Is the sequence $(n\sin n)$ properly divergent i.e. do we have $\lim \space (n \sin n)= +\infty$ , or $\lim \space (n \sin n)= -\infty$ ? ( I think the answer is know because $(n)$ is increasing unbounded but $(\sin n)$ is bounded and divergent i.e. oscillating , but that is just a justification not a proof , I would really appreciate a rigorous proof )
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For natural numbers $n$, the answer is equivalent to http://math.stackexchange.com/questions/665776/convergence-of-the-sequence-frac1n-sinn – JiK Feb 06 '14 at 13:01
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Hint: Every interval of the form $((2k+1)\pi/2-1/2,(2k+1)\pi/2+1/2)$ contains an integer. From this, you can find a subsequence that converges to $\infty$ and a subsequence that converges to $-\infty$. – David Mitra Feb 06 '14 at 13:03
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Was the limit indented over the natural numbers? – Fabian Feb 06 '14 at 13:04
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@Fabian:- Yes , I said sequence , have seen your answer , great! – Feb 06 '14 at 13:05
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My answer does not apply immediately to the sequence. – Fabian Feb 06 '14 at 13:07
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@Fabian: I know because $2k\pi +\pi /2$ is not a natural number. – Feb 06 '14 at 13:11
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Yes, but I have edited it. – Fabian Feb 06 '14 at 13:14
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Set $f(x) = x \sin x$. You find that with $k\in \mathbb{N}$ $$ f(2k \pi+ \pi/2) \geq k$$ and $$ f( 2k \pi - \pi/2) \leq -k.$$
Thus $\limsup_{x\to\infty} f =\infty$ and $\liminf_{x\to\infty} f= -\infty$ and the limit does not exist.
Edit:
The above answer was for the limit over the real numbers. The OP asks for the limit of the sequence $f(n) = n \sin n$. We can proceed very similar.
We have to observe that for $k \in \mathbb{N}$, we have the inequalities $$ f(x) \geq k \text{ for } x \in [2 \pi k + \pi/4, 2 \pi k + 3\pi/4],$$ $$ f(x) \leq -k \text{ for } x \in [2 \pi k- 3\pi/4, 2 \pi k - \pi/4].$$ Furthermore, there is always an integer in the intervals which are of length $\pi/2 \geq 1.5$. Thus, we can again conclude the same thing about limsup and liminf as above.
Fabian
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1An alternative approach with the same conclusion would be to say that since $\sin(\pm 1) \approx \pm 0.841$ and $\sin(\pm 2) \approx \pm 0.909$, in every $7(\gt 2 \pi)$ terms starting from the $n$th, there will be at least one larger than $0.841 n$ and at least one smaller than $-0.841 n$. – Henry Feb 06 '14 at 14:18