Prove that $\sum\limits_{n=1}(-1)^n\frac{x^2+n}{n^2}$ uniformly convergent on $[a,b]$ I used the cauchy criteria, for $p>q$
$$\sup\limits_{x\in [a,b]}\left | S_p(x) -S_q(x)\right |=
\sup\limits_{x}\left | \sum\limits_{k=q+1}^{p}(-1)^n\frac{x^2+n}{n^2} \right |\leq \\ 
\sup\limits_{x}\sum\limits_{k=q+1}^{p}\left | \frac{x^2+n}{n^2} \right |\leq 
\sup\limits_{x}\frac{x^2+q+1}{(q+1)^2}(p-q)=\\
\frac{b^2+q+1}{(q+1)^2}(p-q)$$
Is what I am doing right till now ?
 A: 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.
A: Good way! Now each time let $p=\lfloor\alpha q\rfloor$ with some $\alpha>1$
therefore$$\sup_{x\in [a,b]}\left | S_p(x) -S_q(x)\right |\le {b^2+q+1\over (q+1)^2}(\alpha-1)q$$and by tending $q\to\infty$ as Cauchy's criterion imposes we obtain$$\sup_{x\in [a,b]}\left | S_p(x) -S_q(x)\right |\le \alpha-1$$since this is true for every $\alpha>1$ therefore $$\sup_{x\in [a,b]}\left | S_p(x) -S_q(x)\right |\to 0$$and the proof is complete.

An alternative way

Note that both$$\sum_{n=1}^{\infty}{(-1)^n\over n}=\ln 2\\\sum_{n=1}^{\infty}{(-1)^n\over n^2}$$are convergent, therefore $$\sum_{n=1}^\infty(-1)^n\frac{x^2+n}{n^2}$$ is uniformly convergent by investigating $\sum_{n=N}^\infty(-1)^n\frac{x^2+n}{n^2}$ for large enough $N$. Another way to say it is that $$|f_n(x)-f(x)|{=\left|\sum_{n=N}^\infty(-1)^n\frac{x^2+n}{n^2}\right|\\\le \sup_{x\in[a,b]}\left|\sum_{n=N}^\infty(-1)^n\frac{x^2+n}{n^2}\right|\\=\sup_{x\in[a,b]}\left|\sum_{n=N}^\infty\frac{(-1)^n}{n}+x^2\sum_{n=N}^\infty\frac{(-1)^n}{n^2}\right|\\\le \left|\sum_{n=N}^\infty\frac{(-1)^n}{n}\right|+\sup_{x\in[a,b]}x^2\left|\sum_{n=N}^\infty\frac{(-1)^n}{n^2}\right|\\=
\left|\sum_{n=N}^\infty\frac{(-1)^n}{n}\right|+ \max\{a^2,b^2\}\cdot\left|\sum_{n=N}^\infty\frac{(-1)^n}{n^2}\right|}$$since both $\sum_{n=N}^\infty\frac{(-1)^n}{n}$ and $\sum_{n=N}^\infty\frac{(-1)^n}{n^2}$ tend to zero and $\max\{a^2,b^2\}$ is bounded, then $|f_n(x)-f(x)|$ can be bounded enough and hence the proof is complete.
