Slowing down the divergence Is there any unbounded function $f:\mathbb{N} \rightarrow \mathbb{N}$ with $f(n+1) \geq f(n)>1$, such that for every  $g:\mathbb{N} \rightarrow \mathbb{N}$ with $g(n+1) \geq g(n)>1$ and $$\sum_{n = 1}^\infty \frac{1}{g(n)} = \infty$$ then $$\displaystyle \sum_{n = 1}^\infty \frac{1}{f(n)g(n)} = \infty $$
 A: The answer is no, that is: for every unbounded nondecreasing positive $(f(n))_n$ there exists a nondecreasing $(g(n))_n$ such that the series $\sum\limits_n1/g(n)$ diverges and the series $\sum\limits_n1/(f(n)g(n))$ converges.
Let us show the equivalent statement that for every nonincreasing positive $(a(n))_n$ such that $a_n\to0$ there exists a nonincreasing $(b(n))_n$ such that the series $\sum\limits_nb(n)$ diverges and the series $\sum\limits_na(n)b(n)$ converges.
To do this, assume without loss of generality that $a(n)\leqslant1$ for every $n$ and, for every $k\geqslant0$, consider 
$$
C_k=\{n\mid 2^{-k-1}<a(n)\leqslant2^{-k}\},\qquad c(k)=|C_k|,\qquad d(k)=\sum\limits_{i\leqslant k}c(i).
$$ 
Let $K$ denote the set of $k$ such that $c(k)\ne0$. For every $k$ in $K$ and every $n$ in $C_k$, let 
$$
b(n)=d(k)^{-1}.
$$
Then  $\sum\limits_nb(n)=\sum\limits_{k\in K}c(k)/d(k)$ which is infinite(1) hence the series $\sum\limits_nb(n)$ diverges. Since $a(n)\leqslant2^{-k}$ and $c(k)b(n)\leqslant 1$ for every $k$ in $K$ and every $n$ in $C_k$, $\sum\limits_na(n)b(n)\leqslant\sum\limits_{k\in K}2^{-k}$ hence the series $\sum\limits_na(n)b(n)$ converges.
Note (1): For every $k_0$ choose $k_1\geqslant k_0$ such that $d(k_1)\geqslant2d(k_0)$. This can be done since $d(k)\to+\infty$ when $k\to\infty$. Call $K_0$ the set of $k$ such that $k_0\lt k\leqslant k_1$ and $c(k)\ne0$. Then 
$$
\sum\limits_{k\in K_0}c(k)/d(k)\geqslant\sum\limits_{k\in K_0}c(k)/d(k_1)=(d(k_1)-d(k_0))/d(k_1)\geqslant1/2.
$$
Thus the rests of the series $\sum\limits_{k\in K}c(k)/d(k)$ do not converge to $0$ hence the series diverges.
A: There is no such function $f$. Given $f$, we can construct a counterexample $g$ as follows:
Starting with $g_1=2$ and $n_1=1$, in each step $k$, choose $\Delta_k$ such that $\Delta_k\ge g_k$ and $f(n_k+\Delta_k)\ge2^k$. This is possible since $f$ is unbounded. Now choose $g_{k+1}\ge g_k$ such that $1\le\Delta_k/g_{k+1}\le2$. This is possible, since $\Delta_k/g_k\ge1$. Now choose $n_{k+1}=n_k+\Delta_k$, and $g(n)=g_{k+1}$ for all $n_k\le n\lt n_{k+1}$. This stretch of values adds at least $1$ to the sum of $1/g$ but at most $2^{2-k}$ to the sum of $1/(fg)$, and thus the former diverges while the latter converges.
A: No.   Let $N_k = \min \{n: f(n) \ge 2^k\}$.  I'll leave the case where $N_{k+1} - N_k$ is bounded as an exercise.  So assume $N_{k+1} - N_k$ is unbounded.
 Let $g(n) = \max \{N_{k+1} - N_k: N_k \le n \}$.  Note that if $N_{k+1} - N_k > N_{j+1} - N_j$ for all $j < k$, $g(n) =  N_{k+1} - N_k$ for $N_k \le n < N_{k+1}$, and $\sum_{N_k \le n < N_{k+1}} \frac{1}{g(n)} = 1$.  Since there are infinitely
many of these $k$, $\sum_{n=1}^\infty \frac{1}{g(n)} = \infty$.  On the other hand,
$\sum_{N_k \le n < N_{k+1}} \frac{1}{f(n) g(n)} \le 2^{-k}$, so $\sum_{n=1}^\infty \frac{1}{f(n) g(n)}< \infty$.
