Divergent sequence $(a_n)_{n\in\mathbb{N}}$ such that $(\frac{1}{n} \sum\limits_{j=1}^n a_j)_{n\in\mathbb{N}}$ converges?

Question

I'm searching for a sequence that diverges as such

$(a_n)_{n\in\mathbb{N}}$

but if inserted in

it converges.

See also: Example of $(b_n)$ such that $\lim_{n\to\infty} {\frac1n}\sum_{i=1}^{n-1}b_i$ does not exists and $0\le b_n\le 1$ and Convergence of sequence of averages the other way around. — Martin Sleziak, Dec 18 '19 at 11:48

score 1 · Accepted Answer · answered Oct 30 '17 at 08:29

1

Hint: Try the sequence $a_n=(-1)^n$.

answered Oct 30 '17 at 08:29

Zestylemonzi

I'm not sure I understand. Doesn't $$\lim\limits_{n \rightarrow \infty}{(-1)^n} = 1$$? So it is convergent as I see it? – Lars Oct 30 '17 at 08:38
1

I just watched a proof video why the sequence $$(-1)^n$$ does diverge. Ignore my question above. Thanks. – Lars Oct 30 '17 at 08:51