It is true that if $\lim _n a_n = 0$, then $\lim_n\dfrac{1}{n}\sum_{j=0}^n|a_j|=0$?

Question

It is true that if $\lim _n a_n = 0$, then $\lim_n\dfrac{1}{n}\sum_{j=0}^n|a_j|=0$ ?

Answer 1

Edit: This is an answer to the previous post which has been edited.

False:

Take $a_n=\frac{1}{\log{n}}$

Answer 2

Yes, and I would do this by splitting the sum. Let $\epsilon>0$ and take $N$ large so that $m>N$ implies $|a_m|<\epsilon.$ Then for $n>N$ we have

$$\frac{1}{n}\sum_{j=0}^n|a_j|=\frac{\sum_{j=0}^{N-1}|a_j|}{n}+\frac{1}{n}\sum_{j=N}^n|a_j|\leq\frac{\sum_{j=0}^{N-1}|a_j|}{n}+\frac{1}{n}(n\epsilon)=\frac{\sum_{j=0}^{N-1}|a_j|}{n}+\epsilon.$$ By order limits $$0\leq\lim_{n}\frac{1}{n}\sum_{j=0}^n|a_j|\leq\epsilon.$$ Since $\epsilon$ was arbitary we have $$\lim_{n}\frac{1}{n}\sum_{j=0}^n|a_j|=0$$ as desired.