# Is $(-1/2)^n$ Cesaro summable?

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.

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

$$\frac1n\sum_{k=0}^nS_k=\frac1n\sum_{k=0}^n(-1/2)^k=\frac1n\cdot\frac{1-(-1/2)^{n+1}}{1-(-1/2)}=\frac1{3n}\left(2+\frac1{(-2)^n}\right)\to0$$
• @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) – Simply Beautiful Art Dec 26 '17 at 19:14