$\sum \frac{1}{f(k)}$ converges iff $\sum \frac{f^{-1}(k)}{k^2}$ converges

Let $f$ be a strictly increasing positive continuous function defined on $[1,\infty)$ with limit $\infty$ as $x$ goes towards $\infty$. Then $\sum_{k=0}^{\infty} \frac{1}{f(k)}$ converges if and only if $\sum_{k=0}^{\infty} \frac{f^{-1}(k)}{k^2}$ converges, where $f^{-1}$ denotes the inverse of $f$.

I saw this claim as an exercise in a analysis textbook, linked from this site. Can not remember which one unfortunately. It was listed as a challenging exercise and has proven too challenging for me.

My initial idea was to try to use the integral test. $\sum \frac{1}{f(k)}$ converges exactly when $\int \frac{1}{f(x)}$ converges. I thought I might do some smart change of variable to find that $\int \frac{f^{-1}(x)}{x^2}$ converges. I could not come up with one however and I also realized that I do not know that the sum $\sum \frac{f^{-1}(k)}{k^2}$ satisfies the conditions for using the integral test. Unfortunately I ran out of ideas at that point.

As the OP noted, by the integral test the sum converges iff $\int_0^\infty dx\,\frac{1}{f(x)}$ converges. Integrating by parts, we obtain $$\int_0^\infty dx\,\frac{1}{f(x)} = - \int_0^\infty dx\, \frac{x f'(x)}{f(x)^2}.$$ Changing the integration variable from $x$ to $y=f(x)$ (which is admissible as $f(x)$ is monotonously increasing) yields $$\int_0^\infty dx\,\frac{1}{f(x)}= - \int_{f(0)}^\infty dy\, \frac{f^{-1}(y)}{y^2}.$$ Using the integral test again, we have that the original series converges iff $$\sum \frac{f^{-1}(k)}{k^2}$$ converges.

Edit: (trying to incorporate some comments)

1) the OP did not specify that $f$ is differentiable. One can use a Spline interpolation to make $f \in C^1$. (thaks to user7530)

2) there was a question about the fate of the boundary term $x f(x)$: one should prove the argument in two steps. First assume $\sum f(k)^{-1}$ converges, then the boundary term is zero as $f(k)$ falls off faster than $k^{-1}$ and you can prove that $\sum \frac{f^{-1}(k)}{k^2}$ converges. Then the other way round, assume $\sum \frac{f^{-1}(k)}{k^2}$ converges. The boundary term then reads $f^{-1}(y) y$ which you can prove to be finite by the same method and thereby showing that $\sum f(k)^{-1}$ converges.

Define $A(n)$ to be $\{k: 2^n \leq f(k) < 2^{n+1}\}$, and $|A(n)|$ its cardinality. Let $S_n$ be the sum of the terms of the first sequence over $k$ in $A_n$. Then we have $${1 \over 2^{n+1}}|A(n)| < S_n \leq {1 \over 2^n}|A(n)|$$ Thus the first sum $\sum_n S_n$ is within a factor of $2$ of ${\displaystyle \sum_n {|A(n)| \over 2^n}}$.

Next, let $T_n$ be the sum of the terms of the second sequence over $2^n \leq k < 2^{n+1}$. In the sum $T_n$, since $f^{-1}$ is increasing, $f^{-1}(k)$ is less than $f^{-1}(2^{n+1}) = \sum_{m \leq n+1}|A(m)|$, and at least $f^{-1}(2^n) = \sum_{m \leq n}|A(m)|$. There are $2^n$ terms in the sum defining $T_n$, so we have
$$2^{-2(n+1)}2^n\sum_{m \leq n}|A(m)| \leq T_n < 2^{-2n}2^n\sum_{m \leq n+1}|A(m)|$$ Summing the above in $n$, we have that $\sum_n T_n$ is within a constant factor of $$\sum_n 2^{-n} \sum_{m \leq n} |A(m)|$$ Switching the order of summation, this becomes $$\sum_m |A(m)| \sum_{n \geq m} 2^{-n}$$ $$= 2\sum_m{|A(m)| \over 2^m}$$ This is within a constant factor of $\sum_n {S_n}$ computed above, so one series converges iff the other one does.