TITLE: If $\sum\limits_{n=1}^\infty a_n$ diverges, then $\sum\limits_{n=1}^\infty\exp\left(-\sum\limits_{k=1}^n k a_k\right)$ converges?
QUESTION [6 upvotes]: Let $(a_n)$ be a decreasing sequence of positive numbers and let $$b_n = \exp\left(-\sum_{k=1}^n k\,a_k\right).$$ Is is generally true that
$$
\sum_{n=1}^\infty a_n = +\infty \implies \sum_{n=1}^\infty b_n < +\infty?
$$
On the first hand I am inclined to answer negatively because some series diverge very very slowly, but on the other hand I could not come up with a counterexample.
For instance, $a_n = \dfrac{1}{n(\log n)}$ yields $b_n = \exp\left(-\dfrac{n}{\log n} + o(1)\right)$ so the series is convergent.

REPLY [4 votes]: The following construction gives a counterexample.
Consider a sequence of "jump" times $J_n$ satisfying for all $n$,
$$
    J_n \leq 10^{J_n} \leq 100^{J_n} \leq J_{n+1}
$$
and define $a_k = \dfrac{1}{J_n}$ for all $(k,n)$ such that $J_{n-1} < k \leq J_n$.
The sequence $(a_k)$ is positive, decreasing, and the series $\sum a_k$ is divergent because
$$
    \sum_{k = 2}^\infty a_k = \sum_{n=1}^\infty \frac{J_n-J_{n-1}}{J_n}=\sum_{n=1}^\infty (1+o(1)) = +\infty.
$$
Let $S(n) = \sum_{k=1}^n k a_k$. As $N$ tends to infinity,
$$
    S(J_N) = \sum_{n=1}^N \frac{(J_n-J_{n-1})(J_n + J_{n-1} +1)}{2 J_n} \sim \frac{1}{2}\sum_{n=1}^N J_n \sim \frac{1}{2}J_N.
$$
In addition, since $J_N \leq 10^{J_N} \leq (10^{J_N})^2 \leq J_{N+1}$,
$$
    S(10^{J_N}) = S(J_N) + O(1) \sim \frac{1}{2}J_N.
$$
 Then for all $N$ large enough,
$$
    \sum_{J_N < n \leq J_{N+1}} e^{-S(n)} \geq \sum_{J_N < n \leq 10^{J_N}} e^{-S(10^{J_N})} \geq 10^{J_N} e^{-J_N} \geq 1.
$$
Finally,
$$
\sum_{n=2}^\infty e^{-S(n)} = \sum_{N=0}^\infty \sum_{J_N < n \leq J_{N+1}} e^{-S(n)} = +\infty.
$$


Of course, the constant 10 is not optimal at all.
The construction can be generalized to show that one can find $(a_k)$ such that $\sum_n e^{-e^{S(n)}} = +\infty$, and so on...