Let $(a_n)_{n\in\mathbb{N}}$ be q sequence of real numbers with $\lim_{n\to\infty}a_n=0$. Show that this implies $$ \lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}\lvert a_i\rvert=0. $$

This is my idea how to prove it, unfortunately do not know if it is right:

Let $\varepsilon > 0$ be arbitrary, then there exists a $N(\varepsilon)$ with $\lvert a_n\rvert < \varepsilon$ for all $n\geqslant N(\varepsilon)$.

So it is $$ \lim_{n\to\infty}\sum_{i=0}^{n-1}\lvert a_i\rvert=\sum_{i=0}^{\infty}\lvert a_i\rvert=\sum_{i=0}^{N(\varepsilon)-1}\lvert a_i\rvert+\sum_{i=N(\varepsilon)}^{\infty}\lvert a_i\rvert\leqslant\sum_{i=0}^{N(\varepsilon)-1}\lvert a_i\rvert+\sum_{i=N(\varepsilon)}^{\infty}\varepsilon\leqslant M $$ for a $M\geqslant 0$ for $\varepsilon \to 0$.

So the limits exists. Because $\lim_{n\to\infty}\frac{1}{n}=0$, i.e. the limit exsits, too, one can write the limit as the product of both limits, i.e. $$ \lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}\lvert a_i\rvert=\lim_{n\to\infty}\frac{1}{n}\cdot\lim_{n\to\infty}\sum_{i=0}^{n-1}\lvert a_i\rvert=0\cdot M=0. $$

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    $\begingroup$ It is not true that if $a_n\rightarrow 0$ then $\sum |a_n|$ converges or that $\sum |a_n|\rightarrow M$. Try $a_n=\frac{1}{n}$. $\endgroup$ – d80d2729a352b1366139fc119d3345 Sep 5 '14 at 10:06
  • $\begingroup$ Ok, I think one counterexample is a_n:=1/n. Can you tell me then how to prove it? $\endgroup$ – mathfemi Sep 5 '14 at 10:07
  • $\begingroup$ Try to bound the partial sum $\frac{1}{n}\sum |a_n|$. $\endgroup$ – d80d2729a352b1366139fc119d3345 Sep 5 '14 at 10:09
  • $\begingroup$ For example $$\frac{1}{n}\sum_{i=0}^{n-1}\lvert a_i\rvert\leqslant n\cdot\max_{0\leq i\leq n-1}\lvert a_i\rvert$$ $\endgroup$ – mathfemi Sep 5 '14 at 10:11
  • $\begingroup$ Read stolz-cesaro theorem and proof. $\endgroup$ – d80d2729a352b1366139fc119d3345 Sep 5 '14 at 10:12

Is very easy with Cesàro-Stolz: $$\lim_{n\to\infty}\frac{\sum_{i=0}^{n-1}\lvert a_i\rvert}n=\lim_{n\to\infty}\frac{|a_n|}{1}=\lim_{n\to\infty}|a_n|=|\lim_{n\to\infty}a_n|=0.$$

  • $\begingroup$ We never had that theorem, so I do not think that I shall use that. $\endgroup$ – mathfemi Sep 5 '14 at 11:47
  • $\begingroup$ But you can see the proof and use it in this particular case. $\endgroup$ – Martín-Blas Pérez Pinilla Sep 5 '14 at 12:20

My solution is as follows:

Ignore $a_0$. As $\lim_{n\to\infty} a_n = 0$, for each $ε_n = 1/n^2$, we can find an integer $N(n)$ such that $|a_n| < 1/n^2$ for every $n \ge N(n)$. Then: $$sum_{k=1}^n |a_k| < sum_{k=1}^n 1/k^2 < sum_{k=1}^\infty 1/k^2 =\pi^2/6 $$ Then $$\lim_{n\to\infty} (1/n)\times(sum_{k=1}^n |a_k|) \le \lim_{n\to\infty} 1/n\times\pi^2/6 = 0$$

  • $\begingroup$ But why $\sum_{k=1}^{n}\lvert a_k\rvert < \sum_{k=1}^{n}1/k^2$? $\endgroup$ – mathfemi Sep 5 '14 at 10:52
  • $\begingroup$ From the assumption of $a_n$, I can choose $ε$ to be any arbitrarily small positive number (in this case is $1/n^2$). UNDER THAT CHOICE, I have $|a_n| < 1/n^2$, and thus same for the sum you asked. $\endgroup$ – SiXUlm Sep 5 '14 at 10:58
  • $\begingroup$ But only for $n\geqslant N(n)$. $\endgroup$ – mathfemi Sep 5 '14 at 10:59
  • $\begingroup$ The limit basically tells you that starting from some index ($N(n)$), the sequence will be very close to its limit (0), so it doesn't matter how the sequence behaves before that. $\endgroup$ – SiXUlm Sep 5 '14 at 11:06
  • $\begingroup$ I nvertheless think its not right, because in the second sum it shouldn't be 1/k^2 but 1/n^2 because thats the fixed epsilon and not 1/k^2 with index k in the sum., $\endgroup$ – mathfemi Sep 5 '14 at 11:23

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