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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?

Thanks.

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    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?2012-06-07

3 Answers 3

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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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    Ok, now no question left :)2012-06-08
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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?2012-06-07
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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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    @GenericHuman, I've edit my answer. Now it covers both cases.2012-06-08