real analysis - Is $limlimits_{ntoinfty} sumlimits_{i=1}^n prodlimits_{j=n-i+1}^n a_j=0$?

Let $\{a_n\}_{n\geqslant1}$ be a sequence of numbers in $(0,1)$ such that $a_n\to 0$ but $\sum\limits_{n\geqslant 1}a_n=\infty$ . Is it true that

$$\lim\limits_{n\to\infty}\ \sum\limits_{i=1}^n\ \prod\limits_{j=n-i+1}^n a_j=0\ \ \ ?$$

Had $a_i$'s be decreasing, some proceedings could have been done. But right now, I don't know how to proceed.

Answer

Pick some $p \in (0,1)$.

Choose $n$ large enough so that

1) $a_k < p$ for all $k > n/4$, this is possible since $a_k \to 0$.

2) $p^{n/4}n < p$, this is always possible and doesn't depend on the $a_k$

Now lets look at your sum:

$a_n + a_na_{n-1}+a_na_{n-1}a_{n-2} + \dots + a_na_{n-1}a_{n-2}\dots a_1$

If we look at the first $n/2$ terms, since all the $a_k$ are involved are at most $p$ we can bound this part of the sum by $p+p^2+p^3+\dots = p/(1-p)$.

If we look at the remaining $n/2$ terms they are all divisible by $n/4$ of the $a_k's$ we assumed to be less than $p$. So the total contribution is at most $p^{n/4}n/2 < p/2$ by our second assumption.

So this shows for any $p \in (0,1)$ for all sufficiently large $n$ this sum is bounded by $p/(1-p) + p/2$. But of course we can then choose $p$ to make this as small as we want.