Convergence of Cesàro means for a monotonic sequence If $(a_n)$ is a monotonic sequence and
$$
\lim_{n \to \infty} \frac{a_1 + a_2 + \cdots + a_n}{n}
$$
exists and is finite, does $a_n$ converge?  If so, does it converge to the same limit?
I claimed that this was true in an old answer of mine.  I think I had convinced myself of it at the time but I can't seem to now.

A monotonic sequence $\{a_n\}_{n=1}^\infty$ tends to some limit $A\in[-\infty,+\infty]$.  If $a_n\to\text{some finite number}$ then $\dfrac{a_1+\cdots+a_n}n\to\text{that same number}$.  So the only alternative (assuming, with no loss of generality, that it's nondecreasing) is
$$
a_n \to +\infty\quad \text{ and } \quad \frac{a_1+\cdots+a_n}n \to A<+\infty.
$$
Since $a_n\to+\infty$, we have for all but finitely many $n$, the inequality $a_n>A+1$. Pick $N$ big enough so that if $n\ge N$ then $(a_1+\cdots+a_n)/n>A-1$ and $a_n>A+\frac 9 {10}$.  Now consider
\begin{align}
& \frac{a_1+\cdots+a_{1000N}}{1000N} \\[10pt]
= {} & \frac{N}{1000N} \left( \frac{a_1+\cdots+a_N}{1000N} \right) + \frac{1000N-N}{1000N} \left( \frac{a_{N+1}+\cdots+a_{1000N}}{1000N - N} \right) \\[10pt]
> {} & \frac 1 {1000} (A-1) + \frac{999}{1000} \left(A+\frac 9 {10}\right) = A + 0.8981. 
\end{align}
So there is a contradiction.
