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?
 A: 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.
A: In order to demonstrate that the limiting function is continuous, you can show that it is the uniform limit of a sequence of continuous functions.
The sequence of functions here is straight forward: $$f_N(x) = \sum_{n=1}^N (x+n)^{-2}.$$
We wish to show that these functions converge to our limiting function uniformly.
Note that for each $n$ we have $(x+n)^{-2} \le n^{-2}$ for $x \in [0,1]$. Thus the series is bounded above by $\sum n^{-2}$ at each point and converges by the monotone convergence theorem.
To demonstrate that the convergence is uniform, we need to show that the remainder can be bounded by a function going to zero as a function of $N$ and independent of $x$.
$$|f(x) - f_N(x)| = \sum_{n=N+1}^{\infty} \frac{1}{(x+n)^2}  \le \sum_{n=N+1}^\infty \frac1{n^2}$$
Since $\sum_{n=1}^\infty \frac1{n^2}$ converges, we know that the sum on the right tends to zero as $N\to \infty$. Thus $f$ is the limit of a uniformly convergent sequence of continuous functions and is itself uniformly convergent.

As for the integral, since the sequence $f_N$ is uniformly convergent, we can exchange integration and the limit: $$\int_0^1 f(x) dx = \int_0^1 \lim_{N\to \infty} f_N(x) dx = \lim_{N\to \infty} \int_0^1 f_N(x)dx = \lim_{N\to\infty} \sum_{n=1}^N \int_0^1 \frac{1}{(x+n)^2} dx$$ $$=\lim_{N\to \infty} \sum_{n=1}^N \left(\frac{1}{n} - \frac{1}{n+1}\right).$$
This series telescopes to $1$.
A: Let
$$
f_n(x)=\sum_{k=1}^n\frac{1}{(x+k)^2}.
$$
Then, for every $x\in [0,1]$ we have
$$
|f(x)-f_n(x)|=\sum_{k=n+1}^\infty\frac{1}{(x+k)^2}\le \sum_{k=n+1}^\infty\frac{1}{k^2}.
$$
It follows that
$$
\|f-f_n\|_\infty\le \sum_{k=n+1}^\infty\frac{1}{k^2} \to 0 \mbox{ as } n \to \infty,
$$
i.e. $f_n$ converges uniformly to $f$, and therefore $f\in C([0,1])$ because each $f_n$ is continuous on $[0,1]$. 
Since the convergence $f_n\to f$ is uniform on $[0,1]$, we have:
$$
\int_0^1f(x)\,dx=\lim_{n\to\infty}\int_0^1f_n(x)\,dx.
$$
For every $n$ we have
$$
\int_0^1f_n(x)\,dx=\sum_{k=1}^n\int_0^1\frac{1}{(x+k)^2}\,dx=-\sum_{k=1}^n\frac{1}{x+k}\Big|_0^1=\sum_{k=1}^n\left(\frac{1}{k}-\frac{1}{k+1}\right)=1-\frac{1}{n+1},
$$
and taking the limit we get:
$$
\int_0^1f(x)\,dx=\lim_{n\to\infty}\int_0^1f_n(x)\,dx=\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)=1.
$$
