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.
2026-02-23 17:41:12.1771868472
Is $(-1/2)^n$ Cesaro summable?
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It's not too hard to compute directly:
$$\frac1n\sum_{k=0}^nS_k=\frac1n\sum_{k=0}^n(-1/2)^k=\frac1n\cdot\frac{1-(-1/2)^{n+1}}{1-(-1/2)}=\frac1{3n}\left(2+\frac1{(-2)^n}\right)\to0$$