# Let $\{x_n\}$ be a sequence of real numbers converging to $x$.

Let $\{x_n\}$ be a sequence of real numbers converging to $x$. Show that

$$\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n x_k = x .$$

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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 $n \to \infty$?

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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.

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