Is $1-2+3-4+…$ $2$-Cesàro summable to $1/4$? According to the video https://www.youtube.com/watch?v=jcKRGpMiVTw, the series $s_n=\sum_{k=1}^n (-1)^{k+1} k$ is $2$-Cesàro summable. Because of the not-quite rigorous identity
$$ s_\infty = \frac{1}{(1+1)^2},$$
I expected that $s_n$ would then be $2$-Cesàro summable to $1/4$. However, computing the partial averages of the partial averages of $s_{n\rightarrow \infty}$  with the following Python code yields a sequence that quickly decays past 1/4 and seems to be tending towards 0.
def secondNeg(n):
    if n % 2:
        yield n
    else:
        yield -n
    yield from secondNeg(n+1)

def Cesaro(mean, pos, s):
    result = (mean*pos+next(s))/(pos+1)
    yield result
    yield from Cesaro(result, pos+1, s)

for i in Cesaro(0, 1, Cesaro(0, 1, secondNeg(1))):
    print(i)

As a matter of fact, I obtained that the 985th average of averages is approximately $0.0015618150237966225$.
How do I prove that $s_n$ is (or isn’t) $2$-Cesàro summable, and how do I rigorously compute the value of this sum? And why does the code above not give a sequence converging to $1/4$?
 A: If the limit exists, the Cesaro sum of order $\alpha>0$ of $\sum_{n}a_n$ is equivalent to
$$
C(n,\alpha)  = \lim_{n\to\infty} \sum_{j=0}^{n} \frac{\binom{n}{j}}{\binom{n+\alpha}{j}} a_j
$$We have $a_j = j (-1)^{j+1}$, so
$$
C(n,2) = \lim_{n\to\infty} \sum_{j=0}^n \frac{(-1)^{j+1} j (j-n-2) (j-n-1)}{(n+1) (n+2)}
$$We can factor out the denominator and then the numerator cleans up using the closed-form expressions for sums of powers:
$$
\lim_{n\to\infty} \frac{2 n^2-2 (-1)^n n+6 n-3 (-1)^n+3}{8 (n+1) (n+2)} = \frac{1}{4}
$$If you are willing to take it on faith that the original sum is Cesaro-2 summable, you could also compute its value using Abel summation (as Cesaro implies Abel, see here):
$$
\lim_{r\to 1^-} \sum_{n=0}^{\infty} n(-1)^{n+1} r^n = \lim_{r\to 1^-} \frac{r}{(r+1)^2}= \frac{1}{4}
$$
A: To perform 2-Cesaro summation, sum the partial sums, then sum those sums, and divide by the result obtained by performing the same operation on the partial sums of $1+0+0+0+...$, that is, by the triangular numbers.  Here we transform the partial sums
$1,-1,2,-2,3,-3,4,-4,5,-5,...\to$
$1,0,2,0,3,0,4,0,5,0,...\to$
$\color{blue}{1,1,3,3,6,6,10,10,15,15,...}$
and divide by
$1,1,1,1,1,1,1,1,1,1,...\to$
$1,2,3,4,5,6,7,8,9,10,...\to$
$\color{blue}{1,3,6,10,15,21,28,36,45,55,...}$
The resulting quotient equals the $m$-th triangular number divided by the $n$-th triangular number where $n\in\{2m-1,2m\}$; as $n$ increases without bound this has the limit $1/4$.
