Prove that $\sum\limits_{n=1}(-1)^n\frac{x^2+n}{n^2}$ uniformly convergent on $[a,b]$

Since the series is not absolutely convergent, using the bound,

$$\left|\sum_{n=p+1}^q (-1)^n\frac{x^2+n}{n^2}\right| \leqslant \sum_{n=p+1}^q\frac{x^2+n}{n^2},$$

you will not succeed in proving that partial sums of $\sum_{n \geqslant 1}(-1)^n\frac{x^2+n}{n^2}$ satisfy the Cauchy criterion uniformly because the RHS is the difference of partial sums for a divergent series. Note that $\frac{x^2+n}{n^2} = \mathcal{O}\left(\frac{1}{n} \right)$.

For the bounding series to satisfy the Cauchy criterion, we must have for any $\epsilon > 0$,

$$\sup_{x \in [a,b]}\sum_{n=q+1}^{p} \frac{x^2+n}{n^2} < \epsilon $$

for all sufficiently large $q$ and $p > q$.

However,

$$\sup_{x \in [a,b]}\sum_{n=q+1}^{p} \frac{x^2+n}{n^2} \geqslant \sup_{x \in [a,b]}(p-q) \frac{x^2 +q }{p^2} = (p-q) \frac{\max(a^2,b^2) +q }{p^2} $$

Taking $p = 2q$, we have

$$\sup_{x \in [a,b]}\sum_{n=q+1}^{p} \frac{x^2+n}{n^2} \geqslant q \frac{\max(a^2,b^2) +q }{4q^2} = \frac{1 + \max(a^2,b^2)/q}{4},$$

and since the RHS converges to $1/4$ as $q \to \infty$, the Cauchy criterion is violated.

A correct approach

Uniform convergence can be established using Dirichlet's test. The hypotheses are met since the partial sums $\sum_{n=1}^N(-1)^n$are uniformly bounded and it can be shown that as $n \to \infty$, we have

$$\frac{x^2 + n}{n^2 } \downarrow 0,$$

where the convergence is monotonic and uniform for $x \in [a,b]$.

The uniform convergence is easily established since,

$$\frac{\min(a^2,b^2)+n}{n^2} \leqslant \frac{x^2+n}{n} \leqslant \frac{\max(a^2,b^2) +n}{n^2}$$

See if you can finish by showing that $\frac{x^2 + n}{n^2}$ is decreasing with respect to $n$.

Summation by parts is another approach, but that in essence repeats the steps in a proof of Dirichlet's test for uniform convergence.