# Convergence of an increasing sequence given convergence of the Cesaro sum

I’ve been playing with Cesaro summation and I’m now stuck on a problem. Given an increasing sequence $$(u_n)_{n\in\mathbb{N}}$$ of real numbers, define it’s Cesaro sum as $$C_n = \frac{1}{n} \sum\limits_{k=1}^n u_k$$. I’ve been able to prove a few facts, ie, given any $$(u_n)$$, if $$(u_n)$$ converges to $$l$$ then $$(C_n)$$ does to, $$(u_n)$$ bounded implies $$(C_n)$$ is bounded as well and $$(u_n)$$ increasing implies $$(C_n)$$ increasing.

Here since we assume $$(u_n)$$ increasing we know $$(C_n)$$ is too. I’d like to show that if we further assume that $$(C_n)$$ is convergent, then $$(u_n)$$ is too.

My guess was to try to show that if $$(C_n)$$ is convergent, then $$(u_n)$$ is bounded, and since it’s increasing we can conclude it is convergent. This works well enough for sequences of positive real numbers, but not sure how to make it work when the sequence is necessarily non-negative. I think it’s true, though it’s not obvious why. Any help is appreciated, thank you.

Since $$(u_n)$$ is increasing, we have for $$m < n$$

$$C_n = \frac{1}{n}\sum_{k=1}^mu_k + \frac{1}{n}\sum_{k=m+1}^nu_k\geqslant\frac{1}{n}\sum_{k=1}^mu_k + u_m\left(1 - \frac{m}{n}\right)$$

If $$C_n \to C$$ as $$n \to \infty$$, then for any fixed $$m$$

$$u_m = \lim_{n \to \infty}u_m\left(1 - \frac{m}{n}\right) \leqslant \lim_{n \to \infty}C_n - \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^mu_k = C$$

Therefore, $$(u_n)$$ is bounded ( and convergent since it is an increasing sequence by hypothesis).

• To directly prove $(u_n)$ is bounded ... – RRL Jan 21 at 17:13

All you have to show is that $$u_{n}$$ is bounded. I mean, increasing and bounded sequence is convergent. If $$u_{n}$$ is unbounded so is $$C_{n}$$. Now, convergence of $$C_{n}$$ implies that $$C_{n}$$ is bounded and so is $$u_{n}$$.

• yeah I was trying to prove $(C_n)$ bounded implies $(u_n)$ bounded but maybe what you suggests, the contrapositive, is easier. Thanks. – not an analyst Jan 21 at 16:15