How to show $\sum (-1)^n \frac{x^2+n}{n^2}$ is not uniformly convergent 0n $\mathbb{R}$

Question

How can I show that $$\sum_{n=1}^{\infty} (-1)^n \frac{x^2+n}{n^2}$$ is not uniformly convergent 0n $\mathbb{R}$.
I tried using Abel's test by taking $\sum \frac{(-1)^n}{n}$ as convergent series and $\{\frac{x^2+n}{n}\}$ as monotone decreasing for each $x\in \mathbb{R}$. The problem is showing uniform boundedness of $\{\frac{x^2+n}{n}\}$ on $\mathbb{R}$.
The sequence of function is uniformly bounded on any bounded interval but on the real line as $x$ has no bound so $\{\frac{x^2+n}{n}\}$ cannot be uniformly bounded on $\mathbb{R}$. Thus Abel's test fails. Similarly for the Dirichlet's test.
Then I saw this problem, i.e. using Cauchy sequence to show not uniformly convergent. But I failed to use that technique here.

Answer 1

Hint. For a given $N\geq 1$, what is $$\sup_{x \in \mathbb{R}}\left|\sum_{n=N+1}^{\infty}(-1)^n \frac{x^2+n}{n^2}\right|=\sup_{x \in \mathbb{R}}\left|A_N x^2+B_N\right|?$$ where $A_N=\sum_{n=N+1}^{\infty} \frac{(-1)^n}{n^2}$ and $B_N=\sum_{n=N+1}^{\infty} \frac{(-1)^n}{n}$. Note that $A_N\not=0$.

Answer 2

By contradiction, assume that the series is uniformly convergent on $\Bbb R$ then $$\lim_{n\to\infty }\Vert R_n\Vert_\infty=0$$ where

$$R_n=\sum_{k=n+1}^\infty (-1)^k\frac{x^2+k}{k^2}$$ so since $$R_n-R_{n-1}=(-1)^n\frac{x^2+n}{n^2}$$ we get $$\Vert R_n\Vert_\infty+\Vert R_{n-1}\Vert_\infty\ge\Vert R_n-R_{n-1}\Vert_\infty=\sup_{x\in\Bbb R}\left\vert(-1)^n\frac{x^2+n}{n^2}\right\vert=+\infty$$ which is a contradiction.