# Does the sequence converge with respect to uniform norm?

In the space of continuous functions $C([0,1])$, consider the sequence given by $f_n(x)=\sum_{k=1}^n\frac{\sin(kx)}{k^2}$. I'm trying to show that this is Cauchy with respect to the uniform norm $\|f\|_{\infty}=\sup{|f(x)|}$.

We have that $$\lim_{n\rightarrow \infty}f_n(x)=\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^2}\leq\sum_{k=1}^\infty \frac{1}{k^2}<\infty,$$ and since this gives us a bound independent of $x$, we have that $f_n$ converges with respect to the uniform norm. Since the space is complete with respect to this norm, $f_n(x)$ is indeed Cauchy.

My question is, is this reasoning correct?

• I presume a person tasked with the evaluation of your work may want you to be more precise about your statement of the hypothesis of Weierstrass M-test. – user228113 Mar 30 '18 at 23:29

Use the Weierstrass M-test. Namely $$\left \lVert \sum_{k=m}^n\frac{\sin(kx)}{k^2} \right \rVert \leq \sum_{k=m}^n\left\lVert\frac{\sin(kx)}{k^2}\right\rVert \le\sum_{k=m}^n\frac{1}{k^2}\to0$$ as $m,n\to\infty$. It follows that the partial sums are cauchy in $C([0,1])$. Since $C[(0,1)]$ is complete, the series converges uniformly.

First of all, writing

$$\lim_{n\to \infty} f_n(x) = \sum_{k=1}^\infty \frac{\sin kx}{k^2}$$

in the first place is quite sloppy. You do not know that limit exists (indeed that is what you what to show).

Second, even if you have a sequence $\{f_n\}$ so that

$$\sum_{j=1}^n f_j(x) \le C$$

for some $C$ independent of $x$, it is NOT sufficient to show that $f_n(x)$ converges. This is not true even for numerical sequence: $a_n = (-1)^n$.

Finally, you said: "since this gives us a bound independent of x, we have that $\{f_n\}$ converges with respect to the uniform norm". It seems that this is not what you have shown: it seems that you are trying to use the estimates given by the other answer. Thus you have shown that $$\left\{ \sum_{j=1^n}f_j\right\}$$ is Cauchy. You do not need to use the knowledge that $C[0,1]$ is complete in this situation.