# Cesaro Summation Exercise

This is an exercise with regard to Cesaro summation which I believe should be relatively straightforward, but I can't seem to figure out exactly how to proceed.

For $$\{a_{n}\}_{n \geq 0}$$, we define the partial sums $$s^{0}_{n} = \sum_{k = 0}^{n}a_{k}$$

and cesaro sums $$s^{1}_{n} = \sum_{k = 0}^{n} s_k^0$$

Define then another method of averaging the series by $$s^{2}_{n} = s^{1}_{2n} - s^{1}_{n} = \sum_{k = n + 1}^{2n}s^{0}_k$$

In terms of $$a_{n}$$, simply by looking at the terms in the summations, $$\frac{s^{1}_{n}}{n + 1} = \sum_{k = 0}^{n} \frac{n + 1 - k}{n + 1} a_{k}$$ $$\frac{s^{2}_{n}}{n} = s^{0}_{n} + \sum_{k = n + 1}^{2n} \frac{2n + 1 - k}{n} a_{k}$$

Finally, I wish to show that if $$\frac{s_{n}^1}{n + 1}$$ converges to some limit $$L$$ ($$\{a_{n}\}_{n \geq 0}$$ is Cesaro summable), then $$s^{2}_{n} \rightarrow L$$.

I've attempted to look at the difference between $$s^{2}_{n}$$ and $$s^{1}_{n}$$, but I can't seem to find any nice bounds. Intuitively, this makes sense since the terms near the beginning of the series will have weights close to 1 as $$n \rightarrow \infty$$ in $$s^{1}_{n}$$ while the final weights will be smaller, but I can't seem to figure out exactly how to show this.

• $L$ is not the limit of $s_n^2$ but of $\frac{s_n^2}n$. Sep 22, 2022 at 22:33

## 1 Answer

Since $$\frac{s_n^1}{n+1}\to L$$, $$\frac{s_n^2}n=\frac{s_{2n}^1-s_n^1}n=\frac{s_{2n}^1}{2n+1}\frac{2n+1}n-\frac{s_n^1}{n+1}\frac{n+1}n\to2L-1L=L.$$