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Let $K(x)=\sum_{n=1}^{\infty}\frac{1}{n^r}\cos(nx)$. Find $r$ so that $K(x)$ converges uniformly for $x\in \mathbb{R}$.

Sorry in advance if my question sounds naive. Is Weierstrass M-test applicable here? I think that, if I have used it correctly then $r>1$. I would be grateful if someone could confirm that $r>1$ or could give me any advice on how to tackle this problem.

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    Yes Weierstass M test is useful here and you are right it guarantees uniform convergence for $r > 1$. If the question is to find all such $r$ then I guess you also need to rule out that it can converge uniformly for $r\leq 1$. To do so you only need to consider $x=0$. – 2017-02-11
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    Thank you for your comment. I will certainly consider ruling out that it can converge uniformly for $r\leq1$ – 2017-02-11

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If $r\le 1,$ then

$$\sum_{n=1}^{\infty}f_n(0) = \sum_{n=1}^{\infty}\frac{1}{n^r}\ge \sum_{n=1}^{\infty}\frac{1}{n} = \infty.$$

So the series doesn't even converge pointwise on $\mathbb R,$ much less converge uniformly there. If $r>1,$ then $\sum_{n=1}^{\infty} 1/n^r<\infty.$ Because $|f_n|\le 1/n^r$ on $\mathbb R,$ $\sum_{n=1}^{\infty} f_n$ converges uniformly on $\mathbb R$ by the Weierstrass M test.

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    I could use your help with this problem http://math.stackexchange.com/q/2150771/415301 – 2017-02-26
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Edit: Thanks to zhw's for all the corrections. Following zhw's answer, if $r \leq 1,$ $$ \sum |f_n(0)|=\sum \frac{1}{n^r} \geq \sum \frac{1}{n} $$ diverges, so $\sum f_n$ doesn't converges uniformly on $\mathbb{R}.$

If $r>1,$ you are correct. With $M_n=1/n^r,$ we have $|f_n| \leq M_n$ for all $n \in \mathbb{N}$ and $\sum M_n < \infty.$ Thus, Weiestrass M-test gives the result.

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    Your first two claims are wrong, and your explanation of Weierstrass M is not quite right either. – 2017-02-12
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    You are arguing incorrectly with Weirstrass. See my answer for more on this. – 2017-02-14
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    I don't know what's wrong with that. I wrote the same as you. $|f_n| \leq 1/n^r$ on $\mathbb{R}$ gives $\sum f_n$ is uniformly convergent on $\mathbb{R}$ as $\sum 1/n^r < \infty$ when $r>1.$ – 2017-02-14
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    You are probably thinking the right thing, and not writing it. There is a difference between writing "$\sum |f_n| \le \sum 1/n^r < \infty$" and "$|f_n| \le 1/n^r$ for each $n$ and $\sum 1/n^r<\infty.$" The first statement does not imply uniform convergence. The second one does (by Weierstrass M). – 2017-02-14