Continuity of function consisting of an infinite series. 
Let $f(x) , 0\leq x\leq 1$ be defined by,
$$f(x)=\sum_{n=1}^{\infty}\frac{1}{(x+n)^2}$$.
Show that $f$ is continuous on $[0,1]$ and that,
$$\int_0^1f(x)dx=1$$.

I have never dealt with functions which are infinite series.
How do I approach this type of thing? Can I just ignore the sum and see if $\frac{1}{(x+n)^2}$ is continuous on the interval?
But then again, I wouldn't how to go for the integral with that mindset.
Could anyone help me out here?

The Weierstrass M-Test says that the series $f(x) = \sum_{n=1}^{\infty} \frac{1}{(x+n)^2}$ converges uniformly for $x \in [0,1]$. Moreover, since the partial sums of this series are continuous on $[0,1]$, the limit $f$ must be continuous there as well.
In order to compute the integral, we can interchange the integral and the sum to see that
$$
\int_0^1 f(x)dx = \sum_{n=1}^{\infty} \int_0^1 \frac{dx}{(x+n)^2} = \sum_{n=1}^{\infty} \left( -\frac{1}{(x+n)} \bigg|_0^1 \right) = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) = 1,
$$
where the last sum is telescoping and so it is easily evaluated.
