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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?

Philipp
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3 Answers3

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

Surjeet Singh
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  • When you wrote $\displaystyle \lim\dfrac{b_k}{\frac{1}{n^2}}=\lim n^2\left(1-\cos\left(\frac{1}{k}\right)\right)=\frac{1}{2}\neq 0$, I think it should be $k$ instead of $n$, right? And then you applied L'Hospital's rule twice, right? – Philipp Oct 30 '21 at 10:27
  • Yes we can expand $\cos\left(\frac{1}{n}\right)=1-\frac{1}{2n^2}+\dots$ – Surjeet Singh Oct 30 '21 at 10:32
  • @ÜñîqûêSurjeetSinghania You have to replace $n$ by $k$ in that argument. You are examining $b_k$ and not $b_n$. – Gary Oct 30 '21 at 10:37
  • It is hard to post answer from a mobile – Surjeet Singh Oct 30 '21 at 10:42
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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.

Gary
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Surb
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  • How do you show that $\log\left(\cos\left(\frac{1}{k}\right)\right)=\mathcal O\left(\frac{1}{n^2}\right)$ is true? Can you elaborate a bit? – Philipp Oct 27 '21 at 12:56
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    Just compute the Taylor polynomial of degree 2 of $\log(\cos(x))$ near $0$ @Philipp – Surb Oct 27 '21 at 13:46
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    Ah ok, got it, I calculated the derivatives wrong. So $\log(a_n)=\sum_{k=1}^n \log\left(\cos\left(\frac{1}{k}\right)\right)$ is of $\mathcal O\left(\frac{1}{n}\right)$, right? Btw why the downvotes? Is there something wrong with the answer? In my opinion it's another valid approach. – Philipp Oct 27 '21 at 20:50
  • @Philipp: It's a $\mathcal O(\frac{1}{n^2})$ (not a $\mathcal O(\frac{1}{n})$). For the downvote, I don't know, but it doesn't matter... maybe people think I've done a very complicated answer whereas a much easier argument can be provided... I just wanted to provide a more general answer than just decreasing and bounded from below. – Surb Oct 28 '21 at 07:42
  • Why is it not $\mathcal O(\frac{1}{n})$? If I apply the definition of Big-O-Notation I get: $\sum_{k=1}^m \log\left(\cos\left(\frac{1}{k}\right)\right)= \sum_{k=1}^m O(\frac{1}{n^2})\leq \sum_{k=1}^m C_k\left|\frac{1}{n^2}\right|\leq m\cdot C_{max}\left|\frac{1}{n^2}\right|\leq m\cdot C_{max}\left|\frac{1}{n}\right| $. So it is $\mathcal O(\frac{1}{n})$ and $\mathcal O(\frac{1}{n^2})$. – Philipp Oct 28 '21 at 11:15
  • $\cos(x)=1+\mathcal O(x^2)$ and $\log(1+x)=x+\mathcal O(x^2)$. Combining both yield $\log(\cos(x))=\mathcal O(x^2)$ when $x\to 0$. – Surb Oct 28 '21 at 17:48
  • What is wrong with the calculation in my previous comment? If $x\to 0$ then we always have that $\mathcal O(\frac{1}{n^2}) \subseteq \mathcal O(\frac{1}{n})$, don't we? – Philipp Oct 28 '21 at 18:01
  • It's not wrong, but what do you want to do with the fact that it's an $\mathcal O(1/n)$ ? And yes $\mathcal O(1/n^2)\subset \mathcal O(1/n)$... @Philipp – Surb Oct 28 '21 at 18:08
  • I thought that if $\sum_{k=1}^m \log\left(\cos\left(\frac{1}{k}\right)\right)$ is $\mathcal O(1/n)$ then it also proves that it is convergent. – Philipp Oct 28 '21 at 18:16
  • No, you can't say anything because $\sum_{n\geq 1}\frac{1}{n}=\infty $. – Surb Oct 28 '21 at 19:44
  • I am not sure if I have written it down correctly, or if you've misunderstood it, but I was trying to say that $\left(\sum_{k=1}^m \log\left(\cos\left(\frac{1}{k}\right)\right)\right)$, i.e. each partial sum, is $\mathcal O(1/n)$. So, if each partial sum is $\mathcal O(1/n)$, the series converges. – Philipp Oct 28 '21 at 21:37
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    @Philipp: Thanks for clarifying, it's indeed not what I have understood. Obviously the sum can't be an $\mathcal O(1/m)$ because otherwise the sum would converges to $0$ (which is not the case since it's negative and decreasing). The problem in your previous calculation it's that : what you wrote as $\sum_{k=1}^m \mathcal O(1/n^2)$ should be a $\mathcal O(1/k^2)$, after, when you write it's $\leq \sum_{}C_k|1/n^2|$, I guess that you wanted to write $\sum_{}C_k|1/k^2|$, but this is wrong ! It's true only when $k$ is large, but not for all $k$. – Surb Oct 29 '21 at 06:20
  • To wrap it up, $\log(\cos\left(\frac{1}{k}\right))=\mathcal{O}(\frac{1}{k^2})$ means that for large $k$, i.e. $k\geq n_0$, we have $|\log(\cos\left(\frac{1}{k})\right)|\leq C\frac{1}{k^2}$. So $\sum_{k=n_0}^n|\log(\cos\left(\frac{1}{k}\right))|\leq\sum_{k=n_0}^nC\frac{1}{k^2}$, which shows by the way that $\left(\sum_{k=n_0}^n |\log(\cos\left(\frac{1}{k}\right))|\right)$ is $\left(\sum_{k=n_0}^n\mathcal{O}\left(\frac{1}{k^2}\right)\right)$. As $\sum_{k=1}^nC\frac{1}{k^2}$ converges, $\sum_{k=1}^n\log(\cos\left(\frac{1}{k})\right)$ converges by comparison test, right? – Philipp Oct 31 '21 at 11:47
  • You got it :-) @Philipp – Surb Oct 31 '21 at 13:37
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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.

Gary
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Nope
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