I am required to prove the following Proposition, but despite having spent considerable time thinking about it i have not been able to make much Headway.
Could you please provide some hints and only hints to get me going.
Proposition. $$\text{ If }(x_n)\to x\text{ then }\left(\frac{x_1+x_2+\dots+x_n}{n}\right)\to x$$
PLEASE DO NOT PROVIDE COMPLETE SOLUTIONS.
On
Assume $x_n\to x$. Then for large $n$, most summands in $x_1+\ldots+x_n$ are close to $x$. The remaining (first) few summands are at a bounded distance from $x$. If we divide by $n$, the contribution of those first summands is controllably small, and apart from that, the average is close to $x$.
Hint:
For any $\varepsilon>0$, there exists an $N>0$ such that for all $n>N$, we have $\;|x_n-x|<\varepsilon$ by hypothesis.
On the other hand, you have to check an inequality of this kind: $$\biggl|\frac{x_1+x_2+\dots +x_n}n-x\biggr|=\biggl|\frac{x_1+x_2+\dots +x_n-nx}n\biggr|<\varepsilon$$
Suppose $n>N$ and split the last fraction as \begin{align} \frac{x_1+x_2+\dots +x_n-nx}n&=\frac{x_1+x_2+\dots +x_N-Nx}n+\frac{x_{N+1}+\dots +x_n-(n_N)x}n\\&=\frac{x_1+x_2+\dots +x_N-Nx}n+\frac{(x_{N+1}-x)+\dots +(x_n-x)}n. \end{align} Use the triangle inequality and choose an ad hoc value for $\varepsilon$.