1

In Ron Gordon's answer here, he uses the residue theorem to compute the sum $$\sum_{k=-\infty}^{\infty} \frac{(-1)^{k} (2 k-1)}{k^2-k+1}$$

The theorem for the case of infinite sums states that

Let $f : \mathbb C \mapsto \mathbb C$ be holomorphic on $\mathbb C$ except for finitely many points $p_1, p_2, \dots, p_k$ (which are usually poles), none of which are real integers. Suppose that there exists some $M, R>0$ such that $\left|z^2 f(z)\right| \le M$ is bounded $\forall |z| > R$. If $\forall z$ we let $g(z) = \pi \cot(\pi z) f(z)$ and $h(z) = \pi \csc(\pi z) f(z)$, then $$\sum_{n\in \mathbb Z} f(n) = -\sum_{j=1}^k \mathop{\mathrm{Res}}_{z = p_j} g(z) \quad\text{and}\quad \sum_{n\in \mathbb Z} (-1)^n f(n) = -\sum_{j=1}^k \mathop{\mathrm{Res}}_{z = p_j} h(z)$$

In our case here, $f(x) = \frac{2x-1}{x^{2}-x+1} $, and it is very obvious that $\left|\frac{2x^3-x^2}{x^{2}-x+1}\right|$ is not bounded, but instead grows asymptotically with $\mathcal O(x)$.

Despite this condition not being satisfied (which is the condition that makes the contour integral go to $0$, see the proof and theorem here and here on page 206), the result is still correct.

Why is this? Is there anything I am missing (such as the contour integral going to $0$ regardless from something else)?

Help is appreciated much thanks :) Max0815

Max0815
  • 3,584
  • Perhaps https://math.stackexchange.com/questions/840875/infinite-number-of-poles-and-residue-theorem is of help. – Eric Towers Jun 25 '24 at 02:38
  • The section https://en.wikipedia.org/wiki/Residue_theorem#Evaluating_zeta_functions makes the (correct) argument that one can sneak up on infinitely many poles using a nesting sequence of bounded open sets (on each of which there are only finitely many poles and the function is holomorphic everywhere else in the set). Of course that's what the notation $\sum_{k=-\infty}^\infty \dots$ means : sneak up on the infinite sum by taking a limit through finite sums. – Eric Towers Jun 25 '24 at 02:43
  • @EricTowers The wikipedia article seems to argue that since cot(pi z) is bounded on the complex plane, the integrand is what dominates and since the integrand grows with O(n^-2), the integral goes to 0. Meanwhile on the doc i linked, they say that "Suppose that there exists M > 0 such that ∣z^2 f (z)∣ ≤ M for all |z| > R for some R", and this is what they use to show the integral goes to 0. Should I ignore this requirement then from now on and just look at if the integrand goes to 0 I suppose? – Max0815 Jun 25 '24 at 02:49
  • I suppose it makes sense, since when I'm lazy with rigor i just check if the integrand goes to $0$ when discarding any $z\to \infty$ path integrals, and I can't see why this case would be any different. Instead of requiring $z^2 f(z)$ to be bounded, should I instead just take $f(z) \to 0$ as $z\to\infty$ as the requirement now? (i probably should rigorously work this out but its so much work lol) – Max0815 Jun 25 '24 at 02:52
  • Although the cited answer is over a decade old, Ron Gordon's profile page reads "last seen this week". So, he might be able to offer his own clarification if you'd simply express your concerns in comment on his answer. – Blue Jun 25 '24 at 03:31
  • 1
    My rough feeling is that multiplying by the $\csc$, which decays exponentially as $|z|\to\infty$ in all non-real directions, means you would only need $f(z)\to 0$ (and not faster than linearly), similar to applications of Jordan's lemma. The first formula (multiplying by $\cot$) will be more difficult to extend though. – messenger Jun 25 '24 at 04:10
  • Also the sum of residues in the original question seems to be out by a factor of 2. Should be $2\pi,\mathrm{sech}(\sqrt{3}\pi/2)$. – messenger Jun 25 '24 at 04:13
  • @messenger i think you're right. for cotangent you need $f(z)$ to decay $\mathcal O (z^{-2})$ but for the cosecant you can have it decay harmonically. I suppose if we look at the sum it also makes sense, since it would be an alternating harmonic series which converges ig – Max0815 Jun 25 '24 at 04:37
  • 1
    Yep - in this example, I believe $\lim_{n\to\infty} \sum_{-n}^n f(k)$ would exist, but $\lim_{(m,n)\to\infty} \sum_{-m}^n f(k)$ wouldn't. – messenger Jun 25 '24 at 04:43

0 Answers0