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

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

• Next time please provide what you have tried so far – SK19 Apr 13 '19 at 18:07
• @DavidMitra The question is if the limit is zero, then the Cesaro Mean converges to the same limit. In case zero – Ilovemath Apr 13 '19 at 18:16
• See the answer to this. – David Mitra Apr 13 '19 at 18:26
• Thank you for your help @DavidMitra – Ilovemath Apr 13 '19 at 18:34

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