How do I show that $\sum^{\infty}_{n=1}\frac{x}{n^2}$ is not uniformly convergent on $\mathbb{R}$?

Question

How do I show something is not uniformly convergent? I was thinking maybe showing that the property of uniform convergence(continuity/boundedness preserved, integrity commutes etc) doesn't hold, but the problem is I haven't learned how to calculate $\sum^{\infty}_{n=0}1/n^2$

Answer 1

With $f : x \in \mathbb{R} \mapsto \displaystyle \sum_{k=1}^{+\infty} \dfrac{x}{k^2}$ and for all $n \in \mathbb{N}^*$, $f_n: x \in \mathbb{R} \mapsto \displaystyle \sum_{k=1}^{n} \dfrac{x}{k^2}$. We know that $(f_n)_n$ converges simply to $f$. But for $n \in \mathbb{N}^*$, we have $ f-f_n : x \in \mathbb{R} \mapsto \displaystyle \sum_{k=n+1}^{+\infty} \dfrac{x} {k^2} = x \sum_{k=n+1}^{+\infty} \dfrac{1} {k^2}$ that is not bounded because $\sum_{k=n+1}^{+\infty} \dfrac{1} {k^2} \neq 0$ therefore $(f_n)_n$ doesn't uniformly converge on $\mathbb{R}$.