A question related to the post : On Cesaro convergence.
Let $(a_n)$ an increasing sequence of non-negative real numbers and let $(S_n)$ defined by $$S_n=\frac 1 n\sum_{k=1}^n a_k.$$
Is it true that if $(S_n)$ converges to $L$ as $n \to \infty$ also $(a_n)$ converges to $L$ ?
In the linked post, it is asserted that - in general - this is not true [see Pedro's ansewer, last line : COR If $\lim a_n$ exists and equals $\ell$, so does $\lim S_n$, and it also equals $\ell$. The converse is not true.]
But I suppose that in the case above the "trick" can be that the convergence of $S_n$ impose that $a_n$ is bounded. Is it correct ?
May I have a proof (or a counterexample) ?
Using that $(a_k)_{k \in \mathbb{N}}$ is increasing and non-negative, we get that $$\frac{a_n}{2} \leq \frac{1}{2n} \sum_{k=n}^{2n} a_k \leq S_{2n}.$$ Thus, $S_{2n} \rightarrow L$ already implies that $(a_k)_{k \in \mathbb{N}}$ is bounded and thus convergent (because it is a increasing sequence). Let $\lim_{k \rightarrow \infty} a_k = a$, then we know that the Cesaro mean convergences already to $a$, i.e. $$L= \lim_{n \rightarrow \infty} S_n =a,$$ i.e. $a=L$ as claimed.