Convergence of $a_k:=\cos(\frac{1}{k})a_{k-1}$ 
Does the sequence
$$
a_1=1\\
a_k:=\cos\left(\frac{1}{k}\right)a_{k-1}$$
converge?


Intuitevly I would say yes because we are multiplying numbers $<1$. However, we have infinitely many numbers so I am not sure if this intuition holds?
I was trying to find a sequence, something like $q^n$ where $0<q<1$, which might serve as an upper bound but I failed to get a fix $q$ as $n\to\infty$.
Also the approach using the mean value theorem seemed to be a dead end:
\begin{align*}&\cos\left(\frac{1}{k}\right)=\cos\left(\frac{1}{k}\right)-\cos\left(\frac{\pi}{2}\right)=\sin(\xi_k)\left(\frac{\pi}{2}-\frac{1}{k}\right),\text{ where }\frac{1}{k}<\xi_k<\frac{\pi}{2}\\
&\implies a_k=\sin(\xi_k)\left(\frac{\pi}{2}-\frac{1}{k}\right)\cdot \sin(\xi_{k-1})\left(\frac{\pi}{2}-\frac{1}{k-1}\right)\cdots~\cdot 1\\&<\left(\frac{\pi}{2}\right)^k\cdot\sin(\xi_{k})\cdot\sin(\xi_{k-1})\cdot\sin(\xi_{k-2})\cdots??\end{align*}
Do you have any tips which way to go?
 A: Given that $\displaystyle a_k=\cos\left(\frac{1}{k}\right)a_{k-1}$ with $a_1=1$
$\displaystyle a_n=\prod_{k=2}^n\cos\left(\frac{1}{k}\right)$ for $n\geq 2$
clearly $a_n>0$
Also $a_{n+1}-a_n=a_n\left(\cos\left(\frac{1}{n+1}\right)-1\right)<0\Rightarrow a_n>a_{n+1}$
By $\textit{Monotone convergence theorem}$ sequence $a_n$ Converges
Now $\displaystyle\lim_{n\to\infty} a_n=\prod_{k=2}^{\infty}\cos\left(\frac{1}{k}\right)$
Now we will prove this product not diverges to $0$
We know for $0<b_k<1 $ the product $\prod(1-b_k)$ converges iff
$\sum b_k $ converges
In our orignal problem $\displaystyle1-b_k=\cos\left(\frac{1}{k}\right)\Rightarrow b_k=1-\cos\left(\frac{1}{k}\right)$
clearly $0<b_k<1$ also $\displaystyle \lim\dfrac{b_k}{\frac{1}{k^2}}=\lim k^2\left(1-\cos\left(\frac{1}{k}\right)\right)=\frac{1}{2}\neq 0 $
hence by $\displaystyle\textit{Limit comparison test}    \sum_k b_k$  converges
$\Rightarrow \prod_{k=2}^{\infty}(1-b_k)$ converges$ \Rightarrow \lim a_n\neq 0$
Moreover $\lim a_n\in (0,1)$
A: First take the logarithm. Then the sequence becomes a series of $\log \left(\cos \left(\frac{1}{n}\right)\right).$
Now, by using limit comparison test with $\sum \frac{1}{n^2}$, the series $\sum 
\log \left(\cos \left(\frac{1}{n}\right)\right)$ is  convergent.
Hence, our given sequence  is also convergent.
A: You have that,
$$a_n=\prod_{k=1}^n\cos\left(\frac{1}{k}\right),$$
and thus $$\log(a_n)=\sum_{k=1}^n \log\left(\cos\left(\frac{1}{k}\right)\right).$$
Since $$\log\left(\cos\left(\frac{1}{k}\right)\right)=\mathcal O\left(\frac{1}{k^2}\right),$$
you should be able to conclude.
