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The series $ \ \large \sum_{n=1}^{\infty} \frac{\sin (nx)}{n} \ $

(i) converges uniformly on $ \ [2 \pi-5,5] \ $

(ii) converges uniformly on $ [2 \pi-10,10] \ $

(iii) converges uniformly on $ \ \left[\frac{\pi}{2}, \frac{3 \pi}{2} \right] \ $

(iv) does not converge uniformly .

Answer:

Here we can use Dirichlet's test.

Take $ \ \ a_n=\frac{1}{n} \ $ and $ b_n=\sin (nx) \ $

where $ a_n \ $ is monotonically decreasing .

and

$ \left| \sum_{n=0}^{k} \sin(nx)\right | =\left| \sin (x)+ \sin (2x)+\dots+ \sin (kx) \right| \ \leq \csc \left(\frac{x}{2}\right ) < \infty \ \ \text{if} \ x \neq 0,\pm 2\pi,\pm 4\pi, \dots $

Thus the series converges uniformly 0n $ \ \left[\frac{\pi}{2}, \frac{3 \pi}{2} \right] \ $

The option (i) is wrong because $ [5, 2 \pi-5]=[5, 1.14] \ $ , which is absurd .

Thus option (iii) is true.

But I need better reason why the other options are not correct .

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  • $\begingroup$ Where does the inequality $\sin x+\sin2x+\cdots+\sin kx\leqslant\csc\left(\frac x2\right)$ come from? $\endgroup$ – José Carlos Santos Nov 18 '17 at 14:31
  • $\begingroup$ It is actually absolute value. $ | \sin x+\sin 2x+...........+\sin kx | =| \frac{\cos (x/2)- \cos (n+1/2) x}{2 \sin (x/2)} | \leq cosec (x/2) \ $ $\endgroup$ – M. A. SARKAR Nov 18 '17 at 14:49
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    $\begingroup$ Concerning (i), one could argue that $[5,2\pi - 5] = \varnothing$, and every sequence of functions vacuously converges uniformly on $\varnothing$. The same for (ii). $\endgroup$ – Daniel Fischer Nov 18 '17 at 14:56
  • $\begingroup$ option (i) and (ii) should be incorrect because only one answer can be correct and option (iii) is already correct $\endgroup$ – M. A. SARKAR Nov 18 '17 at 15:03
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Dirichlet's test is pretty useful, but in dealing with pointwise convergence only.

It is well known that $$ f(x)=\sum_{n\geq 1}\frac{\sin(nx)}{n} $$ is the Fourier series of a sawtooth wave, a $2\pi$-periodic function which equals $\frac{\pi-x}{2}$ on the interval $(0,2\pi)$. We may notice that $f(x)$ is not continuous at the origin, since $f(0)=0$ but $\lim_{x\to 0^\pm}f(x)=\pm\frac{\pi}{2}$ (do you need a formal proof of this fact? If this is the case, please notify me in the comments). The same applies at every element of $2\pi\mathbb{Z}$. A uniformly convergent series of continuous functions is convergent to a continuous function, hence the given Fourier series cannot be uniformly convergent over any closed interval containing $[2\pi k-\varepsilon,2\pi k+\varepsilon]$ for some $k\in\mathbb{Z}$ and $\epsilon>0$. On the other hand the convergence is uniform over any compact set with a positive distance from $2\pi\mathbb{Z}$.

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