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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 $1/4$?

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  • $\begingroup$ Can't speak to the code, but showing the regularized sum converges to $1/4$ isn't bad. $\endgroup$ – FearfulSymmetry Sep 5 '20 at 23:19
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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): $$ \lim_{r\to 1^-} \sum_{n=0}^{\infty} n(-1)^{n+1} r^n = \lim_{r\to 1^-} \frac{r}{(r+1)^2}= \frac{1}{4} $$

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  • $\begingroup$ I see in Shawyer & Watson (the reference cited in Wikipedia) that $A_n^\alpha = \sum_{k=0}^nE_{n-k}^\alpha a_k$ and $E_n^\alpha=\binom{n+\alpha}{\alpha}$. But how does that lead to the first equation in your answer? $\endgroup$ – Rodrigo Sep 7 '20 at 16:34
  • $\begingroup$ It's just the binomial theorem. $\endgroup$ – FearfulSymmetry Sep 7 '20 at 17:00
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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 $1/4$.

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    $\begingroup$ Fixed the code and showed a solution exists; nice work. $\endgroup$ – FearfulSymmetry Sep 6 '20 at 0:24

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