Inspired by Patrick Corn, again

Calculus Level 5

1 − 2 + 3 − 4 + 5 − 6 + ⋯ \large 1-2+3-4+5-6+\cdots

For a series ∑ n = 1 ∞ a n \displaystyle \sum_{n=1}^{\infty} a_n we define its partial sums s n = ∑ k = 1 n a k \displaystyle s_n=\sum_{k=1}^n a_k , their average values b n = 1 n ∑ k = 1 n s k \displaystyle b_n=\frac{1}{n}\sum_{k=1}^n s_k , and the average values of those, c n = 1 n ∑ k = 1 n b k \displaystyle c_n=\frac{1}{n}\sum_{k=1}^n b_k .

Let's say that the series ∑ n = 1 ∞ a n \displaystyle \sum_{n=1}^{\infty} a_n is Corn summable if lim ⁡ n → ∞ c n \displaystyle \lim_{n\to\infty}c_n exists. In this case, lim ⁡ n → ∞ c n \displaystyle\lim_{n\to\infty}c_n is called the Corn sum of the series.

Find the Corn sum of the series 1 − 2 + 3 − 4 + 5 − 6 + ⋯ 1-2+3-4+ 5 - 6+\cdots

If you come to the conclusion that no such sum exists, enter 0.666.

Inspiration .


The answer is 0.25.

Otto Bretscher by Otto Bretscher , 65, USA

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1 solution

Patrick Corn
Feb 29, 2016

As I noted in the solution to the earlier version, we have b 2 m = 0 b_{2m} = 0 and b 2 m + 1 = m + 1 2 m + 1 = 1 2 + 1 4 m + 2 b_{2m+1} = \frac{m+1}{2m+1} = \frac12 + \frac1{4m+2} . So c 2 k = 1 2 k ∑ m = 0 k − 1 ( 1 2 + 1 4 m + 2 ) = k / 2 2 k + 1 4 k ∑ m = 0 k − 1 1 2 m + 1 \begin{aligned} c_{2k} &= \frac1{2k}\sum_{m=0}^{k-1} \left( \frac12 + \frac1{4m+2} \right) \\ &= \frac{k/2}{2k} + \frac1{4k} \sum_{m=0}^{k-1} \frac1{2m+1} \\ \end{aligned} and that second term certainly goes to 0 0 as k → ∞ k \to \infty , because the sum is less than the harmonic number H k H_k , which is less than ln ⁡ ( k ) + 1 \ln(k)+1 for sufficiently large k k (see Euler-Mascheroni constant for details).

So c 2 k → 1 / 4 c_{2k} \to 1/4 as k → ∞ k \to \infty , and c 2 k − 1 = 2 k 2 k − 1 c 2 k c_{2k-1} = \frac{2k}{2k-1} c_{2k} , so it goes to 1 / 4 1/4 as well. So the answer is 0.25 \fbox{0.25} .

The "Corn sum" just seems to be a ( C , 2 ) (C,2) Cesaro sum (I think?), and there is some general theorem (that I don't know much about) that if the Cesaro sum exists, then so does the Abel sum, and they're equal. I think this works for higher-order Cesaro sums as well, not just ( C , 1 ) (C,1) . I might be wrong about this--do you have a good reference, Otto?

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Yes indeed! (+1)

I enjoyed reading your fancy proof of the convergence of c n c_n . I thought of it this way: We know that if a sequence converges to a a , then so does its Cesàro average. Consequently, if the even and odd terms of a sequence converge to a a and b b , respectively, then its Cesàro average will converge to a + b 2 \frac{a+b}{2} . Applied to b n b_n , with the even terms converging to 0 and the odd ones to 1 2 \frac{1}{2} , we see that lim ⁡ n → ∞ c n = 1 4 \lim_{n\to\infty}c_n=\frac{1}{4}

The best reference for this kind of stuff may still be Hardy's "Divergent Series". The relationship between Cesàro and Abel summability is discussed in Section 5.12, Theorems 55 and 56. You are right: Cesàro (or Corn) summability does imply Abel summability, with the same limit (the proof is actually pretty straightforward).

It is interesting that this limit of 1 4 \frac{1}{4} can be obtained by a discrete method. I find this approach more intuitive than the Abel sum, but this may just be a matter of taste.

Again, I want to thank you for bringing up this interesting topic!

Otto Bretscher - 5 years, 3 months ago

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