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True or False: The series

$$\sum_{n=1}^\infty \frac{\sin x}{1+n^2x^2}$$

converges uniformly on $[-\pi,\pi]$.

Appreciate comments on attempt below, especially if it's incorrect!

Attempt at Solution: False.

An infinite series $\sum_{n=1}^\infty f_n$ of real-valued functions on a metric space $E$ converges uniformly if and only if, given any $\epsilon \gt 0$, there exists an integer $N$ such that for any $n \gt m \ge N$,

$$|f_{m+1}(x)+f_{m+2}(x)+\dots+f_n(x)|\lt\epsilon$$

for all $x\in E$. Now let

$$f_n(x) = \frac{\sin x}{1+n^2x^2}$$

For small enough $\epsilon \gt 0,\ \sin \epsilon$ ~ $ \epsilon$. So in particular, $\sin \frac{\epsilon}{m} \gt \frac{\epsilon}{2m}$ for any positive integer $m$.

Choose $\epsilon$ small enough such that the above holds and also $$\frac{1}{1+16\epsilon^2}\gt \frac{2}{3}$$

Now let $N$ be any positive integer and let $x = \frac{\epsilon}{N} \lt \pi$. For this fixed $x$, $f_n(x)$ is strictly decreasing as $n \to \infty$. So we get

$$|f_{N+1}(x)+f_{N+2}(x)+\dots+f_{4N}(x)|$$ $$\gt 3N (f_{4N}(x))$$ $$= 3N \frac {\sin \frac{\epsilon}{N}}{1+16N^2(\frac{\epsilon}{N})^2}$$ $$\gt 3N \frac {\frac {\epsilon}{2N}}{1+16\epsilon^2}$$ $$= (\frac{3}{2})(\frac{1}{1+16\epsilon^2})(\epsilon)$$ $$\gt \epsilon$$

Thus, the series does not converge uniformly on $[-\pi,\pi]$.

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1 Answer

up vote 3 down vote accepted

Your question is simple: Is my proof correct? I have read your proof, and my answer is also simple: Yes!

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Thanks John :-) –  Conan Wong Dec 11 '12 at 16:10
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