I have recently familiarized myself with the peculiar result of:
$1-1+1-1+1\ldots=\frac{1}{2}$
Following this enlightment I was now interested in finding out whether the following infinite series has the same Cesaro sum as the latter:
$-1+1-1+1-1\ldots=$
I did a few calculations and manipulations and ultimately arrived at a Cesaro sum of $-\frac{1}{2}$. However I have no method of verifying my result as I am not familiar with the technicalities of this subject.
The manipulations I performed to arrive at this result are:
let $(-1+1-1\ldots)=S$
$-1-S=-1-(-1+1-1\ldots)$
$-1-S=-1+1-1\ldots$
$-1-S=S$
$2S=-1$
$S=-\frac{1}{2}$
The Cesaro sum is what the running sums average out to. The running sums of $1,-1,1,-1,\ldots$ are $1,0,1,0,\ldots,$ which averages out to 1/2.
Likewise, the running sums of $-1,1,-1,1,\ldots$ are $-1,0,-1,0,\ldots,$ which averages out to $-1/2.$
If you want a more rigorous definition of "averages out", then what we mean is that as $n\to \infty$, $\frac{s_1+\ldots+s_n}{n}$ goes to a limit. Here (in the first series) $$ s_1/1 = 1\\(s_1+s_2)/2 = (1+0)/2 = 1/2 \\(s_1+s_2+s_2)/3 = 2/3\\(s_1+s_2+s_3+s_4)/4 = 1/2$$ and then $3/5, 1/2,4/7, 1/2,\ldots.$ That converges to $1/2.$
And, as we've seen, the running sums of the second series are the exact opposite, so it's the same thing for the second series, only everything has the opposite sign.