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Let$$P_n=\prod_{k=1}^n k\sin\frac{x}{k}$$ For which $x$ this sequence converges?

I have done trivial cases only:

For $x\in\pi\mathbb{Z}$ all $P_n$ are 0's.

For $x\in(-1,1)$ it converges to 0 by the use of $|\sin t|\le|t|$.

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    Your first trivial case is not true, and your second trivial case is not that trivial! $$ \begin{align} P_n=\prod_{k=1}^{n}k\,\sin(x/k) &= x^n\,\prod_{k=1}^{n}\frac{\sin(x/k)}{(x/k)} \\ |x|\gt1 \quad &: \quad P_n\rightarrow\infty \\ x\in[0,+1[ \quad &: \quad P_n=0 \\ x=+1 \quad &: \quad P_n=c \\ x=-1 \quad &: \quad P_n\rightarrow\{+c,-c\} \\ x\in]-1,0[ \quad &: \quad P_n\rightarrow\,\color{red}{?} \end{align} $$ – 2017-02-13
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    @HazemOrabi The first case is trivially true. You've also written a bunch of things that are not correct, such as $P_n=0,P_n = c.$ Also, you've given no reasons. – 2017-02-13
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    $$ \prod_{k=1}^{\infty}k\,\sin(\pi/k)=0\times\infty $$ – 2017-02-13
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    @HazemOrabi Therefore the first factor is $0$ in all the $P_n.$ So all $P_n = 0$! – 2017-02-13
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    For a chosen $\,n\,$, yes $\,P_n=0\,$. But for $\,n\to\infty\,$, the story is different. [Check Here](https://www.wolframalpha.com/input/?i=Product%5Bk+Sin%5BPi%2Fk%5D,+%7Bk,+1,+Infinity%7D%5D). – 2017-02-13
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    @HazemOrabi This is getting silly. If $x\in \pi \mathbb Z,$ then all $P_n=0.$ – 2017-02-13
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    @zhw.: With all respect, for $\,n\to\infty\,$, it is an indefinite case of $\,0\times\infty\,$ which require a limit treatment to remove the indefinite. That is my opinion, and maybe I am wrong. Apologizing. – 2017-02-13

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Hint for $x=1.$ We can write

$$P_n= \exp \left (\sum_{k=1}^{n} \ln (k\sin (1/k)) \right).$$

Thus $P_n$ converges if $\sum_{k=1}^{\infty} \ln (k\sin (1/k))$converges. Use $\sin u = u + O(u^3)$ and $|\ln(1+u)| \le 2|u|$ for $|u|$ small to see this sum does indeed converge.

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    Where did $x$ go? – 2017-02-13
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    @martycohen $x=1.$ – 2017-02-13
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    For $x=1$, indeed $P_n$ converges. (+1) – 2017-02-13