Problem: Prove that the series
\[
\sum_{n=1}^{\infty} \frac{\sin nx}{n^2},
\]
converges uniformly to a continuous function on the interval $[0,1]$.
Solution: We will use the
Weierstrass $M$-test to prove the uniform convergence. Let us recall the Weierstrass $M$-test.
(Weierstrass $M$-test) Let $f_n$ be a sequence of real valued functions on the set $X$. Assume that there exists $M_n \geq 0$ such that $\left\vert f_n(x) \right\vert \leq M_n$ for all $n \in \mathbb{N} $ and $x \in X$. Further assume that $\sum_{n=1}^{\infty} M_n$ is convergent. Then the series $\sum_{n=1}^{\infty} f_n$ is absolutely and uniformly convergent on $X$.
Here in the given problem,
\[
f_n(x) = \frac{\sin nx}{n^2}.
\]
Take $M_n = \frac{1}{n^2}$. We have
\[
\left\vert \sin nx \right\vert \leq 1 \implies \left\vert \frac{\sin nx}{n^2} \right\vert \leq \frac{1}{n^2}.
\]
By the $p$-test, the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is convergent and therefore, by Weierstrass $M$-test, the given series is uniformly convergent.
In order to conclude the limit function is continuous, we recall that convergence of a series means the sequence of partial sum is convergent. That is,
\[
\sum_{n=1}^{\infty} f_n \text{ converges } \iff S_N = \sum_{n=1}^{N} f_n \text{ converges } .
\]
As we proved that the given series is uniformly convergent, this implies the sequence of function
\[
F_N(x) = \sum_{n=1}^{N} \frac{\sin nx}{n^2}
\]
converges uniformly. Since $F_N$ is a finite sum of continuous function, $F_N$ is continuous. As the convergence is uniform, the limit function will also be continuous. This proves that the given series converges to a continuous function on $\mathbb{R} $, in particular, on the interval $[0,1]$.