Given sequence converges to $c \in \overline{\mathbb{R}}$ extended reals, prove limit of average = c Problem: Given $a_n \to c \in \overline{\mathbb{R}}$, prove that  $\displaystyle \lim_{n \to \infty} \frac{a_1 + a_2 + \cdots + a_n}{n} = c$.
I have read many of the related posts on this topic:


*

*Convergence of the arithmetic mean

*Mean of a Convergent Sequence

*Two Limits Equal - Proof that limn→∞an=Llimn→∞an=L implies limn→∞∑n1akn=Llimn→∞∑1nakn=L

*Prove convergence of the sequence (z1+z2+⋯+zn)/n(z1+z2+⋯+zn)/n of Cesaro means
These all attempt to split the summation in the numerator like so: $\sum_{i=1}^N + \sum_{i=N}^n$ and then bring in the c, so that eacn term is $\frac{(a_1 - c) + (a_2 - c)}{n}$.
However, all of these previous proofs assume that the sequence is convergent, which means that $\frac{(a_1 - c) + (a_2 - c)}{n} = 0$. In my case, the sequence converges to a number in the extended reals and can thus diverge to $-\infty$ or $\infty$, meaning $c$ could be $-\infty$ or $\infty$, and we can't assume $\lim_{n \to \infty}\frac{(a_1 - c) + (a_2 - c)}{n} = 0$.
Are the methods in the above links still applicable? If not, how should I approach this proof?
 A: We can do one-sided limits in a uniform way:

Lemma.
  Let $a_1,a_2,\dots\in\mathbb R$ and define $m_n=(a_1+\dots+a_n)/n$. Then
  $$\limsup_{n\to\infty} m_n \leq \limsup_{n\to\infty} a_n\tag{1}$$
  where we allow limits in $\overline{\mathbb R}$.
Proof. For every finite $U>\limsup_{n\to\infty} a_n$ there exists $N$ such that for all $n> N$ we have $a_n\leq U$, so
  $$m_n = \frac{Nm_N + a_{N+1}+\dots+a_n}{n} \leq \frac {Nm_N} n + \frac{n-N}{n} U \to U\text{ as $n\to\infty$}.$$
  taking $\limsup_{n\to\infty}$ of both sides we get $\limsup_{n\to\infty} m_n\leq U$ for all $U>\limsup_{n\to\infty} a_n$, which gives (1).
Note that if the right-hand-side of (1) is $+\infty$, this argument is vacuous - there is no such $L$ - but it's still a valid argument.

We can apply this lemma to $a_n$ and the negated sequence $-a_n$ to get:
$$\liminf_{n\to\infty} a_n \leq \liminf_{n\to\infty} m_n\leq \limsup_{n\to\infty} m_n \leq \limsup_{n\to\infty} a_n$$
In particular if $\lim_{n\to\infty} a_n=c\in\overline{\mathbb R}$ then $\lim_{n\to\infty} m_n$ exists and equals $c$.
