prove $\sum_{n=1}^{\infty}a_n$ converges iff $\sum_{n=1}^{\infty}f(a_n)$ converges 
let $f$ be differentiable on $x=0$.  $f(0)=0, f'(0) \neq 0$
$a_n$ is a positive sequence converging to $0$.
prove $\sum_{n=1}^{\infty}a_n$ converges iff $\sum_{n=1}^{\infty}f(a_n)$ converges.

I was thinking we know that $f$ has the same sign on $(0, \delta)$ for some $\delta>0$, assuming it is positive for example we can try showing its partial sum sequence is bounded, or other stuff that works for a sequence with the same sign, but with no success  so far.
I'm looking for a tiny hint, just for the direction assuming $\sum_{n=1}^{\infty}a_n$ converges
 A: Because you said you only wanted a tiny hint. 
HINT. Use the definition of the derivative at $x=0$.
Perhaps a stronger hint, 'use' the Mean Value Theorem. [Note it does not apply here. But if it did, what would you use it to do?  You will observe that the same things you use the MVT to produce can be obtained from the definition of the derivative alone in this case. Except I think it is simpler to see what you want in the MVT case before trying to deal with the limit definition of the derivative case.] 
A: As you already pointed out, $f$ keeps the same sign in a small interval $I=(0, \delta)$. Without loss of generality we may assume that $f|_I>0$. 
Assuming the series $\sum a_n$ converges, first note that $a_n \to 0$. Therefore there exists $N>0$ such that for all $n>N$, $f(a_n) \le c \times a_n $ where $c>0$ can be chosen independently of $n$. Let us prove that this is indeed the case.

Hint: By the differentiability of $f$ at $0$ we have $0<f'(0) = \lim_n\frac{f(a_n)- f(0)}{a_n}$
   (why?). Hence there exists $N_1>0$ such that for
  all $n>N_1$ we have $\frac{f(a_n)- f(0)}{a_n}<2f'(0)$. Rearrange the
  terms, remember that $f(0)=0$ and conclude.

I hope this helps. If you have any further question/remark, feel free to ask.
A: In fact you may prove both directions simultaneously.
If we assume, without loss of generality, that $f'(0) > 0$, then there exist $0 < \alpha < \beta$ such that $\alpha x \leq f(x) \leq \beta x$ for sufficiently small (positive) $x$.
Is this enough for a hint?
