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

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

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

False:

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

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  • $\begingroup$ The statement has been edited. This is an answer to the former post $\endgroup$ – DINEDINE Apr 13 '19 at 20:30
  • $\begingroup$ OK, I hadn't noticed that. – Perhaps you should add that information to the answer, as it is currently misleading. $\endgroup$ – Martin R Apr 13 '19 at 20:31

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