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I know that the series $\sum_{n=1}^\infty \sin(nx) / n^\alpha$, with $0 < \alpha < 1$, converges for $x \in [0,2\pi]$.

I'm trying to understand if the function $f(x) = \sum_{n=1}^\infty \sin(nx) / n^\alpha$ is continuous on $[0,2\pi]$. I think not. In particular, I think that $\lim_{x \to 0^+} f(x) = +\infty$ while, obviously, $f(0) = 0$. But unfortunately, I'm not even able to find a sequence $x_n \to 0^+$ such that $f(x_n) \not\to 0$.

Does anyone have any suggestions? Thank you very much.

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4  
If $\alpha <1/2$ then the $L^2$ norm of your function is $\sum _{n=1}^{\infty} \frac{1}{n^{2\alpha}}$, which is infinite. Hence your function is not $L^2$ and therefore not continuous. –  Venkataramana Jun 10 '13 at 10:43
    
How about a (perhaps) more delicate question. Is it continuous on open interval $(0,2\pi)$ ?? –  Gerald Edgar Jun 10 '13 at 13:26
    
@Gerald, My answer below answers that question also. Yes, It is continuous on the open interval $(0,2\pi)$- –  Johan Andersson Jun 10 '13 at 13:28

3 Answers 3

The comment of Aakumadula is enough to see that is not continuous for $ \alpha < 1/2 $. By expressing $ \sin (nx) $ in terms of complex exponentials by Euler's formula we notice that this can be expressed as a difference between two periodic zeta-functions (special cases of the Lerch zeta-function), and by applying the functional equation of the periodic zeta-function (or Lerch) this can be expressed as a sum $a(\alpha) \zeta(1-\alpha,1-x/(2 \pi))+ b(\alpha) \zeta(1-\alpha,x/(2 \pi) ) $ where $\zeta(s,y)=\sum_{n=0}^\infty (n+y)^{-s}$ denote the Hurwitz zeta-function and $a(\alpha),b(\alpha)$ are some gamma-factors. The discontinuity can now be seen from the fact that $\zeta( 1 - \alpha,y)=y^{\alpha-1}+O(1)$ for $ 0 < y < 1 $ and any fixed $ 0 < \alpha < 1 $.

The choice of which functional equation we want to use is somewhat arbitrary (it all amounts down to the Poisson summation formula), for example we can also use the functional equation for the Hurwitz zeta-function, see e.g eq 8 in the link below, and the fact that cos is an even function and sin is an odd function.

http://mathworld.wolfram.com/HurwitzZetaFunction.html

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@Aakumadul & Andersson: Thank you for you answers. However, I am puzzled. Because the question comes from a first course in Fourier series... –  user34842 Jun 10 '13 at 15:33

For small $x>0$ we can write $$ f(x) = x^{\alpha-1} \cdot x \sum_{n=1}^\infty \frac{\sin nx}{(nx)^\alpha}, $$ which is $x^{\alpha-1}$ times a Riemann sum for $$ I_\alpha := \int_0^\infty \sin u \frac{du}{u^\alpha}. $$ This suggests that $x^{1-\alpha} f(x) \rightarrow I_\alpha$ as $x \rightarrow 0^+$. We shall see that $I_\alpha > 0$, so $x^{1-\alpha} f(x) \rightarrow I_\alpha$ means that not only is $f$ discontinuous at $x=0$ but $f(x) \rightarrow \pm \infty$ as $x \rightarrow 0^\pm$.

The fact that the Riemann sum approaches the integral is not immediate, because the interval of integration is infinite; but it can be shown by partial summation: the partial sums of $x \sum_n \sin nx$ are bounded, and $\{(nx)^{-\alpha}\}_{n=1}^\infty$ is a decreasing positive sequence, so $x \sum_{n=n_0}^\infty \sin nx / (nx)^\alpha = O((n_0 n)^{-\alpha})$ which approaches zero as $n_0 n \rightarrow \infty$.

Positivity of $I_\alpha$ is obtained as usual by showing that the integral is positive on each interval $(0,2\pi)$, $(2\pi,4\pi)$, $(4\pi,6\pi)$, etc., which we do by writing each $\int_{2\pi m}^{2\pi m + 2\pi} \sin u \, du/u^\alpha$ as $$ = \int_{2\pi m}^{2\pi m + \pi} \sin u \frac{du}{u^\alpha} + \int_{2\pi m + \pi}^{2\pi m + 2\pi} \sin u \frac{du}{u^\alpha} = \int_{2\pi m}^{2\pi m + \pi} \sin u \, \left(\frac1{u^\alpha} - \frac1{(u+\pi)^\alpha}\right) \, du $$ with the integrand positive on $2\pi m < u < 2\pi m + \pi$ (and small enough for the sum over $m$ to converge).

[In fact $I_\alpha$ can be evaluated in closed form $-$ it's $\Gamma(1+\alpha) \cos(\alpha\pi/2)$ $-$ but we do not need this.]

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Let $x_n=\frac{\pi}{2n}$. You have $x_n\to 0$ and $\sum_{n=1}^\infty \frac{\sin(nx_n)}{n^\alpha}=\sum_{n=1}^\infty \frac{\sin(\frac{\pi}{2})}{n^\alpha}=\sum_{n=1}^\infty \frac{1}{n^\alpha}$.

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1  
Yes, but what about it? (What is this good for?) –  Włodzimierz Holsztyński Oct 15 '13 at 22:26
    
You're getting confused by two different $n$s. –  Scott Morrison Oct 15 '13 at 22:48

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