If sequence converges only uniformly, could any of the mean serieses converge absolutly?

Question

TIL that if a series converges absolutely, the arithmetic, geometric, and harmonic means' serieses also converge to the same limit. Mathematically speaking:

If

$$\lim_{n \to \infty} a_n = L $$

Then:

$$\lim_{n \to \infty} \frac{a_1 + a_2 + ... + a_n}{n} = \lim_{n \to \infty} \sqrt[n]{a_1 a_2 a_3 ... a_n} = \lim_{n \to \infty} \frac{n}{a_1^{-1}+a_2^{-1}+\ldots+a_n^{-1}} = L$$

Partial proofs:

• Prove convergence of the sequence $(z_1+z_2+\cdots + z_n)/n$ of Cesaro means
• On Cesàro convergence: If $ x_n \to x $ then $ z_n = \frac{x_1 + \dots +x_n}{n} \to x $

Would this be true if $L = \infty$? If not, I'm seeking an example that will satisfy:

$$lim_{n \to \infty} a_n = \infty $$

While:

$$lim_{n \to \infty} \frac{a_1 + a_2 + ... + a_n}{n} = S $$

Answer 1

No, such example,does not exist.

If $x_n \to +\infty$ then $b_n:=\frac{x_1+...+x_n}{n} \to +\infty$

$\textbf{Proof}$

Let $M>0$

Then exists $m \in \Bbb{N}$ such that $x_n>M,\forall n \geq m$

Thus $b_n>\frac{x_1+...+x_{m-1}}{n}+\frac{[n-(m-1)]M}{n},\forall n \geq m$

So $\liminf_nb_n \geq M$

Since $M$ is arbitrary then $\liminf_nb_n=+\infty$