How do I find the Cesaro sum of the series $\{1, -1, 1, -1, ...\}$? I've seen that the Cesaro sum is $1/2$ but haven't been able to find the steps for figuring that out.
 A: 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 $(t_{n}) \to 1/2$.
A: 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
$$
\begin{align}
&\lim_{n\to\infty}\frac1n\sum_{k=1}^n(n-k+1)(-1)^{k-1}\\
&=\lim_{n\to\infty}\frac1n\left[\vphantom{\frac{()}2}\right.(n+1)\overbrace{\frac{1-(-1)^n}2}^{\text{sum of }(-1)^{k-1}}-\overbrace{\left(-\lfloor n/2\rfloor+n\frac{1-(-1)^n}2\right)}^{\text{sum of }k(-1)^{k-1}}\left.\vphantom{\frac{()}2}\right]\\
&=\lim_{n\to\infty}\left[\frac1n\frac{1-(-1)^n}2+\frac{\lfloor n/2\rfloor}n\right]\\[3pt]
&=\frac12
\end{align}
$$
A: One such proof:
$$S=1-1+1-1+1-...$$$$\therefore 1-S=1-(1-1+1-1+1-...)=1-1+1-1+1-...=S$$$$\therefore 1=2S$$$$\therefore S=\frac{1}{2}$$
Correction - using Cesaro method:
If $a_n=(-1)^{n+1}$ for $n\ge1$ we get your sequence of $1,-1,1,-1,...$. The sequence of partial sums $S_n$ for this will thus be: $1,0,1,0,...$. So, whilst this does not converge, if we calculate the terms of this sequence: $\frac{S_1+S_2+...+S_n}{n}$, we get:$$\frac{1}{1},\frac{1}{2},\frac{2}{3},\frac{2}{4},\frac{3}{5},\frac{3}{6},...$$You can see here that the terms tend to $\frac{1}{2}$ and, therefore, taking the limit of this sequence as $n\to\infty$ we get:$$\lim_{n\to\infty}\frac{S_1+S_2+...+S_n}{n}=\frac{1}{2}$$We can therefore say that the original series is Cesaro summable since the average value of its partial sums tend to a fixed value, in this case $\frac{1}{2}$.
In other words, the Cesaro sum of an infinite series is the limit of the arithmetic mean (average) of the first $n$ partial sums of the series, as $n$ goes to infinity.
