TITLE: Is the expression $\sum_{n=1}^{\infty}\prod_{k=1}^{n}\left(1-\frac{2}{3k}\right)$ bounded?
QUESTION [6 upvotes]: Today I have encounter an integral:
$$\int_0^{\infty}\left[\frac{1}{3}\frac{\sin x}{x}+\cdots+\frac{1\times4\times\cdots\times(3n-2)}{3^nn!}\left(\frac{\sin x}{x}\right)^n+\cdots\right]\text{d}x$$
since $$\int_0^{\infty}\frac{\sin x}{x}=\frac{\pi}{2}$$
so I want to estimate $$\sum_{n=1}^{\infty}\prod_{k=1}^{n}\left(1-\frac{2}{3k}\right)$$
Is the above expression bounded? Thanks very much :)

REPLY [3 votes]: N.S.'s idea can be adapted to yield the exact asymptotics. To wit,
$$
\prod_{k=1}^n\left(1-\frac{2}{3k}\right)=\frac13\prod_{k=2}^n\left(1-\frac{2}{3k}\right)\geqslant\frac13\prod_{k=2}^n\left(\frac{k-1}k\right)^{2/3}=\frac1{3n^{2/3}}.
$$
This uses the fact that, by convexity, for every $a$ and $t$ in $(0,1)$, $1-at\geqslant(1-t)^a$. 
Likewise, for every $a$ in $(0,1)$, 
$$
\prod_{k=1}^n\left(1-\frac{a}{k}\right)=(1-a)\prod_{k=2}^n\left(1-\frac{a}{k}\right)\geqslant(1-a)\prod_{k=2}^n\left(\frac{k-1}k\right)^{a}=\frac{1-a}{n^a},
$$
hence
$$\sum_{n=1}^{N}\prod_{k=1}^n\left(1-\frac{a}{k}\right)\geqslant\sum_{n=1}^{N}\frac{1-a}{n^a}\geqslant\int_1^{N+1}\frac{1-a}{x^a}\mathrm dx=(N+1)^{1-a}-1,
$$
and, in particular,
$$\sum_{n=1}^{+\infty}\prod_{k=1}^n\left(1-\frac{a}{k}\right)=+\infty.
$$
In the other direction, for every $a$ and $t$ in $(0,1)$, $1-at\leqslant(1+t)^{-a}$, hence
 $$
\prod_{k=1}^n\left(1-\frac{a}{k}\right)\leqslant\prod_{k=1}^n\left(\frac{k}{k+1}\right)^{a}=\frac{1}{(n+1)^a},
$$
and
$$\sum_{n=1}^{N}\prod_{k=1}^n\left(1-\frac{a}{k}\right)\leqslant\sum_{n=1}^{N}\frac{1}{(n+1)^a}\leqslant\int_0^{N}\frac{1}{x^a}\mathrm dx=\frac{N^{1-a}}{1-a}.
$$