Investigate the convergence of $\sum_{n=1}^{\infty} u_{n}(x)$, where $u_{n}(x)=(-1)^n \frac{x^2+n}{n^2}$ Let $u_{n}(x)=(-1)^n \dfrac{x^2+n}{n^2}$. Show that the series $\sum_{n=1}^{\infty} u_{n}(x)$ converges uniformly on any interval $[a;b]$ but does not converge absolutely at any point $x\in R$.
For the absolute convergence, we have $\lim_{n\to\infty} \dfrac{\frac{x^2+n}{n^2}}{\frac{1}{n}}=1$ and the series $\sum_{n=1}^{\infty} \dfrac{1}{n}$ diverges, thus the given series does not converge absolutely at any point $x\in R$. 
For the uniform convergence, I have no idea. Can anyone help me please?
 A: (Rough Sketch of Proof): Fix an interval $[a, b]$, then there exists $N_1$ such that for all $n>N_1$ we have
\begin{align}
\frac{x^2+n}{n^2} \leq \frac{b^2+a^2+n}{n^2}
\end{align}
which tends to $0$ as $n\rightarrow \infty$. Hence $|u_n(x)|$ tends to $0$ uniformly for all $x \in [a, b]$. Thus, by the alternating series theorem, we have that
\begin{align}
\sum^\infty_{k=0}(-1)^k\frac{x^2+k}{k^2}
\end{align}
converges for all $x \in [a, b]$. Moreover, we have
\begin{align}
\left|\sum^n_{k=m} (-1)^k \frac{x^2+k}{k^2} \right|\leq \sum^n_{k=m}\frac{x^2}{k^2} + \left|\sum^n_{k=m}(-1)^k\frac{1}{k} \right| \leq \sum^n_{k=m}\frac{a^2+b^2}{k^2} + \left|\sum^n_{k=m}(-1)^k\frac{1}{k} \right| 
\end{align}
for all $x \in [a, b]$. 
Here's the proof. Fix $\epsilon>0$. Since
\begin{align}
\sum^\infty_{k=1}(-1)^k\frac{1}{k} \ \ \text{ and } \ \ \sum^\infty_{k=1}\frac{a^2+b^2}{k^2}
\end{align}
are convergent then there exists $N$ such that for all $m, n>N$, we have
\begin{align}
\left|\sum^n_{k=m} (-1)^k \frac{1}{k} \right|<\epsilon \ \ \text{ and } \ \ \left|\sum^n_{k=m} \frac{a^2+b^2}{k^2} \right|<\epsilon
\end{align}
which means
\begin{align}
\left|\sum^n_{k=m} (-1)^k \frac{x^2+k}{k^2} \right|<2\epsilon
\end{align}
i.e. the partial sums are uniformly Cauchy. Hence the series converges uniformly on $[a, b]$. 
A: This may be not an answer to the question.
$$u_n=\frac{(-1)^n \left(n+x^2\right)}{n^2}=\frac{(-1)^n }{n}+x^2\frac{(-1)^n }{n^2}$$ $$\sum_{n=1}^{\infty} u_{n}=\sum_{n=1}^{\infty}\frac{(-1)^n }{n}+x^2\sum_{n=1}^{\infty}\frac{(-1)^n }{n^2}=-\sum_{n=1}^{\infty}\frac{(-1)^{n +1}}{n}+x^2\sum_{n=1}^{\infty}\frac{(-1)^n }{n^2}$$ The first summation is the alternating harmonic series and its value is $-\log(2)$. Concerning the second summation (which corresponds almost to the Basel problem), we can write $$\sum_{n=1}^{\infty}\frac{(-1)^n }{n^2}=-\sum_{n=1}^{\infty}\frac{1 }{n^2}+2\sum_{n=1}^{\infty}\frac{1 }{(2n)^2}=-\frac 12\sum_{n=1}^{\infty}\frac{1 }{n^2}=-\frac{\pi ^2}{12}$$  All of the above makes $$\sum_{n=1}^{\infty} u_{n}=-\frac{\pi ^2}{12}  x^2-\log (2)$$
