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This post shows laurent series of $(\sin z) ^{-1}$ using the Cauchy Product. I have seen many compute thisby applying this argument $\frac{1}{1 - (z^3/3!-z^5/5^1+...)} = \sum_{k=1}^\infty (z^3/3! -z^5/5^+...)^k$.

Can this be rigorously justified? (a) Can we rearrange the terms? (b) Is this still the inverse of $\sin z$?

I believe these are the steps. Replacing the terms given by $a_{n}$:

(i) We begin with $\sum _m ( \sum_n a_n )^m $. As $\sum_n |a_n| < \infty$ by hypothesis. From Cauchy's Product, $(\sum_n a_n)^m = \sum_{n} b_{mn}$ and $ \sum_n |b_{mn}| < \infty$. Thus, $\sum_m \sum_n |b_{mn}| < \infty$.

(ii) By rearrangment of positive series (Tonellis), $\sum_m \sum_n |b_{m,n}| = \sum_{m,n} |b_{mn}| = \sum_{k} |b_{f(k)}| < \infty$

(iii) Absolute convergence implies $\sum_k b_{f(k)} = \sum_{m} \sum_n b_{mn} < \infty$. The power series has radius of convergence $\ge $ the radius of which $\sum a_n$ converges ($a_n$ is function of $z^n$) with $|\sum a_n | <1$.

Bryan Shih
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1 Answers1

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The left-hand side of your expansion starts with $1$ instead of $z$, but if you proceed as in robjohn's answer to the linked question, you have $$ \sin z = z\sum_{k=0}^{\infty} \frac{(-1)^{k} z^{2k}}{(2k+1)!}, $$ so that the series $$ \frac{1}{\sin z} = \frac{1}{z} \cdot \frac{1}{1 - \sum\limits_{k=1}^{\infty} \frac{(-1)^{k-1} z^{2k}}{(2k+1)!}} = \frac{1}{z} \cdot \sum_{n=0}^{\infty} \biggl(\sum_{k=1}^{\infty} \frac{(-1)^{k-1} z^{2k}}{(2k+1)!}\biggr)^{n} $$ converges absolutely in the largest disk about $0$ such that $\left|\frac{\sin z}{z} - 1\right| < 1$, and so can be rearranged.

Andrew D. Hwang
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  • May you elaborate the details on why it can be rearranged - we are taking powers of power series? I believe we first apply Cauchy Product in the terms, then apply this theorem: https://www.math.ku.edu/~lerner/m500f09/Rearrangements.pdf , is this right? – Bryan Shih May 26 '17 at 00:39
  • The important thing is, the double series is absolutely convergent, see for example Rearrangement of double infinite sums. (If you like, pick an arbitrary bijection from $\mathbf{N} \times \mathbf{N}$ to $\mathbf{N}$ to arrange the terms in a single absolutely convergent sequence, then apply the theorem you linked.) – Andrew D. Hwang May 26 '17 at 10:37
  • I have updated the post; and I believe the argument is correct, what do you think? – Bryan Shih May 26 '17 at 22:17
  • Yes, that looks right. :) – Andrew D. Hwang May 26 '17 at 23:30