Given a real sequence $(a_n)_n$ converging to a finite value $a$, a property of the Cesàro mean, defined as the arithmetic mean

$$ b_n=\frac{a_1+\ldots+a_n}{n}, $$


$$ \lim_{n\to\infty}b_n=a,\tag1 $$

so that, supposing $a_n\neq0$ $\forall\,n$ and $a\neq0$, we can also deduce

$$ \lim_{n\to\infty}\frac{b_n}{a_n}=1.\tag2 $$

Is result $(2)$ also valid for $a=0$?
And are results $(1)$ and/or $(2)$ also valid for $a=+\infty$?


The result (2) fails in both cases. For $a=0$, let $a_n=1/n$. Then $b_n=(\ln(n)+O(1))/n$ and so $$\lim\limits_{n\to\infty}\frac{b_n}{a_n}=\lim_{n\to\infty} \ln(n)+O(1)=+\infty.$$ For $a=+\infty$ let $a_n=n$ so $b_n=(n+1)/2$ thus $$\lim_{n\to\infty} \frac{b_n}{a_n}=\lim_{n\to\infty} \frac{n+1}{2n}=\frac{1}{2}.$$

However, (1) is true when $a=+\infty$. To see this, suppose $a_n\to+\infty$. We want to show that for any $x$, there is some $N$ such that $n\geq N\implies b_n>x$. Fix $x$. Note that we have some $N_1$ such that $n\geq N_1\implies a_n>0$. Let $y=\max\left\{-\sum_{n=1}^{N_1} a_n,0\right\}$. We have some $N_2$ such that $n\geq N_2\implies a_n>2x+y$. Let $N=2N_2$. Then $$n\geq N\implies b_n>\frac{\sum_{n=1}^{N_1} a_n +(n-N_2)y+(n-N_2)2x}{n}\ge\frac{(n-N_2)2x}{n}\ge x$$ and so $\lim\limits_{n\to\infty} b_n=+\infty$.

  • $\begingroup$ Interesting, but cannot follow the last reasoning. $\endgroup$ – enzotib Sep 23 '12 at 18:46
  • $\begingroup$ @enzotib I added more details. $\endgroup$ – Alex Becker Sep 23 '12 at 19:01
  • $\begingroup$ The case $a_n=1/n$ could be better explained, being $b_n/a_n=\sum_{n=1}^{\infty}1/n=\infty$. For the proof of $(1)$ in the case $a=+\infty$, I have found a proof in a book, it is similar, maybe a little bit simpler. $\endgroup$ – enzotib Sep 24 '12 at 16:11

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