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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.

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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.

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  • $\begingroup$ I didn't get what you did with the second $n/2$ terms. Can you please explain better? $\endgroup$ – Qwerty Oct 13 '16 at 19:48
  • $\begingroup$ Lets assume $n$ is divisible by 4 (otherwise I should have some floors in the indices). The second $n/2$ terms are all of the form $a_{n/2+1}a_{n/2+2}...a_{n/4} \times \text{ (something less than 1)}$ and are hence are each less than $p^{n/4}$ by the first assumption. There are $n/2$ such terms so thats a total of at most $p^{n/4}n/2$, which we then bounded by $p/2$ using the second assumption. $\endgroup$ – Nate Oct 13 '16 at 19:57
  • $\begingroup$ Okay . So you claim that The second $n/2$ terms all have $a_na_{n-1}\cdots a_{3n/4}$ in them. And since all of these $a_i$ 's are $<p$ (because $i\geqslant 3n/4>n/4 $) , and rest of the $a_i$ 's are $<1\ \ \therefore$ the above is true. Is this right? $\endgroup$ – Qwerty Oct 13 '16 at 20:01
  • $\begingroup$ Haha right that should be $p^{3n/4}$ rather than $p^{3n/4}$, oops. In any case the same estimates work, I was just not being efficient. $\endgroup$ – Nate Oct 13 '16 at 21:20
  • $\begingroup$ No problem! Still works after all.. Thats the goal..;-D $\endgroup$ – Qwerty Oct 14 '16 at 5:18

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