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: You have that, in general, if $b_n$ is monotone and unbounded, then
$$\liminf_{n\to\infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}\leq \liminf_{n\to\infty}\frac{a_n}{b_n}\leq\limsup_{n\to\infty}\frac{a_n}{b_n}\leq\limsup_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
(this is Stolz-Cesaro theorem)
Apply this for $b_n = n$, and $a_n = a_1+...+a_n$ to get
$$\liminf_{n\to\infty} a_n \leq \liminf_{n\to\infty}\frac{a_1+...+a_n}{n}\leq \limsup_{n \to \infty}\frac{a_1+...+a_n}{n}\leq\limsup_{n\to\infty}a_n$$
If $a_n$ tends to $\pm \infty$, then the convergence of your sequence to a finite limit is contradicted. Thus $a_n$ is bounded, and the monotonicity implies convergence. 
A: $a_n$ monotonic $\implies \lim a_n $ exists in $[-\infty,\infty] \implies$ Cesaro means $\to$ same limit. Since these means converge to a finite limit, so does $a_n.$
Added on 10 April 2015: I don't know why, but there's something in my first answer that seems to be controversial. Perhaps it's because it's lacking in details. I apologize for any confusion, but it's about as simple as anything can be, given the following well known result: Let $a_n$ be a real sequence, and let $C_n$ be its Cesaro means. Then $a_n \to L \in [-\infty,\infty]\implies C_n\to L.$ Corollary: If $a_n$ is monotone and $C_n \to L \in [-\infty,\infty], $ then $a_n \to L.$ Proof: $a_n$ monotone implies $a_n\to$ some $M\in [-\infty,\infty].$ Therefore, $C_n \to M.$ But it is given $C_n \to L.$ Hence $M=L$ by the uniqueness of limits, and therefore $a_n \to L.$
A: The Sequence Converges Implies the Mean Converges
Suppose that
$$
\lim_{n\to\infty}a_n=L
$$
Then for any $\epsilon\gt0$, there is an $N$ so that if $n\ge N$, we have $|a_n-L|\le\epsilon$. Then
$$
\begin{align}
&\left|\,\lim_{M\to\infty}\frac1M\left(\sum_{n=1}^{N-1}a_n+\sum_{n=N}^Ma_n\right)-L\,\right|\\
&=\left|\,\lim_{M\to\infty}\frac1M\sum_{n=1}^{N-1}(a_n-L)+\lim_{M\to\infty}\frac1M\sum_{n=N}^M(a_n-L)\,\right|\\
&\le0+\lim_{M\to\infty}\frac{M-N+1}M\epsilon\\[9pt]
&=\epsilon
\end{align}
$$
Thus, the mean converges to the same limit.
Mean Converges Implies the Sequence Converges
We will prove the contrapositive. If $a_n$ does not converge, then the mean does not converge.
If $a_n$ is monotonic and bounded, then Monotone Convergence says $a_n$ converges. Thus, if $a_n$ does not converge, $a_n$ is not bounded.
Without loss of generality, suppose $a_n$ is increasing, but not bounded above. Then, for any $L$, there is an $N$ so that if $n\ge N$, we have $a_n\ge L$. This implies that
$$
\begin{align}
\lim_{M\to\infty}\frac1M\left(\sum_{n=1}^{N-1}a_n+\sum_{n=N}^Ma_n\right)
&\ge\lim_{M\to\infty}\frac1M\sum_{n=1}^{N-1}a_n+\lim_{M\to\infty}\frac{M-N+1}{M}L\\
&=L
\end{align}
$$
Thus, for any $L$ we have the mean is at least $L$.
A: 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.
