Study the convergence of $\sum_{n=1}^{\infty}{\prod_{k=1}^{n}{\sin (k)}}$

Question

Can you help me to study the convergence of the following series:

$$\sum_{n=1}^{\infty}{\prod_{k=1}^{n}{\sin (k)}}$$

Thanks.

Answer 1

The series is convergent since it is absolutely convergent. Since the sequence $\{\{n/\pi\}\}_{n\in\mathbb{N}}$ is equidistributed $\pmod 1$, we have that $|\sin n|\leq\frac{\sqrt{3}}{2}$ holds for about $\frac{2N}{3}$ integers in the range $[1,N]$, assuming that $N$ is large enough. This gives: $$\left|\prod_{n=1}^{N}\sin n\right|\leq \left(\frac{\sqrt{3}}{2}\right)^{2N/3}.$$

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Since it is well-known that $$\prod_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{2n}{2^n}, $$ exploiting the concavity of $\sin x$ over $[0,\pi]$ and the Karamata's inequality we get the true order of magnitude of $|a_n|$: $$\left|\prod_{n=1}^{N}\sin n\right|\approx\frac{1}{2^N}.$$ This also follows from Weyl's equidistribution theorem and the fact that: $$\int_{0}^{\pi}\log\sin x\,dx = -\pi\log 2.$$

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Moreover, since $$ \sum_{n=113}^{+\infty}\prod_{\substack{k\in[1,n]\\k\neq 113}}\sin k$$ is bounded by some small constant by similar arguments, and $\frac{355}{113}$ is an extremely good approximation of $\pi$, we can compute the first four significant figures of the value of the series by just summing the first $113$ terms:

$$\sum_{n=1}^{+\infty}\prod_{k=1}^{n}\sin k = 1.\color{red}{6583}\ldots.$$

Answer 2

$|a_{n+2}/a_n|$ is bounded above by a number $x<1$, so $$\sum_n |a_n|<C\sum_n x^{n/2}$$