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I'd like to use the Cauchy criterion to show that

$$f_n(x)=\sum\limits_{k=1}^n\frac{\sin(kx)}{k^2} \mbox{convergences uniformly} $$

Here is what I did:

We want to show that $\forall \, \varepsilon_1>0 \, \exists \, n_1 \in \mathbb N$ such that $\forall \, n,m ≥ n_1 \, \left\lvert \, f_n(x)-f_m(x) \right\rvert < \varepsilon_1 \, \forall \, x \in \mathbb R $.

Let $\varepsilon_1>0$.

At this point we recall that $\sum\limits_{k=1}^\infty\frac{1}{k^2}$ is convergent. That means it's Cauchy. So we know that $\forall \, \varepsilon_2 > 0 \, \exists \, n_2 \in \mathbb N$ such that $\forall \, n,m ≥ n_2 \, \left\lvert \sum\limits_{k=m+1}^n\frac{1}{k^2} \right\rvert< \varepsilon_2 $. Very nice, maybe we could use it.

Let $\varepsilon_2 = \varepsilon_1$, and then $n_1=n_2$. Now $\forall \, n,m ≥ n_1 , \, \forall \, x \in \mathbb R$:

\begin{align} |f_n(x)-f_m(x)| &= \Big|\sum\limits_{k=m+1}^n\frac{\sin(kx)}{k^2}\Big| \\ &\leq \Big|\sum\limits_{k=m+1}^n\frac{1}{k^2}\Big| \\ &< \varepsilon_2 = \varepsilon_1 \end{align}

Any comments on the correctness of the proof are greatly appreciated.

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That is correct. – ougao Apr 1 '13 at 1:54
Note that, $ \Big|\sum\limits_{k=m+1}^n\frac{1}{k^2}\Big|< \Big|\sum\limits_{k=m+1}^{\infty}\frac{1}{k^2}\Big|< \epsilon.$ – Mhenni Benghorbal Apr 1 '13 at 2:06
up vote 6 down vote accepted

Your work is correct. In fact, note that the core of the proof is that $$\tag 1 \sum_{k=1}^\infty k^{-2}$$ converges, thus it is Cauchy (or fundamental). With this fact, and the fact that $$|\sin(kx)|\leq 1$$ you can make you function small by using the majorant convergent series in $(1)$ as an upper bound, regardless of $x$.

This is a particular observation of the general Weiertrass test, which gives sufficient conditions for a series $f=\sum f_n $ to converge uniformly over a domain $D$.

THM Let $\langle f_n:n\in \Bbb N\rangle$ be a sequence of functions defined on a domain $D$ such that for each $n$ and $x\in D$, $$|f_n(x)|\leq M_n$$ for some sequence $\langle M_n:n\in\Bbb N\rangle$. If $\sum M_n$ exists (that is, the sequence is summable) then the function $f=\sum f_n$ converges uniformly on $D$.

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