TITLE: How do I find the Cesaro sum of the series $\{1, -1, 1, -1, ...\}$?
QUESTION [9 upvotes]: I've seen that the Cesaro sum is $1/2$ but haven't been able to find the steps for figuring that out.

REPLY [3 votes]: The Cesàro Sum of a series $\sum\limits_{k=1}^\infty a_k$ is
$$
\begin{align}
\lim_{n\to\infty}\frac1n\sum_{m=1}^n\sum_{k=1}^ma_k
&=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\sum_{m=k}^na_k\\
&=\lim_{n\to\infty}\frac1n\sum_{k=1}^n(n-k+1)a_k
\end{align}
$$
Since $a_k=(-1)^{k-1}$, we get
$$
\begin{align}
&\lim_{n\to\infty}\frac1n\sum_{k=1}^n(n-k+1)(-1)^{k-1}\\
&=\lim_{n\to\infty}\frac1n\left[\vphantom{\frac{()}2}\right.(n+1)\overbrace{\frac{1-(-1)^n}2}^{\text{sum of }(-1)^{k-1}}-\overbrace{\left(-\lfloor n/2\rfloor+n\frac{1-(-1)^n}2\right)}^{\text{sum of }k(-1)^{k-1}}\left.\vphantom{\frac{()}2}\right]\\
&=\lim_{n\to\infty}\left[\frac1n\frac{1-(-1)^n}2+\frac{\lfloor n/2\rfloor}n\right]\\[3pt]
&=\frac12
\end{align}
$$