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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.

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

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

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