Convergence of $a_k:=\cos(\frac{1}{k})a_{k-1}$

Question

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

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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?

Answer 1

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 $\lim a_n\in (0,1)$

Answer 2

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.

Answer 3

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.