Let $\{x_n\}$ be a sequence of real numbers converging to $x$. Show that
$\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n x_k = x .$
Let $\{x_n\}$ be a sequence of real numbers converging to $x$. Show that
$\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n x_k = x .$
Hint: $\{x_n\} \to x$ means that for all $\varepsilon > 0$, there exists $N$ such that $n > N$ implies $|x_n - x| < \varepsilon$.
In this case, let $n > N$. Then,
$ \sum_{k=1}^n x_k = \sum_{k=1}^N x_k + \sum_{k=N+1}^n x_k $
Now, divide by $n$, and let $n \to \infty$. What happens to $\displaystyle\frac{1}{n}\sum_{k=1}^N x_k$ as $n \to \infty$?
Hint: It may be easier to show that the limit of the difference is zero:
$\lim_n |(\frac{1}{n}\sum_{k=1}^nx_k)-x|=0$.
We can rewrite the interior term as $\frac{1}{n}|x_1+\ldots+x_n-nx|$. Apply the triangle inequality to this, and you will see this is less than the sum of two terms that become arbitrarily small as $n\to\infty$ for different reasons.