Exercise on convergent series I am stumped by the following exercise (3.24 in Biler--Witkowski's book "Problems in mathematical analysis"):

Let $f$ be a continuous, increasing function from $[0,+\infty]$ to itself. Show that $$\sum \frac{1}{f(n)}$$ converges iff 
  $$\sum \frac{f^{-1}(n)}{n^2}$$ converges.

Initially, I thought one could apply the integral test and do a change of variables in the integral (assuming $f$ to be $C^1$, say, which does not change anything) to move from one series to the other. This does not quite work - one of the issues being that $x \mapsto \frac{f^{-1}(x)}{x^2}$ does not need to be decreasing.
I think I'm probably missing something obvious, as the exercise does not even deserve a hint in the book - would anyone have an idea?
Note: this is not homework! I have not used series in a while and need to brush up on them for a class I'll be teaching over the summer, so I thought I'd spend the weekend doing exercises - so far, so good, except for this particular exercise.
Thanks for your help!
 A: Suppose that $f$ is $C^1$ and that the series $$\sum_{n=1}^\infty \frac{1}{f(n)}$$ is convergent then by test integral, the integral
$$\int_1^\infty \frac{dt}{f(t)}\tag{1}$$ is also convergent, then we integrate by parts
$$\int_1^\infty \frac{dt}{f(t)}=\lim_{t\to\infty}\frac{t}{f(t)}-\frac{1}{f(1)}+\int_1^\infty \frac{tf'(t)}{f^2(t)}dt$$
and by variable change
$$\int_1^\infty \frac{dt}{f(t)}=\lim_{t\to\infty}\frac{t}{f(t)}-\frac{1}{f(1)}+\int_{f(1)}^\infty \frac{f^{-1}(t)}{t^2}dt$$
The convergence of the improper integral $(1)$ imply 
$$\lim_{t\to\infty}\int_t^{2t}\frac{dt}{f(t)}=0 $$
hence we can deduce by monotonicity of $f$ that
$$\lim_{t\to\infty}\frac{t}{f(t)}=0$$
hence we can see that the integral
$$\int_{f(1)}^\infty \frac{f^{-1}(t)}{t^2}dt$$
is convergent. Moreover we have
$$\sum_{n=1}^\infty \frac{f^{-1}(n)}{(n+1)^2}\leq \sum_{n=1}^\infty\int_n^{n+1}\frac{f^{-1}(t)}{t^2}dt=\int_{f(1)}^\infty\frac{f^{-1}(t)}{t^2}dt<+\infty$$
hence the series
$$\sum_{n=1}^\infty \frac{f^{-1}(n)}{n^2}$$ 
is convergent.
By the same manner we can prove the other implication.
