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 .


Here we can use Dirichlet's test.

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

where $ a_n \ $ is monotonically decreasing .


$ \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 .

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $2\pi\mathbb{Z}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.