continuity of a function series I need help finding for which $x$ the function:
$$\sum_{n=1}^{\infty}\frac{x+n(-1)^n}{x^2+n^2}$$
is continuous. 
I first need to show that the series converges uniformly, the thing is I don't know how to deal with the $(-1)^n$. I tried comparing to the series $\sum_{n=1}^{\infty}\frac{x+n}{x^2+n^2}$ but I think I hit a dead end there.
any suggestions? 

$f(x) = \sum_{n=1}^{\infty}\frac{x+n(-1)^n}{x^2+n^2}$
We need to show that for any $\epsilon$ there exists a $\delta$ such that for all $x,y$ in the domain, when $|x-y|< \delta,$ then $|f(x) - f(y)| < \epsilon$.
The first thing we are going to need is some upper bound on $f(x)-f(y)$
$f(x) -f(y) =$$ \sum_{n=1}^{\infty}\frac{x+n(-1)^n}{x^2+n^2}-\frac{y+n(-1)^n}{y^2+n^2}\\
\sum_{n=1}^{\infty}\frac{x^2y - y^2x+n(-1)^n(y^2-x^2)}{x^2y^2+(x^2+y^2)n^2 + n^4}\\
(y-x)\sum_{n=1}^{\infty}\frac{xy+n(-1)^n(x+y)}{x^2y^2+(x^2+y^2)n^2 + n^4}$
Now we need to show that $\sum_{n=1}^{\infty}\frac{xy+n(-1)^n(x+y)}{x^2y^2+(x^2+y^2)n^2 + n^4}$ converges.
The denominator grow at the order of $n^4$ and numerator grow with the order of $n$, so the series converges. 
What we really want to be able to say is that:
There exists an $M$ such that $\sum_{n=1}^{\infty}\frac{xy+n(-1)^n(x+y)}{x^2y^2+(x^2+y^2)n^2 + n^4} < M$
If $x,y$ are big the denominator grows more quickly than the numerator, and if $x,y$ are near $0,$ the numerator collapses to $0,$ while the denominator does not.
It seems obvious, but I suggest you tighten that up.  Anyway, once that has been established you will be able to say.
$|f(x)-f(y)|< M|y-x|$
$\delta = \frac \epsilon M$
Update
You are looking for something along these lines?
let $\{f\}$ be sequence of functions such that
$f_k(x) = \sum_{n=1}^{k}\frac{x+n(-1)^n}{x^2+n^2}$
We want to show that this sequence converges uniformly to $f(x)$
for any $\epsilon>0$ there exists a $K>0$ such that $k,m>K \implies \sup|f_k(x) - f_m(x)|<\epsilon$ 
$f_k(x) - f_m(x) = \sum_{n=k}^{m}\frac{x+n(-1)^n}{x^2+n^2}$
Unfortunately, I have to run... and can't figure out the last bit.
