TITLE: A divergent sequence whose average sequence converges
QUESTION [8 upvotes]: Can we find a sequence $\{a_n\}$ such that $\lim_{n\to\infty}a_n=\infty$, but
$$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n a_i=a<\infty?$$
Does the sequence $\{\ln(\ln(n))\}$ satisfy this condition? If yes, how can we show that? Thank you!

REPLY [7 votes]: Sequence $(\ln \ln n)$ does not satisfy your condition. No sequence can satisfy your condition if $\lim a_n = +\infty$.
Suppose the sequence $(a_n)$ satisfies your condition and converges to positive infinity. Let $b_n = \frac{1}{n}\sum_{i=1}^n a_i$. We shall prove that $(b_n)$ is unbounded.
Let $A$ be an arbitrary positive number. There is an $n_0 \in \mathbb{N}$ such that for any $n \geq n_0$ we have $a_n > A$. Then, for any $n \geq n_0$,
$$
b_n = \frac{1}{n}\left(\sum_{i=1}^{n_0}a_i + \sum_{i=n_0+1}^n a_i\right) \geq
\frac{A(n-n_0)+const}{n}
$$
As $n $ goes to $\infty$, the RHS converges to $A$, which means that for a sufficiently large $n$ we have $b_n > A - 1$. Since $A$ is arbitrary, sequence $(b_n)$ is unbounded and doesn't have a finite limit.
Still, if $a_n \to \infty$ means unsigned infinity, then one can build such an example. For instance:
$$
a_n = (-1)^n \ln \left[\frac{n+1}{2}\right].
$$

REPLY [3 votes]: In fact, we have the following Abelian theorem:
Theorem. If $a_n \to \alpha$ for some $\alpha \in [-\infty. \infty]$, then its average also tends to the limit $\alpha$. That is,
$$\lim_{n\to\infty} a_n = \alpha\in [-\infty. \infty] \quad\Longrightarrow\quad \lim_{n\to\infty}\frac{1}{n} \sum_{k=1}^{n} a_k = \alpha. $$
The proof for the case $\alpha = \pm\infty$ does not differ so much from that for the finite case.