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I am struggling to prove that a sequence $(a_n)$ diverges to infinity if its sequence of averages $s_n=\frac1n\sum_{k=1}^n a_n$ diverges to infinity.

My attempt:

We prove the contrapositive. Let $N\in\mathbb N$. Pick $n=N+1$. Suppose $(a_n)$ does not diverge to infinity, i.e. $a_n<M$ for some $M\in\mathbb R$. Then,

$$s_n=\frac1n\sum_{k=1}^N a_k+\frac{a_n}{n}<\sum_{k=1}^N a_k + M.$$

Let $M'=\sum_{k=1}^N a_k + M$, and we have $s_n<M'$ for all $N\in\mathbb N$. Hence, $(s_n)$ does not diverge to infinity. QED.

The problem with my proof is that $M'$ is dependent on the value of arbitrary $N$. I am trying to find a way to eliminate $M'$'s dependency on $N$. I am thinking about using the fact that $\sum_{k=1}^N a_k<\sum_{k=1}^\infty a_k$ but I don't know whether this infinite sum converges.

John Davies
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    If $\vert a_k\vert \leq M$, then $$ \left\vert \sum_{k=1}^n a_k \right\vert \leq \sum_{k=1}^n \vert a_k\vert \leq Mn.$$ Dividing by $n$ yields that $\vert s_n\vert\leq M$. – Severin Schraven Sep 30 '23 at 03:15

1 Answers1

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For an arbitrary sequence $(a_n)$ of real numbers and the sequence $(s_n)$ of the corresponding means $s_n=\frac1n\sum_{k=1}^n a_n$ hold the relationships $$ \tag{$*$} \liminf_{n \to \infty} a_n \le \liminf_{n \to \infty} s_n \le \limsup_{n \to \infty} s_n \le \limsup_{n \to \infty} a_n \, . $$ That is a special case of the general form of the Stolz–Cesàro theorem, see also If $\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$ then $\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n$ or Show that $\lim \inf a_n\le\lim\inf s_n.$.

This relationship holds even if a $\liminf$ or $\limsup$ is plus or minus infinity.

If $(s_n)$ diverges to $+\infty$ then $(*)$ implies that $\limsup_{n \to \infty} a_n = +\infty$, i.e. the sequence $(a_n)$ is not bounded above.

But it does not follow that $(a_n)$ diverges to $+\infty$. A counterexample is the sequence $a_n = 0, 1, 0, 2, 0, 3, 0, 4,\ldots$ with $s_n \sim n/8$.

Martin R
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