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?
 A: The proof is quite wrong and almost nonsensical (Sorry!).
1) You cannot assume $\lim_{n \to \infty} S_n$ exists.
2) Since $n$ is the variable which tends to infinity, you cannot repeat $n$ times like that and get it out of the limit.
3) The statement $na = \lim_{n \to \infty} S_n$ is nonsensical.
A: As mentioned in answer from Aryabhata your method of proof has serious errors. I am providing a standard proof below.
Let $y_{n} = x_{n} - a$ then $\lim_{n \to \infty}y_{n} = 0$ and $$\lim_{n \to \infty}\frac{y_{1} + y_{2} + \cdots + y_{n}}{n} = \lim_{n \to \infty}\frac{x_{1} + x_{2} + \cdots + x_{n}}{n} - a$$ We will establish that $$\lim_{n \to \infty}\frac{y_{1} + y_{2} + \cdots + y_{n}}{n} = 0$$ and this will solve our problem. Invoking a new sequence $y_{n}$ tending to zero simplifies some of the logic below in a minor way.
Let $\epsilon > 0$ be arbitrary. Since $\lim_{n \to \infty}y_{n} = 0$, we have a positive integer $m$ such that $|y_{n}| < \epsilon$ whenever $n > m$. Let $M = \max\{|y_{1}|, |y_{2}|, \ldots, |y_{m}|\}$. Now we can see that
$\displaystyle \begin{aligned}\left|\frac{y_{1} + y_{2} + \cdots + y_{n}}{n}\right| &= \left|\frac{y_{1} + y_{2} + \cdots + y_{m} + y_{m + 1} + \cdots + y_{n}}{n}\right|\\
&\leq \left|\frac{y_{1} + y_{2} + \cdots + y_{m}}{n}\right| + \left|\frac{y_{m + 1} + \cdots + y_{n}}{n}\right|\\
&\leq \frac{|y_{1}| + |y_{2}| + \cdots + |y_{m}|}{n} + \frac{|y_{m + 1}| + |y_{m + 2}| + \cdots + |y_{n}|}{n}\\
&\leq \frac{m}{n}\cdot M + \frac{n - m}{n}\epsilon\end{aligned}$
In the above the value of $m$ is dependent on $\epsilon$ and value $M$ depends on $m$ so as far as the variable $n$ is concerned we can treat $m, M$ as constants depending on $\epsilon$. If $n \to \infty$ then $\left(\dfrac{m}{n}\cdot M + \dfrac{n - m}{n}\epsilon\right) \to \epsilon$. It follows that there is a positive integer $p$ such that $$\left|\frac{y_{1} + y_{2} + \cdots + y_{n}}{n}\right| \leq \epsilon$$ for $n > p$. This means that $$\lim_{n \to \infty}\frac{y_{1} + y_{2} + \cdots + y_{n}}{n} = 0$$
If you read the above proof carefully you will find that we have split $(y_{1} + y_{2} + \cdots + y_{n})/n$ into two parts. In one part the terms are not small but bounded and the number of terms is fixed. Considering the denominator $n$ this part tends to zero as $n \to \infty$. In the other part the number of terms depends on $n$ and becomes infinite as $n \to \infty$, but each term itself is small (less than $\epsilon$) so that the whole part is less than $\epsilon$. It follows that the expression $(y_{1} + y_{2} + \cdots + y_{n})/n$ is small when $n \to \infty$.
A: 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.
