I wanted to prove the following:
if $x_n \to x$ then $y_n \to x$ where $$ y_n = {x_1 + \dots + x_n \over n}$$
Please can you tell me if my proof is correct? My proof is this:
Let $\varepsilon > 0$. Fix $N$ such that $n > N$ implies $|x_n - x| < {\varepsilon \over 2}$.
Then $\left |{1 \over n} \sum_{k=N+1}^{N + n} x_k - x \right | < {\varepsilon \over 2}$. Now let $M$ be such that ${|x_1 + \dots + x_N | \over M} < {\varepsilon \over 2}$ and $M > N$. Then $$ \left | \sum_{k=1}^M {x_k \over M} - x \right | \le \left | \sum_{k=1}^N {x_k \over M} \right | + \left | \sum_{k=N+1}^M {x_k \over |M-N|} - x \right | < \varepsilon $$ Here the proof is finished. But one can observe:
It is possible that $y_n$ converges even if $x_n$ doesn't: If $x_{2n} = 0$ and $x_{2n + 1} = 1$ then $x_n$ does not converge but $y_n \to {1 \over 2}$.
