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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) ?

Mauro ALLEGRANZA
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    You can use that an increasing sequence has a limit (eventually $+\infty$), and Cesaro's theorem is true also for a sequence converging to $+\infty$. – Kelenner Apr 24 '18 at 08:56
  • What I say is that as $a_n$ is increasing, then $a_n\to l$ (with $l=+\infty$ allowed). But then $S_n\to l$ by Cesaro's theorem (it is true also for $l=+\infty$). It is easy to finish. – Kelenner Apr 24 '18 at 09:47
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    Your hypothesis is that $S_n\to L$. By my previous comment, $S_n\to l$. Hence $L=l$, and $a_n\to L$. – Kelenner Apr 24 '18 at 09:55

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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.

p4sch
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