Regularizing a divergent sum

Question

I have a sum of an infinite series

$$ S = \frac{1}{3} - 4 + \frac{196}{15} - 21 + 27 - 33 + 39 - 45 + 51 - 57 + 63 + ... $$

which appears to diverge. This can be separated as such

$$ S = (\frac{1}{3} - 4 + \frac{196}{15}) + \sum_{n=1}^{\infty} (-1)^{n} (6n +15) = A + B$$

My question is, is it true that reg($S$) = reg($A$) + reg($B$), where reg($S$) refers to the regularised $S$? Apparently the Abel summation method follows this linearity in regularisation property, even if $A$ or $B$ are divergent. The way I have therefore calculated it is: $\text{reg}(S) = \text{reg}(A) + \text{reg}(B) = A + \text{reg}(B) = \frac{47}{5} - 9 = \frac{2}{5}$, which according to a different calculation I have done (another infinite sum of a few non-zero terms, the rest zero), seems to be correct. $\text{reg}(B) = \text{reg} \bigg( \sum_{n=1}^{\infty} (-1)^{n} (6n +15) \bigg)$ was computed with Mathematica's "Sum" function using option Regularisation -> "Abel".

If anyone could point out why this is indeed the case, or a place/source I can consult more regularisation of a divergent series like this, that would be great. Thanks.

Answer 1

Yes, what you did is correct because Abel summation is stable. (Note that while Abel summation is linear, we are not using linearity here.)

Definition (Abel summability)

$\sum\limits_{n = 0}^{+\infty} a_n$ is Abel summable iff $\sum\limits_{n = 0}^{+\infty} a_n z^n$ has radius of convergence $R = 1$ and $\lim\limits_{x \to 1^-} \sum\limits_{n = 0}^{+\infty} a_n x^n = S$ for some $S \in \mathbb{R}$. In this case, we define the Abel sum $\operatorname{reg}_{\mathcal{A}} \sum\limits_{n = 0}^{+\infty} a_n = S$.

Proposition (Stability of Abel summation)

Let $S \in \mathbb{R}$. The following are equivalent:

1. $\sum\limits_{n = 1}^{+\infty} a_n$ is Abel summable and $\operatorname{reg}_{\mathcal{A}} \sum\limits_{n = 1}^{+\infty} a_n = S$
2. $\sum\limits_{n = 0}^{+\infty} a_n$ is Abel summable and $\operatorname{reg}_{\mathcal{A}} \sum\limits_{n = 0}^{+\infty} a_n = a_0 + S$

Proof of proposition

We will show (1) $\implies$ (2). The other direction can be shown with a similar argument.

Since $\sum\limits_{n = 1}^{+\infty} a_n$ is Abel summable, $\sum\limits_{n = 0}^{+\infty} a_{n + 1} z^n$ has radius of convergence $R_1 = 1$. By the Cauchy–Hadamard theorem, $$\frac{1}{\limsup\limits_n \sqrt[n]{|a_{n + 1}|}} = R_1 = 1$$ This implies that $\sum\limits_{n = 0}^{+\infty} a_n z^n$ has radius of convergence $$R_2 = \frac{1}{\limsup\limits_n \sqrt[n]{|a_n|}} = \frac{1}{\limsup\limits_n \sqrt[n]{|a_{n + 1}|}} = 1$$ since $\limsup$ depends only the tail of the sequence.

Next, observe that for $|z| < 1$, we have $$a_0 + z \sum\limits_{n = 0}^{+\infty} a_{n + 1} z^n = \sum\limits_{n = 0}^{+\infty} a_n z^n$$ since both power series have radius of convergence $R_1 = R_2 = 1$ by previous argument. In particular, for $0 < x < 1$, we have $$a_0 + x \sum\limits_{n = 0}^{+\infty} a_{n + 1} x^n = \sum\limits_{n = 0}^{+\infty} a_n x^n$$ By taking limit as $x \to 1^-$, we see that the LHS converges to $a_0 + S$. Therefore, the RHS must converge to $a_0 + S$ as well by the squeeze theorem.

Hence we have shown that $\sum\limits_{n = 0}^{+\infty} a_n z^n$ has radius of convergence $R_2 = 1$ and $\operatorname{reg}_{\mathcal{A}} \sum\limits_{n = 0}^{+\infty} a_n = \lim\limits_{x \to 1^-} \sum\limits_{n = 0}^{+\infty} a_n x^n = a_0 + S$ which is exactly what we need to show for (2).