$f:[-1,1] \to \mathbb{R}$, $f(0)=0$, $f'(0)\ne0$. Show that $\sum_{k=2}^\infty f \left( \frac{1}{n \ln n} \right)$ diverges. 
$f:[-1,1] \to \mathbb{R}$,  $f(0)=0$, $f'(0)\ne 0$. 
Show that $\sum_{k=2}^\infty f \left( \frac{1}{n \ln n} \right)$ diverges.

I can't figure out how the information about the derivative helps me. Hint, please?
 A: Assume $f'(0)=k>0.$ That is,
$$\forall \epsilon>0\exists \delta >0: 0<x<\delta \implies \frac{f(x)}{x}>k-\epsilon.$$ Thus, for $\epsilon=k/2$ we have 
$$\exists \delta >0: 0<x<\delta \implies \frac{f(x)}{x}>k/2.$$ Thus, for $n$ big enough we have
$$f\left(\frac{1}{n\log n}\right)\ge \frac k{2 n\log n}.$$ Thus the series is not convergent. We can argue in a similar way if $f'(0)=k<0.$
A: By definition
$$
0\neq f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{f(x)}{x}=
\lim_{n\to+\infty}\frac{f(a_n)}{a_n}
$$
which is true for every sequence $(a_n)_n\subset[-1,1]$ converging to $0$.
Thus let's take 
$$
a_n:=\frac1{n\log n}
$$
from which you have that
$$
f\left(\frac1{n\log n}\right)\sim\frac1{n\log n}\;\;\mbox{as}\;\;n\to+\infty
$$
and you can conclude since $\sum_n\frac1{n\log n}$ diverges by Cauchy condensation criteria.
A: As $f$ is differentiable at $0$, we know that $f$ is continuous at $0$. So for a small enough $\delta$, we have that $f$ does not change sign on $[0, \delta]$. Suppose without loss of generality that $f$ is positive on $[0,\delta]$.
Then we apply the standard limit comparison test, and compare our sum to $\frac{1}{x\ln x}$. Then
$$ \lim_{n \to \infty} \frac{f\left(\frac{1}{n \ln n}\right)}{\frac{1}{n \ln n}} = \lim_{n \to \infty} \frac{f\left(\frac{1}{n \ln n}\right) - f(0)}{\frac{1}{n \ln n} - 0} = f'(0) \neq 0.$$
So the sum $\sum f(\frac{1}{n \ln n})$ behaves like the sum $\sum \frac{1}{n \ln n}$, and therefore diverges.

[Thank you to Jack D'Aurizio for pointing out an error, and then to Hetebrij for indicating how to correct that error]
