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I want to prove that

$$ \nexists \lim_{n \to \infty} \frac{\cosh (ny) \sin (nx)}{n^3} $$

for any $x \notin \pi\mathbb{Z}$, $y \neq 0$.

I think it is obvious, but I don't find a way to prove it rigurously. This is what I've done so far:

Suppose $\lim_{n \to \infty} \frac{\cosh (ny) \sin (nx)}{n^3} = L \in \mathbb{R}$. Now, we have

$$ \frac{\cosh (ny) \sin (nx)}{n^3} = \frac{1}{2} \frac{e^{ny}}{n^3}\sin (nx) + \frac{\sin(nx)}{2n^3 e^{ny}}$$

We can assume wlog that $y>0$. Now, because $(\sin (nx))_n$ is a bounded sequence and $\lim_{n \to \infty} 2n^3e^{ny}=0$ (if $y <0$ we would have taken $\lim_{n \to \infty} 2n^3e^{-ny}=0$), we know that

$$ \lim_{n \to \infty}\frac{\sin(nx)}{2n^3 e^{ny}}=0 $$

So

$$ \lim_{n \to \infty} \frac{e^{ny}}{n^3} \sin (nx) = 2L $$

My intuition now says that the product of a sequence that diverges to infinity with a divergent sequence that is bounded is always divergent. But I've searched a lot and I haven't found any similar result.

The most closely related results I've found use subsequences, but the only case I think it can be applied is to $x= \frac{\pi}{2}$, since $(\sin (\frac{n\pi}{2}))_n=(0,1,0,-1,0,1,\dots)$ so you can take a subsequence that diverges.

Edit:

As someone commented, it can be proven that $\sin (nx)$ cannot tend to zero if $n \notin \pi \mathbb{Z}$ but it remains unclear to me where to find the result that states that if $\lim_{n \to \infty} f(n)= \infty$ and $\lim_{n \to \infty} g(n) \neq 0$ then $\nexists \lim_{n\to \infty} f(n)g(n)$

Juan
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    This answers your question: Find all $x$ such that lim $\sin (nx)$ exists it proves that if $x\notin\pi\Bbb Z,$ $\sin(n\pi)$ cannot tend to $0$ and therefore $\frac{e^{ny}}{n^3} \sin (nx)$ cannot converge, thereby ending your proof. Note however that this sufficient condition $x\notin\pi\Bbb Z$ is necessary (otherwise your sequence is $0$). – Anne Bauval May 14 '23 at 22:34
  • @AnneBauval Thank you, I didn't realize $\sin (nx) =0$ for $x \in \pi \mathbb{Z}$. I've edited the question with the new information, but it is a established result why then $\frac{e^{ny}}{n^3} \sin (nx)$ cannot converge? – Juan May 15 '23 at 11:03
  • @Gonçalo Sorry, I thought it was standard notation that $\lim_n$ is $\lim_{n \to \infty} $ when it is not specified, I've edited the question. – Juan May 15 '23 at 11:05
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    You established it yourself! ;-) Just look at what you wrote: $\lim_{n \to \infty} \frac{e^{ny}}{n^3} \sin (nx) = 2L$. This clearly implies (for $y>0$) $\lim_{n\to\infty}\sin(nx)=0$, since $\lim_{n \to \infty} \frac{e^{ny}}{n^3} =+\infty$ . And @StasVolkov I think the 1st comment above, with a better link (and vote to close as duplicate), was sufficient. – Anne Bauval May 15 '23 at 11:36

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Since $e^{ny}/n^3\to \infty$ when $y>0$ (for $y<0$ the argument is similar) we only need to show that $sin(nx)\not\to 0$; hot establish this, see why sin(n) is dense in[-1,1] and here the argument is similar.

van der Wolf
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  • Maybe it is a dumb question, but while it is evident for me, where can I find a proof that shows that if $\sin (nx) \not\to 0$ then the limit doesn't exist? – Juan May 15 '23 at 11:07
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    if $\sin(nx)\not\to 0$ it means that for infinitely many $n$'s we have $|\sin(nx)|\ge\epsilon$ for some $\epsilon>0$. For those $n$s, $|e^{ny}/n^3 \sin(nx)|\ge e^{ny}/n^3 \epsilon\to\infty$. – van der Wolf May 15 '23 at 11:27