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Let $\alpha$ be a real number. I'm asked to discuss the convergence of the series $$ \sum_{k=1}^{\infty} \frac{\sin{(kx)}}{k^\alpha} $$ where $x \in [0,2\pi]$.

Well, I show you what I've done:

  • if $\alpha \le 0$ the series cannot converge (its general term does not converge to $0$ when $k \to +\infty$) unless $x=k\pi$ for $k=0,1,2$. In other words, if $\alpha \le 0$ there is pointwise convergence only in $x=0,\pi,2\pi$.

  • if $\alpha \gt 1$, I can use the Weierstrass M-test to conclude that the series is uniformly convergent hence pointwise convergent for every $x \in [0,2\pi]$. Moreover the sum is a continuous function in $[0,2\pi]$.

Would you please help me in studying what happens for $\alpha \in (0,1]$? Are there any useful criteria that I can use?

Does the series converge? And what kind of convergence is there? In case of non uniform but pointwise convergence, is the limit function continuous?


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Perhaps an analysis of the $\zeta(x)$ series may help reveal some answers. – Argon Jun 7 '12 at 20:10
try this's_test – Norbert Jun 7 '12 at 20:12
Dirichlet's test seems a very beautiful tool! Thanks! Using Dirichlet, I have pointwise convergence for every $x \in [0,2\pi]$ also when $\alpha \in (0,1]$. What about the uniform convergence? – Romeo Jun 7 '12 at 20:16
up vote 3 down vote accepted

The main problem (at least for me) is to prove that sum of the series $$ f_\alpha(x)=\sum\limits_{k=1}^\infty\frac{\sin kx}{k^\alpha},\quad x\in[0,2\pi] $$ is discontinuous for $\alpha\in(0,1]$.

Lemma 1. For $\alpha\in(0,1]$ we have $$ \sum\limits_{k=1}^\infty\int\limits_{k-1/2}^{k+1/2}\frac{\sin(xt)}{k^\alpha}dt= \int\limits_{1/2}^\infty\frac{\sin(xt)}{t^\alpha}dt+\varphi(x) $$ where $|\varphi(x)|\leq 2^{1-\alpha}$ for all $x\in[0,2\pi]$.

Proof. It is enough to show that difference between this sum and this integral is bounded by some constant. Now, we make estimation $$ \left|\sum\limits_{k=1}^\infty\int\limits_{k-1/2}^{k+1/2}\frac{\sin(xt)}{k^\alpha}dt- \int\limits_{1/2}^\infty\frac{\sin(xt)}{t^\alpha}dt\right|= \left|\sum\limits_{k=1}^\infty\int\limits_{k-1/2}^{k+1/2}\left(\frac{\sin(xt)}{k^\alpha}- \frac{\sin(xt)}{t^\alpha}\right)dt\right|\leq $$ $$ \sum\limits_{k=1}^\infty\int\limits_{k-1/2}^{k+1/2}|\sin(xt)|\left|\frac{1}{k^\alpha}-\frac{1}{t^\alpha}\right|dt\leq \sum\limits_{k=1}^\infty\int\limits_{k-1/2}^{k+1/2}\left|\frac{1}{k^\alpha}-\frac{1}{t^\alpha}\right|dt= $$ $$ \sum\limits_{k=1}^\infty\left(\int\limits_{k-1/2}^{k}\left(\frac{1}{t^\alpha}-\frac{1}{k^\alpha}\right)dt+\int\limits_{k}^{k+1/2}\left(\frac{1}{k^\alpha}-\frac{1}{t^\alpha}\right)dt\right)= \sum\limits_{k=1}^\infty\left(\int\limits_{k-1/2}^{k}\frac{1}{t^\alpha}dt-\int\limits_{k}^{k+1/2}\frac{1}{t^\alpha}dt\right)\leq $$ $$ \sum\limits_{k=1}^\infty\left(\int\limits_{k-1/2}^{k}\frac{1}{(k-1/2)^\alpha}dt-\int\limits_{k}^{k+1/2}\frac{1}{(k+1/2)^\alpha}dt\right)\leq \sum\limits_{k=1}^\infty\left(\frac{1}{2(k-1/2)^\alpha}-\frac{1}{2(k+1/2)^\alpha}\right)=2^{\alpha-1} $$

