Solve This Limit Problem The Question: Show that if the $lim_{n \to\ \infty} x_{n} = x$ Then $lim_{n\to\ \infty} \frac{1}{n} \sum_{i=1}^{n} x_{i} = x$. 
The Attempt: I tried proving this using the epsilon-delta proof technique. 
For $\epsilon > 0$, there is a $\delta >0$ such that $x_{n} > \delta$ implies $|\frac{1}{n} \sum_{i=1}^{n} x_{i} -x | < \epsilon$. I am not sure where to go from here. I have to manipulate this equation in a way which to find a delta which satisfies the inequality. Can you give me a hint in which I can try to work on this problem. 
Thank you for your help and support. I really appreciate for your time and patience. 
 A: First off, what you wrote makes no sense. We have to find $n_0 \in \mathbb N$, such that for all $n\ge n_0$, etc.
I'll give a hint. Write:
$$x = \frac1{n} \sum_{i=1}^n x$$
A: Hint: 
Separate $\sum_{i=1}^{n} x_{i}$ into two parts such that one is close enough to $x$ and another is small enough.
A: $|\frac{1}{n}\sum\limits_{i=1}^{n}x_{i}-x|=|\frac{1}{n}\sum\limits_{i=1}^{n}x_{i}-\frac{1}{n}\sum\limits_{i=1}^{n}x|=|\frac{1}{n}||\sum\limits_{i=1}^{n}(x_{i}-x)|$. By convergence, of $x_{n}$ to $x$, given $\epsilon>0$ we can find $N \in \mathbb{N}$ such that $n \geq N \Rightarrow |x_{n}-x|<\frac{\epsilon}{2}$. So if $n \geq N,|\frac{1}{n}\sum\limits_{i=N}^{n}(x_{i}-x)|\leq  |\frac{1}{2n}||({n-N}){\epsilon}<\frac{\epsilon}{2}$. As for the sum $|\sum\limits_{i=1}^{N-1} \frac{1}{n}(x_{i}-x)|\leq \sum\limits_{i=1}^{N-1} \frac{1}{n}|x_{i}-x|$, it's a sum of sequences converging to $0$ so it converges to $0$, hence we can find $M$ such that $n \geq M \Rightarrow |\sum\limits_{i=1}^{N-1} \frac{1}{n}(x_{i}-x)|<\frac{\epsilon}{2}$. Let $T=max(M,N)$ and I will leave it for you!
