Convergence of the sequence $a_{n}=n\left(\prod\limits_{k=1}^{n} \sin\frac{1}{k}\right)^{1 / n}$ Check the Convergence of the sequence $a_{n}=n\left(\prod\limits_{k=1}^{n} \sin\frac{1}{k}\right)^{1 / n}$.
Find the limit if $a_{n}$ is convergent.
I tried to implement Cauchys Theorem for the Product of first n tems of a Convergent Sequente, But I Stuck.
Can logarithm be helping us?
 A: Asymptotically the sequence $(a_n)$ behaves as
$$b_n = n \left(\prod_{k=1}^n\frac{1}{k}\right)^{1/n}\,.$$
I haven't tried to prove it rigorously but it shouldn't be hard, since for $x\in(0,1]$ $\sin x$ is always strictly positive and it's increasing, so that the first factors with low $k$ do not bring any problem and due to the exponent $1/n$ they should bring a negligible contribution to the product.
Now $\left(\prod_{k=1}^n\frac{1}{k}\right)=\frac{1}{n!}$. So you have
$$b_n = \left(\frac{n^n}{n!}\right)^{1/n}\,.$$
Using the Stirling approximation $\frac{n!}{n^n}\sim \frac{\sqrt{2\pi n}}{e^n}$. So eventually you get
$$a_n\sim b_n \sim \frac{e}{(2\pi n)^{1/(2n)}}\to e\,.$$
A: Applying Stolz–Cesàro theorem, but let's do a few transformations first
$$a_n=n\left(\prod\limits_{k=1}^{n} \sin\frac{1}{k}\right)^{1 / n}=
\left(\prod\limits_{k=1}^{n} n \cdot\sin \frac{1}{k}\right)^{1 / n}$$
and consider $b_n=\ln{a_n}$ (which is defined, since $a_n>0,\forall n$) or
$$b_n=\frac{\sum\limits_{k=1}^{n} \ln{\left(n\cdot\sin{\frac{1}{k}}\right)}}{n}$$
Now
$$\frac{\sum\limits_{k=1}^{n+1} \ln{\left((n+1)\cdot\sin{\frac{1}{k}}\right)}-\sum\limits_{k=1}^{n} \ln{\left(n\cdot\sin{\frac{1}{k}}\right)}}{n+1 - n}=\\
\sum\limits_{k=1}^{n+1} \ln{\left((n+1)\cdot\sin{\frac{1}{k}}\right)}-\sum\limits_{k=1}^{n} \ln{\left(n\cdot\sin{\frac{1}{k}}\right)}=\\
n\ln{(n+1)}-n\ln{n}+\ln\left((n+1)\cdot\sin{\frac{1}{n+1}}\right)=\\
\ln{\left(1+\frac{1}{n}\right)^n}+\ln\left(\frac{\sin{\frac{1}{n+1}}}{\frac{1}{n+1}}\right)\to \color{red}{\ln{e}+\ln{1}=1}, n\to\infty$$
According to Stolz–Cesàro theorem
$$\lim\limits_{n\to\infty}b_n=1$$
and, as a result
$$\lim\limits_{n\to\infty}a_n=e$$
