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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 :)

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It isn't. A quick computation is that the inner product is of size roughly $\frac{1}{n^{2/3}}$. – André Nicolas Feb 2 '13 at 4:46
I've edited your title so that it looks rather neat. You have better use single "$" sign, so the title doesn't block much space. – Michael Li Feb 2 '13 at 4:51
thanks. next time I will use "$" instead. – Laura Feb 2 '13 at 4:53
Got something from an answer below? – Did Nov 24 '13 at 9:51

$$\prod_{k=1}^{n}\left(1-\frac{2}{3k}\right) \geq \frac13\prod_{k=2}^{n}\left(1-\frac{3}{3k}\right)=\frac13\prod_{k=2}^{n}\left(\frac{k-1}{k}\right)=\frac{1}{3n}$$

and $\sum \frac{1}{3n}$ diverges.

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Nice, no asymptotics! – Byron Schmuland Feb 2 '13 at 5:01
Definitely a powerful answer (+1) – user 1618033 Feb 2 '13 at 12:11

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.

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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 $$\sum_{n=1}^{N}\prod_{k=1}^n\left(1-\frac{a}{k}\right)\leqslant\sum_{n=1}^{N}\frac{1}{(n+1)^a}\leqslant\int_0^{N}\frac{1}{x^a}\mathrm dx=\frac{N^{1-a}}{1-a}. $$

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$$a_n = \prod_{k=1}^n \left(1 - \dfrac2{3k}\right) = \dfrac{1 \times 4 \times 7 \times 10 \times \cdots \times (3n-2)}{3 \times 6 \times 9 \times 12 \times \cdots \times 3n} = \dfrac{\Gamma(n+1/3)}{\Gamma(1/3) \Gamma(n+1)}$$ 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.

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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.

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