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I'm wondering how to go about proving or disproving pointwise and uniform convergence of $\sum_{n=1}^\infty \sin(x/n^2)$ on $\mathbb R$ and on $[0,1]$.

My idea is to use the fact that $|\sin x| < |x|$, and as $x \to 0$ $\sin x \to x$, but I'm not quite sure how. I can't even see how to prove or disprove pointwise convergence in these cases, let alone uniform.

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Using those facts, the series is dominated by the convergent series $|x|\sum_n\frac1{n^2}$, hence the series converges pointwise. We also radily see from this that we have uniform convergence on bounded domains.

However, we don not have uniform convergence on all of $\Bbb R$, for if $\epsilon<\sin 1$, then the $n$th summand is not $<\epsilon$ at $x=n^2$.

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    Can you clarify? It is not true that for any x in R, my series is dominated by your series. For x = 1000, my series sums to a value greater than 30 whereas yours remains much smaller. – 2017-01-29
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    We (or at least I) say that $\sum a_n$ is dominated by $\sum b_n$ if there $a_n=O(b_n)$, that is $|a_n|\le cb_n$ for some $c$ and for almost all $n$. In stricter interpretation, it is dominated by $|x|\sum_n\frac1{n^2}$, so still convergent. – 2017-01-29