Proof that the limit of a sequence is equal to the limit of its partial sums divided by n Let $\{ x_n \}_n$ be a sequence of real numbers. Suppose $ \lim_{n \to \infty}x_n=a.$
Show that
$$\lim_{n \to \infty} \frac{x_1+x_2+...+x_n}{n}=a$$
As it is my first proof I'm not really sure whether I am allowed to do the following steps:
$$S_n = \sum_{i=1}^{n}{x_n} = S_{n-1}+x_n$$
Therferore we get $x_n=S_n-S_{n-1}$
Taking $\lim_{n \to \infty} x_n=\lim_{n \to \infty}(S_n-S_{n-1})=\lim_{n \to \infty}S_n-\lim_{n \to \infty}S_{n-1}$
We know that: $\lim_{n \to \infty} x_n=a$ and we also know that $S_{n-1}=S_{n-2}+x_{n-1}$
Therefore we get: $a=\lim_{n \to \infty}S_n-\lim_{n \to \infty}S_{n-2}-\lim_{n \to \infty}x_{n-1}$
Now $\lim_{n \to \infty}x_{n-1}=\lim_{n \to \infty}x_{n}=a$ and $a=\lim_{n \to \infty}S_n-\lim_{n \to \infty}S_{n-2}-a$
Repeating these steps we get: $a=\lim_{n \to \infty}S_n-\lim_{n \to \infty}S_{n-n}-(n-1)a=\lim_{n \to \infty}S_n-\lim_{n \to \infty}S_{0}-(n-1)a$
As $\lim_{n \to \infty}S_{0}=0$ we get $a=\lim_{n \to \infty}S_n-(n-1)a$ or if we rearrange $n*a=\lim_{n \to \infty}S_n$
Using the product rule for limits we get: $a=\frac{\lim_{n \to \infty}S_n}{n}=\lim_{n \to \infty}\frac{S_n}{n}$ q.e.d.
Is this proof consistent and complete?

HINT
You can assume $(x_n) \to 0$ WLOG, since you can look at $x_n' = x_n - a$ else; this will allow you to even tackle only $x_n \geq 0$ (why?) and then proceed with a lot less restraints.