Lemma 2. For $\alpha\in(0,1]$ we have $$ \sum\limits_{k=1}^\infty\frac{\sin kx}{k^\alpha}= \frac{x^\alpha}{2\sin(x/2)}\int\limits_{x/2}^\infty\frac{\sin y}{y^\alpha}+\frac{x\varphi(x)}{2\sin(x/2)} $$ where $|\varphi(x)|\leq 2^{1-\alpha}$ for all $x\in[0,2\pi]$.

Proof. Note that $$ \int\limits_{k-1/2}^{k+1/2}\sin(xt)dt= -\frac{1}{x}\cos(xt)\biggl|_{k-1/2}^{k+1/2}= \frac{2\sin(kx)\sin(x/2)}{x} $$ so, $$ \sin(kx)=\frac{x}{2\sin(x/2)}\int\limits_{k-1/2}^{k+1/2}\sin(xt)dt $$ Hence from lemma 1 we conclude $$ \sum\limits_{k=1}^\infty\frac{\sin kx}{k^\alpha}= \sum\limits_{k=1}^\infty\frac{x}{2k^\alpha\sin(x/2)}\int\limits_{k-1/2}^{k+1/2}\sin(xt)dt= \frac{x}{2\sin(x/2)}\sum\limits_{k=1}^\infty\int\limits_{k-1/2}^{k+1/2}\frac{\sin(xt)}{k^\alpha}dt= $$ $$ \frac{x}{2\sin(x/2)}\left(\int\limits_{1/2}^\infty\frac{\sin(xt)}{t^\alpha}dt+\varphi(x) \right) $$ Making substitution $y=tx$ we get $$ \sum\limits_{k=1}^\infty\frac{\sin kx}{k^\alpha}= \frac{x}{2\sin(x/2)}\left(\frac{1}{x^{1-\alpha}}\int\limits_{x/2}^\infty\frac{\sin y}{y^\alpha}dy+\varphi(x) \right)= \frac{x^\alpha}{2\sin(x/2)}\int\limits_{x/2}^\infty\frac{\sin y}{y^\alpha}dy+\frac{x\varphi(x)}{2\sin(x/2)} $$

Corollary 3. For $\alpha\in(0,1]$ the function $f_\alpha$ is discontinuous at $0$.

Proof. Obviously $f_\alpha(0)=0$. Let $\alpha\in(0,1)$, then from fromula proved in lemma 2 we see that $$ \lim\limits_{x\to +0}f_\alpha(x)=\lim\limits_{x\to +0}\left(\frac{x^\alpha}{2\sin(x/2)}\int\limits_{x/2}^\infty\frac{\sin y}{y^\alpha}dy+\frac{x\varphi(x)}{2\sin(x/2)}\right) $$ Since $\varphi$ is bounded then the second term is bounded while the first tends to infinity. Hence the last limit is $\lim\limits_{x\to +0}f_\alpha(x)=+\infty$.

If $\alpha=1$, then $|\varphi(x)|\leq 1$ and since $$ \lim\limits_{x\to+0}\frac{x}{2\sin(x/2)}=1, $$ then $$ \left|\frac{x\varphi(x)}{2\sin(x/2)}\right|<\frac{\pi}{3} $$ for some $\delta_1>0$ and $x\in(0,\delta_1)$. Since $$ \lim\limits_{x\to+0}\frac{x}{2\sin(x/2)}\int\limits_{x/2}^\infty\frac{\sin y}{y}dy= \int\limits_{0}^\infty\frac{\sin y}{y}dy=\frac{\pi}{2} $$ then $$ \frac{x}{2\sin(x/2)}\int\limits_{x/2}^\infty\frac{\sin y}{y}dy>\frac{2\pi}{5} $$ for some $\delta_2>0$ and all $x\in(0,\delta_2)$. Thus for all $x\in(0,\min(\delta_1,\delta_2))$ we see that $$ f_\alpha(x)=\frac{x^\alpha}{2\sin(x/2)}\int\limits_{x/2}^\infty\frac{\sin y}{y^\alpha}dy+\frac{x\varphi(x)}{2\sin(x/2)}>\frac{2\pi}{5}-\frac{\pi}{3}>0 $$

In both cases $\lim\limits_{x\to+0}f_\alpha(x)\neq 0$. hence $f_\alpha$ is discontinuous at $0$.

