# 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?

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.

• This is one of those times where using $\exp$ (\exp) is better than using $e^{\text{whatever}}$. – Asaf Karagila Nov 1 '14 at 19:11
• What about a comparison test using the fact that e.g. you must have $a_n\gt\dfrac{1}{n\ln^2n}$ for all sufficiently large $n$? $a_n\gt\dfrac1{n^2}$ isn't enough but a more slowly convergent series might be... – Steven Stadnicki Nov 1 '14 at 19:55
• @StevenStadnicki Yeah, intuitively it seems that $a_n$ must be larger than $1/(2n\sqrt n) \approx (\sqrt n-\sqrt{n-1})/n$, which makes $b_n$ smaller than roughly $\exp(-\sqrt n)$, hence summable. But I don't see how to do the comparison — we can't get that $a_n\ge 1/(2n\sqrt n)$ in the tail, just that it's not $\le$ in any tail, right? – user21467 Nov 1 '14 at 20:16

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, $$\sum_{n=2}^\infty e^{-S(n)} = \sum_{N=0}^\infty \sum_{J_N < n \leq J_{N+1}} e^{-S(n)} = +\infty.$$

1. Of course, the constant 10 is not optimal at all.
2. 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...
• Nice, +1. I think that the first $S(10^{J_N})$ should read $S(J_N)$ and that the computation on this exact line is not used afterwards. – Did Nov 5 '14 at 9:56
• @Did: thanks, the typo is now fixed. The estimate of $S(J_N)$ is actually used in order to estimate $S(10^{J_N})$. – Siméon Nov 5 '14 at 15:31

I think it's true. ($a_n$) is a decreasing sequence of positive numbers, hence it converges.

If $(a_n) \rightarrow L \neq 0$, very easy to prove that $b_n$ converges, with comparison of diverging series.

So, let's say that L=0.

Now there are two possibilities:

1) $(v_n)$ = $(na_n)$ unbounded : There is a subsequence $(g_k = v_{f(k)})$, with f a strictly increasing function of N$\rightarrow N$, such as: $g_k \geq k$.

$(v_k)$ is a positive sequence. Hence : $\sum_{k=0}^{f(n)} v_k \geq \sum_{k=0}^n v_{f(k)} = \sum_{k=0}^n k$ ; since there is more terms in the sum on the left.

From this, you get: $b_{f(n)} \leq e^{-\sum_{k=0}^n k} = G_n$

You get : $(b_{f(n)})$ convergent, and with the right framing with $(b_n)$ you can prove that the latter converges as well.

2) $(v_n)$ bounded; Bolzano-Weirstrass theorem applied to ($v_n$) give a converging sub-sequence that I'll note : ($c_n$) = ($v_{f(n)}$) , with f a strictly increasing function of N$\rightarrow N$. Let L ($\neq 0$ , for now) be that limit.

Likewise : $\sum_{k=0}^{f(n)} v_k \geq \sum_{k=0}^n v_{f(k)}$

From this, you get: $b_{f(n)} \leq e^{-\sum_{k=0}^n c_k} = C_n$

$c_n = L + o(1)$ => $\sum_{k=0}^n c_k = nL + o(n)$ ; You get $C_n$ convergent, and hence $(b_{f(n)})$ convergent. From this you can get $(b_n)$ convergent by framing the two sequence $(b_n)$ and $(b_{f(n)})$ together.

It remains the case: L=0

Edit : I tried one other way:

We have, using the decreasing of ($a_n$) :$k*a_k \geq k*a_n$ , for k=1..n

=> $\sum_{k=1}^n k*a_k \geq a_n*\sum_{k=1}^n k = \frac{a_n*n*(n+1)}{2}$

Hence: $b_n = \exp\left(-\sum_{k=1}^n k\,a_k\right) \leq \exp\left(-\frac{a_n*n*(n+1)}{2}\right) \leq \exp\left(-\frac{a_n*n^2}{2}\right) = \exp\left(-n^a*\frac{a_n*n^{2-a}}{2}\right)$

With: 0 < a < 1

Now I'm pretty much convinced that : $(a_n)$ decreases $\rightarrow 0, \sum a_n$ diverges => $(a_n*n^{2-a}) \rightarrow +\infty$ . That of course being intuitively because $a_n$ converges to 0 slower than $n^{2-a}$ diverging to $+\infty$ since $\sum a_n$ diverges. But I still haven't figured a proper proof yet..

But if this is true, you get : $b_n \leq \exp\left(-n^a*K\right)$ for $n \geq N_o$ , $N_o$ being defined by the divergence of $(a_n*n^{2-a})$.

We have: $\exp\left(-n^a*K\right) = O(\frac{1}{n^2})$ , hence : $b_n \leq O(\frac{1}{n^2})$ , so $(b_n)$ converges.

Of course this holds only if the property above is true... What do you think?

second edit: Well @siméon found a counter-example that makes it collapse, so forget it..

• Unfortunately, the sequence $(a_n)$ in my answer satisfies $\liminf_{n\to\infty} n^\epsilon a_n = 0$ for all $\epsilon > 0$. – Siméon Nov 5 '14 at 15:37
• Yes I just saw that :). I was in the middle of this, and I thought about a sequence $(a_n)$ that would be steady during a certain time and then decrease again (what you built essentially, nice one ) which may crush my proposal but I did not try to go any further. My mistake :), but my edit sounded nice enough to give it a try – mvggz Nov 5 '14 at 15:42