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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?

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for n>N, such that x/N < pi/2, we have sin(x/(n+1)^2) < sin(x/(n)^2) .. perhaps –  Santosh Linkha Aug 10 '12 at 22:23
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2 Answers 2

up vote 3 down vote accepted

Let $\displaystyle M_n=\frac{\pi}{2n^2}$. Suppose your series is $\sum u_n(x)$. From what you showed, $$|u_n(x)|\leq \frac{x}{n^2}\leq \frac{\pi}{2n^2}$$ on all of $[0,\frac{\pi}{2}]$. Since the seires $\sum M_n$ converges, the Weierstrass M-test shows that the series converges uniformly, and since the partial sums are all continuous the limit is continuous as well.

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Your series' convergence is uniform. Hence the limit is a continuous function.

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Can you show explicitly that the series converges uniformly? –  neelp Aug 10 '12 at 20:57
    
But your inequality is uniform with respect to x! –  aska langley Aug 10 '12 at 21:05
    
Or $\sum_{k=N}^{\infty}sin(x/k^2)<C\sum_{k=N}^{\infty}1/k^2\to 0$ –  aska langley Aug 10 '12 at 21:07
    
I still don't really see what you are saying. So say we have $\alpha > 0$. Then we can choose $N$ such that if $j,k\geq N$, then $\sum_{n=j}^{k}{sin(\frac{\pi/2}{n^2})} < \alpha$ and so $\sum_{n=j}^{k}{sin(x/n^2)} < \sum_{n=j}^{k}{sin(\frac{\pi/2}{n^2})} < \alpha$ since the latter sum converges pointwise? –  neelp Aug 10 '12 at 21:13
    
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