Explanation of Laurent Series

Question

In this pdf: http://sym.lboro.ac.uk/resources/Handout-Laurent.pdf, it says that $$\frac{1}{1-z} = \begin{cases} \displaystyle\sum_{n=0}^\infty z^n, & |z|<1 \\ -\displaystyle\sum_{n=1}^\infty \frac{1}{z^n}, & |z|>1 \end{cases}$$ I know how they get the first one, but how do you get $\displaystyle\sum_{n=1}^\infty \frac{1}{z^n}, |z|>1$? Given this information, how do we find the Laurent Series of something like $\frac{1}{n^3(n+1)^3}$ or $\frac{1}{(n+1)^5}$?

Answer 1

For the second: if $\lvert z\rvert > 1$, then $$ \frac{1}{1-z}= -\frac{1}{z}\frac{1}{1-\frac{1}{z}} = -\frac{1}{z}\sum_{n=0}^\infty \left(\frac{1}{z}\right)^n = -\sum_{n=1}^\infty \frac{1}{z^n} $$ using the first result applied to $x\stackrel{\rm def}{=} \frac{1}{z}$, since $\lvert x\rvert < 1$. Note that the index in the rightmost sum starts at $1$, not $0$.

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As an application, let us look at $\frac{1}{n^3(n+1)^3}$. We can write $$ \frac{1}{n^3(n+1)^3} = \frac{1}{n^6}\frac{1}{(1+\frac{1}{n})^3}. $$ Recall that $$ \frac{2}{(1+z)^3} = \frac{d^2}{dz^2}\frac{1}{1+z} = \frac{d^2}{dz^2}\frac{1}{1-(-z)} $$ so that, as long as $\lvert z\rvert < 1$, we have by properties of power series within their radius of convergence Recall that $$\begin{align*} \frac{2}{(1+z)^3} &= \frac{d^2}{dz^2} \sum_{k=0}^\infty (-1)^k z^k = \sum_{k=0}^\infty (-1)^k \frac{d^2}{dz^2} z^k = \sum_{k=2}^\infty (-1)^k k(k-1) z^{k-2} \\&= \sum_{k=0}^\infty (-1)^{k+2} (k+2)(k+1) z^k \end{align*}$$ so that $$ \frac{1}{n^3(n+1)^3} = \frac{1}{2n^6}\sum_{k=0}^\infty (-1)^{k+2} (k+2)(k+1) \frac{1}{n^k}. $$