# 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 $$1/4$$?

• Can't speak to the code, but showing the regularized sum converges to $1/4$ isn't bad. – FearfulSymmetry Sep 5 '20 at 23:19

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): $$\lim_{r\to 1^-} \sum_{n=0}^{\infty} n(-1)^{n+1} r^n = \lim_{r\to 1^-} \frac{r}{(r+1)^2}= \frac{1}{4}$$

• 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? – Rodrigo Sep 7 '20 at 16:34
• It's just the binomial theorem. – FearfulSymmetry Sep 7 '20 at 17:00

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 $$1/4$$.

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