Is the expression $\sum_{n=1}^{\infty}\prod_{k=1}^{n}\left(1-\frac{2}{3k}\right)$ bounded? 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 :)
 A: Informal solution: Take the logarithm of the product. We have $\log(1-t)=-t+O(1/t^2)$. Summing, we get 
$$-\frac{2}{3}\left(1+ \frac{1}{2}+\cdots+\frac{1}{n}\right)+O(1/n)=-\frac{2}{3}H_n+O(1/n),$$
where $H_n$ is the $n$-th harmonic number. 
But the difference between $H_n$ and $\log n$ is bounded. 
It follows that our product behaves like $n^{-2/3}$, so the sum diverges. 
A: 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
$$\sum_{n=1}^{N}\prod_{k=1}^n\left(1-\frac{a}{k}\right)\leqslant\sum_{n=1}^{N}\frac{1}{(n+1)^a}\leqslant\int_0^{N}\frac{1}{x^a}\mathrm dx=\frac{N^{1-a}}{1-a}.
$$
A: $$a_n = \prod_{k=1}^n \left(1 - \dfrac2{3k}\right) = \dfrac{1 \times 4 \times 7 \times 10 \times \cdots \times (3n-2)}{3 \times 6 \times 9 \times 12 \times \cdots \times 3n} = \dfrac{\Gamma(n+1/3)}{\Gamma(1/3) \Gamma(n+1)}$$
Further the asymptotic for $\Gamma(z)$ is given by
$$\Gamma(z+1) \sim \sqrt{2 \pi z} \left(\dfrac{z}{e} \right)^z$$
Hence,
\begin{align}
a_n & \sim \dfrac1{\Gamma(1/3)} \dfrac{\sqrt{2 \pi (n-2/3)} \left(\dfrac{n-2/3}{e} \right)^{n-2/3}}{\sqrt{2 \pi n} \left(\dfrac{n}{e} \right)^{n}}\\
& \sim \dfrac{e^{2/3}}{\Gamma(1/3)} \left(1 - \dfrac2{3n}\right)^{n-2/3} \dfrac1{n^{2/3}}\\
& \sim \dfrac1{\Gamma(1/3)} \dfrac1{n^{2/3}}
\end{align}
Hence, $$\sum_{n=1}^{N} a_n \sim \dfrac{3 N^{1/3}}{\Gamma(1/3)}$$ which obviously shows that your series diverges.
A: $$\prod_{k=1}^{n}\left(1-\frac{2}{3k}\right) \geq \frac13\prod_{k=2}^{n}\left(1-\frac{3}{3k}\right)=\frac13\prod_{k=2}^{n}\left(\frac{k-1}{k}\right)=\frac{1}{3n}$$
and $\sum \frac{1}{3n}$ diverges.
A: Since  $$\prod_{k=1}^{n}\left(1-\frac{2}{3k}\right) = \left|{-1/3\choose n}\right|\sim {n^{-2/3}\over\Gamma(1/3)},$$  the sum diverges. 
