How do I find the Cesaro sum of the series $\{1, -1, 1, -1, ...\}$?

Here's a fairly detailed sketch: The partial sums of the original series are $s_{2k-1} = 1$, $s_{2k} = 0$, for $k \geq 1$. The averaged partial sums of this sequence are $$ t_{n} = \frac{1}{n} \sum_{k=1}^{n} s_{k} = \begin{cases} \frac{m}{2m-1} & \text{if $n = 2m-1$ is odd,} \\ \frac{m}{2m} = \frac{1}{2} & \text{if $n = 2m$ is even.} \end{cases} $$ Clearly $(t_{n}) \to 1/2$.

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 $$ \begin{align} &\lim_{n\to\infty}\frac1n\sum_{k=1}^n(n-k+1)(-1)^{k-1}\\ &=\lim_{n\to\infty}\frac1n\left[\vphantom{\frac{()}2}\right.(n+1)\overbrace{\frac{1-(-1)^n}2}^{\text{sum of }(-1)^{k-1}}-\overbrace{\left(-\lfloor n/2\rfloor+n\frac{1-(-1)^n}2\right)}^{\text{sum of }k(-1)^{k-1}}\left.\vphantom{\frac{()}2}\right]\\ &=\lim_{n\to\infty}\left[\frac1n\frac{1-(-1)^n}2+\frac{\lfloor n/2\rfloor}n\right]\\[3pt] &=\frac12 \end{align} $$