(x_1)}{3!n^3}\right)-\left(f(0)-\frac{f'(0)}{n}+\frac{f''(0)}{2!n^2}-\frac{f'''(x_2)}{3!n^3}\right)-2f'(0)\right]$$
$$=\frac{f'''(x_1)+f'''(x_2)}{6n^2},$$
where $x_1\in(0,\frac{1}{n})$ and $x_2\in(-\frac{1}{n},0)$ for each $n\geq 1$. Then since $f\in C^3(\mathbb{R})$, we have
$$n \left[f \left(\frac{1}{n}\right)- f \left(-\frac{1}{n}\right)\right]-2 f'(0)=O\left(\frac{1}{n^2}\right)$$
and the given series converges by the comparison test with $\sum_{n=1}^{\infty}\frac{1}{n^2}$
