# Does $\sum_{n=1}^\infty \frac{1}{n! \sin(n)}$ diverge or converge?

Does the series $$\sum_{n=1}^\infty \frac 1 {n!\sin(n)}$$ converge or diverge? Even the necessary condition of the convergence is difficult to verify.

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Since at least one person seemed uncomfortable with the details of Hagen von Eitzen’s answer, I thought it would be a nice exercise to flesh it out.

As he says, the irrationality measure says that there are only finitely many positive integers $p,q$ such that $|\pi – \frac{p}{q}| < q^{-8}$. Since the left side is never 0 ($\pi$ is irrational), we may find a constant $c$ such that for all positive integers $p,q$, we have $|\pi – \frac{p}{q}| \ge c q^{-8}$.

To elaborate, let $\{(p_1, q_1), \dots, (p_m, q_m)\}$ be all the pairs of positive integers such that $|\pi – \frac{p_i}{q_i}| < q_i^{-8}$. Then set $$c := \min\left\{1, q_i^8 \left|\pi – \frac{p_i}{q_i}\right| : i=1, \dots, m\right\}.$$ Since $c$ is the minimum of finitely many strictly positive numbers, we have $c>0$.

Now given a positive integer $n \ge 3$, let $k_n$ be the nearest integer to $\frac{n}{\pi}$. Then $|n – k_n \pi| \le \frac{\pi}{2}$. Noting that for any $|x| < \frac{\pi}{2}$, we have $|\sin x| \ge \frac{|x|}{2}$, we thus have
$$|\sin (n)| = |\sin(n-k_n\pi)| \ge \frac{|n-k_n\pi|}{2} = \frac{k_n}{2} \left| \frac{n}{k_n} – \pi \right| \ge \frac{k_n}{2} \cdot c k_n^{-8} = \frac{c}{2 k_n^7}.$$

I claim for sufficiently large $n$ we have $k_n \le \frac{n}{3}$. Specifically, since $k_n$ is the nearest integer to $\frac{n}{\pi}$, we have $$k_n \le \frac{n}{\pi} + \frac{1}{2} = \frac{n}{3} + \frac{1}{2} – \left(\frac{1}{3}-\frac{1}{\pi}\right)n.$$ Therefore, we will have $k_n \le \frac n 3$ for any $n$ such that $\frac{1}{2} – \left(\frac{1}{3}-\frac{1}{\pi}\right)n \le 0$. Solving this inequality for $n$, we find that we get $k_n \le \frac n 3$ if $$n \ge \frac{1}{2 \left(\frac{1}{3} – \frac{1}{\pi}\right)} \approx 33.28;$$ in particular, for all $n \ge 34$.

Hence for all $n \ge 34$ we have
$$|\sin(n)| \ge c’ n^{-7}$$
where $c’ = \frac{3^7 c}{2}$, and therefore
$$\left|\frac{1}{n! \sin(n)} \right| \le c’ \frac{n^7}{n!}.$$
Since $\sum \frac{n^7}{n!}$ converges, we have that $\sum \frac{1}{n! \sin(n)}$ converges absolutely.

The question is: How good can $\pi$ be approximated by rationals? From here we learn that the irrationality measure of $\pi$ is $<8$. That is: There are at most finitely many fractions $\frac nk$ with $$\tag1\left|\pi-\frac nk\right|<\frac1{k^8}.$$
For the $n$th summand in your series, $\sin n$ can only be small if $n\approx k\pi$ for some $k$.
By $(1)$, we have $|k\pi-n|>\frac1{k^7}\approx \frac{\pi^7}{n^7}\gg\frac2{n^7}$ for almost all summands. Since for small $x$ we have $|\sin x|\approx|x|$, then also $|\sin n|\gg\frac1{n^7}$. Hence we can compare the absolute series
$$\sum_{n=1}^\infty\left|\frac1{n!\sin n}\right|$$
with
$$\sum_{n=1}^\infty\frac{n^7}{n!}$$
which clearly converges. Therefore your series converges absolutely.