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I want to show that $\sum^{\infty}_{n=1}\frac{\sin(nx)}n$ converges uniformly on $[a, 2\pi - a]$ for $0<a<2\pi$

Actually, I know $\sum^{\infty}_{n=1}\frac{\sin(nx)}n$ converges (by using Dirichlet test).

However, it is difficult for me to prove converge "uniformly".

How can I prove this?

Do I have to use Weierstrass M-test?

Then how?

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Dirichlet's test can actually prove not only a mere pointwise convergence, but also uniform convergence! –  sos440 Sep 10 '12 at 14:36
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And if you want to avoid a direct reference to this theorem, you may just take summation by parts and apply Weierstrass $M$-test. –  sos440 Sep 10 '12 at 14:43

1 Answer 1

Hint: The Dirichlet test says that $$ \left|\sum_{k=n}^\infty\frac{\sin(kx)}{k}\right| \le\sup_{m\ge n}\left|\sum_{k=n}^m\sin(kx)\right|\frac1n\tag{1} $$ and $$ \begin{align} \left|\sum_{k=n}^m\sin(kx)\right| &=\left|\mathrm{Im}\left(\frac{e^{i(m+1)x}-e^{inx}}{e^{ix}-1}\right)\right|\\ &\le\frac1{|\sin(x/2)|}\tag{2} \end{align} $$

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