Corollary 4. For $\alpha\in(0,1]$ the series $$ \sum\limits_{k=1}^\infty\frac{\sin kx}{k^\alpha} $$ doesn't converges uniformly on $[0,2\pi]$.

Proof. Assume that this series converges uniformly. Since this series is the sum of continuous functions and its converges uniformly, then its sum $f_\alpha$ must be continuous function on $[0,2\pi]$. This contradicts corallary 3, hence our series is not uniformly convergent on $[0,2\pi]$.

Remark 5. Despite the above, this series converges uniformly on $[\delta,2\pi-\delta]$ for all $\delta\in(0,\pi]$. You can use Dirichlet test to prove this.

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Great, your proof is beautiful! Thanks! I have still one more question: what about the continuity of the limit function (for $\alpha \in (0,1]$)? – Romeo Jun 8 '12 at 6:56
@Romeo, see edits to my answer. – Norbert Jun 8 '12 at 12:18
Your proof doesn't work when $\alpha=1$: the RHS of Lemma 2 remains bounded. – Generic Human Jun 8 '12 at 19:22
@GenericHuman, I've edit my answer. Now it covers both cases. – Norbert Jun 8 '12 at 21:03

Note that $$\sum_{k=1}^{N} \sin(kx) = \dfrac{\sin(Nx/2)}{\sin(x/2)} \sin \left( \left(\dfrac{N+1}2 \right)x\right)$$ Hence, for each given $x$, the sum is bounded by $\dfrac1{\sin(x/2)}$.

Hence by generalized alternating series test (also known as Dirichlet's test) the sum converges.

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I thank you all for your kind replies. The fact that the bound depends on $x$ suggests me that the convergence is not uniform when $\alpha \in (0,1]$... Am I right? How can I prove this formally? – Romeo Jun 7 '12 at 20:23

There is an easier proof that $f_\alpha$ is discontinuous at 0.

Let $x=\pi/n$ for some even $n$. Then for $1\le i\le n$, group terms $a_{2nk+i}$ and $a_{2nk+i+n}$ together: $$\frac{\sin (2nk+i)x}{(2nk+i)^\alpha}+\frac{\sin (2nk+i+n)x}{(2nk+i+n)^\alpha}\ge\frac{n\alpha}{(2nk+i)(2nk+i+n)^\alpha}\sin \frac{i\pi}{n} \ge 0$$ We need only use $k=0$ and $i\le n/2$ for the lower bound: $$\begin{align} f_\alpha(x)\ge&\alpha \sum_{i=1}^{n/2} \frac{n}{i(i+n)^\alpha}\sin \frac{i\pi}{n}\\ \ge&\alpha\sum_{i=1}^{n/2} \frac{2}{(i+n)^\alpha}\qquad{\text{($\sin t\ge 2/\pi\cdot t$)}}\\ \ge&2\alpha\log\frac{3n/2+1}{n+1}\\ \rightarrow&2\alpha\log 3/2 \end{align}$$ So $f_\alpha$ is bounded from below by a positive number as $x\rightarrow 0$, for all $\alpha\in(0,1]$. Because $f_\alpha(0)=0$, the convergence cannot be uniform around 0.

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Right, I was bounding $\cos t$ instead of $\sin t$. Thanks! Note that actually $f_1(x)=\frac{\pi-x}{2}$. – Generic Human Jun 8 '12 at 20:52
It seems that your proof shows that $\lim\limits_{x\to+0} f_1(x)=+\infty$. But it is known that $f_1(x)=(\pi-x)/2$ for $x\in(0,\pi)$. – Norbert Jun 8 '12 at 21:04
Ok, now no question left :) – Norbert Jun 8 '12 at 21:06

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