Question regarding function series $\sum _{n=1}^{\infty }\sin\left(\frac{x}{n^2}\right)$ - continuity?

Question

$$\sum _{n=1}^{\infty }\:\sin\left(\frac{x}{n^2}\right)$$
I know how to check if a series of functions is continuous.
The problem is, I need to check if series above is continuous in $(-\infty, \infty)$.
So I tried doing the $M$-test: I did $$\frac{\left|\sin\left(\frac{x}{n^2}\right)\right|}{\frac{|x|}{n^2}}\leq 1$$ for all $x \in (-\infty,\infty)$. But: $$\sum_{n=1}^{\infty}\frac{|x|}{n^2}$$ is not convergent for $|x|\geq n$ But still, the answer says its convergent Uniformly... How so? if $x=n$ its a contradiction to it. This is the most problematic question for me in calculus 2, I dont know why its uniformly continuity.

Answer 1

HINT: Continuity is a local property. This means that to check continuity it is enough to fix $x \in \Bbb R$, and see if the function is continuous at $x$.

Now, having fixed $x \in \Bbb R$, there is some integer $N$ big enough such that $$\frac{|x|}{N^2}<1$$ Hence the M-Test can be done using $$M_n = \cases{1 & if $n <N$ \\ \frac{|x|}{n^2} & if $n \ge N$}$$ Since the series $\sum_n M_n$ is convergent, the M-Test applies, and in a neighbourhood of $x$ you have uniform convergence.