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Let $\{a_n\}$ be a sequence of positive real numbers such that $\lim_{n\to \infty} \sigma_n=0$ where $\sigma_n=\left(\sum_{k=1}^n a_k\right)/n$.

Does $\lim_{n} \sigma_n=0\implies \lim_n a_n=0$?

I feel that this should be true but I cannot find a proof for this nor can I find a counterexample. I tried my hand at the proof in the following way:

Since $a_n=n\sigma_n-(n-1)\sigma_{n-1},\ n\ge 2$, we have $\lim_n a_n/n=0$. Also, it is not difficult to prove that $\lim\inf_n a_n=0,\ L:=\lim\sup_n\ge 0$. But then I am stuck.

I will really appreciate short hints, rather than a full answer. Thanks in advance.

Martin Sleziak
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Samrat Mukhopadhyay
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2 Answers2

7

Hint:

Consider $a_n = 1$ when $n$ is a perfect square, and $a_n=0$ when $n$ is not a perfect square.

Show that in this case you have $$\sigma_n =\frac{\lfloor \sqrt{n} \rfloor}{n}$$

Note that if by positive you mean strictly positive, then you can add $a_n$ to any positive sequence converging to 0 : $\frac{1}{n}+a_n$ will do.

N. S.
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This is called Cesàro convergence. For a sequence $a_n$, define $\sigma_n = \frac{1}{n}\sum_{k=1}^n a_k$. Then (I) implies (II): $$ {\rm(I)}\qquad a_n \to L \\ {\rm(II)}\qquad \sigma_n \to L $$ But "in general" (II) does not imply (I). [See the other answers.] But under "some conditions", (II) does imply (I). Such a condition may be known as a Tauberian condition.

Ref: https://en.wikipedia.org/wiki/Abelian_and_tauberian_theorems

GEdgar
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