Let $f(x)=\sum_{n=1}^{\infty}{\sin\left(\frac{x}{n^2}\right)}.$
a) Show that the series converges for $x\in [0,\pi/2]$.
b) Show that $f$ is monotone and continuous on this interval.
This is what I have for (a). Is it right? I am stuck on continuity for part (b).
For (a), I showed that since $\frac{x^n}{n!}>\frac{x^{n+2}}{(n+2)!},$ we get that for $x\in[0,\pi/2]$, $0\leq \sin x = x-\sum_{n\in I}{\frac{x^n}{n!} - \frac{x^{n+2}}{(n+2)!}}$ where $I=\{3,7,11,\ldots\}$. Since the last term is negative or zero, $\sin x\leq x$ for $x\in [0,\pi/2]$.
Thus, $\sum_{n=1}^{\infty}{\left|\sin\left(\frac{x}{n^2}\right)\right|} \leq \sum_{n=1}^{\infty}{\frac{x}{n^2}}$ which converges. So the series converges.
b) I was able to show $f$ is monotone. Is there a nice way to show $f$ is continuous?