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It is easy if $S_n=(-1)^n$; it is Cesaro summable to $0$.

But I am unable to find if the sequence $S_n=(-1/2)^n$ is Cesaro summable or not.

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    $\begingroup$ Fact: if $a_n \to a$, then $\frac{a_1+\dots+a_n}{n} \to a$. $\endgroup$ – mathworker21 Dec 26 '17 at 17:02
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    $\begingroup$ In other terms, summable $\to$ Cesàro summable. Cesàro-summability would be of little use, if it were not an extension of usual summability. $\endgroup$ – Jack D'Aurizio Dec 26 '17 at 17:13
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It's not too hard to compute directly:

$$\frac1n\sum_{k=0}^nS_k=\frac1n\sum_{k=0}^n(-1/2)^k=\frac1n\cdot\frac{1-(-1/2)^{n+1}}{1-(-1/2)}=\frac1{3n}\left(2+\frac1{(-2)^n}\right)\to0$$

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    $\begingroup$ For an absolutely convergent series shouldn't the regular sum be equal to the Cesàro sum? $\endgroup$ – Michael McGovern Dec 26 '17 at 19:12
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    $\begingroup$ @MichaelMcGovern Yes, here $S_\infty=\lim\limits_{n\to\infty}(-1/2)^n=0$ is the regular sum. (agreed if you think this $S_n$ is confusing) $\endgroup$ – Simply Beautiful Art Dec 26 '17 at 19:14

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