TITLE: $f:[-1,1] \to \mathbb{R}$, $f(0)=0$, $f'(0)\ne0$. Show that $\sum_{k=2}^\infty f \left( \frac{1}{n \ln n} \right)$ diverges.
QUESTION [5 upvotes]: $f:[-1,1] \to \mathbb{R}$,  $f(0)=0$, $f'(0)\ne 0$. 
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?

REPLY [5 votes]: 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 $f'(0)=k<0.$