Suppose that a sequence is Cesaro summable. Prove....

Question

Suppose that a sequence $a_{n}$ is Cesaro summable. Prove that

$$\lim_{n \to \infty }\frac{a_{n}}{n}=0$$

Answer 1

Suppose $(a_1 + \cdots +a_n)/n \to L \in \mathbb R.$ Then

$$\frac{a_n}{n} = \frac{a_1 + \cdots + a_n}{n} - \frac{n-1}{n}\frac{a_1 + \cdots + a_{n-1}}{n-1} \to L-L = 0.$$

Let's apply this to the Cesaro sums: Suppose $(S_1 + \cdots +S_n)/n \to L\in \mathbb R,$ where $S_n = a_1 + \cdots +a_n.$ By the above, $S_n/n \to 0.$ Applying the above again shows $a_n/n \to 0.$

(NOTE: This was edited from the original, where I only included the first result. Thanks to @Bungo for pointing this out.)

Answer 2

Let $S_m$ denote the $m$th partial sum of the series: $$S_m = \sum_{k=1}^{m}a_k$$ Let $T_n$ denote the average of the first $n$ partial sums: $$T_n = \frac{1}{n}\sum_{m=1}^{n}S_m$$ Since $\sum a_n$ is Cesaro summable, $T_n$ converges to some limit $L$, so $T_n - T_{n-1} \to L - L = 0$ as $n \to \infty$.

Now, $$\begin{aligned} T_n - T_{n-1} &= \frac{1}{n}\sum_{m=1}^{n}S_m - \frac{1}{n-1}\sum_{m=1}^{n-1}S_m\\ &= \left(\frac{1}{n}\sum_{m=1}^{n}S_m - \frac{1}{n}\sum_{m=1}^{n-1}S_m\right) + \left(\frac{1}{n}\sum_{m=1}^{n-1}S_m - \frac{1}{n-1}\sum_{m=1}^{n-1}S_m\right)\\ &= \frac{1}{n}S_n - \frac{1}{n}\left(\frac{1}{n-1} \sum_{m=1}^{n-1}S_m\right)\\ &= \frac{1}{n}S_n - \frac{1}{n}T_{n-1} \\ \end{aligned}$$ Since $T_n - T_{n-1} \to 0$ and $T_{n-1}/n \to 0$, this means that $S_n / n \to 0$. Therefore also $$\frac{S_{n-1}}{n} = \left(\frac{S_{n-1}}{n-1}\right) \left(\frac{n-1}{n}\right) \to (0)(1) = 0$$ and so we conclude that $$\frac{a_n}{n} = \frac{S_n - S_{n-1}}{n} \to 0$$