TITLE: $\sum \frac{1}{f(k)}$ converges iff $\sum \frac{f^{-1}(k)}{k^2}$ converges
QUESTION [11 upvotes]: 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.

REPLY [4 votes]: 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.