If $a_i \in {\mathbb R}_+$ for all $i \in {\mathbb Z_+}$, and
\begin{align} \limsup_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n} a_i < \infty, \end{align}
then does the $\lim$ exist? In other words, is there any $c \in {\mathbb R}$ such that
\begin{align} \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n} a_i =c. \end{align}
Notes: $(1/n)\sum_{i=1}^{n}a_i$ does not increase with $n$, even though $\sum_{i=1}^{n}a_i$ does. Hence, the Monotone Convergence Theorem cannot be applied.
On
No, there are plenty of examples where this fails. Let $a_i=1$ whenever $i$ does not have a 1 in it's decimal expansion. You'll see plenty of such $i$ for $i<1000$ but then suddenly no $i$ in $1000<i<2000$. Then it will jump again. The limsup exists since it is bounded above by $1$. But the average will oscillate forever. To prove this, notice that the average drops below $1/2$ whenever $i\in [10^x,2\cdot 10^x]$ but then is above $1/2$ when $i\in [2\cdot 10^x,3\cdot 10^x]$.
Firstly, thanks to you all!
With your help, I think I can find an counterexample such that $c_n := 1/n \sum_{i=1}^n a_i$ goes between $1/2$ and $1$ infinitely times.
$a_1 = 1$, $a_2=a_3=\varepsilon$, $a_4 = 1-2\varepsilon$, $a_5 =3$, $a_6=a_7=a_8=a_9=a_{10}=a_{11}=\varepsilon$,$a_{12}=1-6\varepsilon$,...
The main idea is
\begin{align} \frac {1}{1} = 1,c_2,c_3,\frac {2}{4} = \frac {1}{2},\frac {5}{5} =1,c_6,c_7,c_8,c_9,c_{10},c_{11},\frac {6}{12},\ldots \end{align}