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I need some help finding the Laurent expansion and residue of $$\dfrac{\exp \left(\frac1z \right)}{(1-z)}$$

So far I've done $$\sum_{j=0}^\infty \frac{z^{-j}}{j!} \sum_{k=0}^\infty z^k = \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{z^{k-j}}{j!}$$

but don't know where to go from here. And is it also possible to use Cauchy product when one of the powers is $<0$ and the other is $>0$?

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See also math.stackexchange.com/questions/122368/… for a very similar question. –  mrf Jun 5 '12 at 10:52
    
If a summation converges absolutely, then you can rearrange its terms however you like without changing its value. –  Hurkyl Jun 5 '12 at 12:43

2 Answers 2

up vote 2 down vote accepted

You can rewrite the series in the form: $$\sum_{j=0}^\infty \sum_{k=0}^\infty \frac{z^{k-j}}{j!}=\sum_{n=1}^{\infty}\left(\sum_{j=n}^{\infty}\frac{1}{j!}\right)z^{-n}+\sum_{n=0}^{\infty}\left(\sum_{j=0}^{\infty}\frac{1}{j!}\right)z^n=\sum_{n=1}^{\infty}\left(e-\sum_{j=0}^{n-1}\frac{1}{j!}\right)z^{-n}+\sum_{n=0}^{\infty}ez^n$$ So the residue at $0$ is the coefficient of $z^{-1}$, which is $\sum_{j=1}^\infty\frac{1}{j!}=e-1$.

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As for the residue: going "naive" may be a good idea$$\frac{1}{1-z}\,e^{\frac{1}{z}}=\left(1+z+z^2+z^3+...+z^n+...\right)\left(1+\frac{1}{z}+\frac{1}{2z^2}+\frac{1}{6z^3}+...+\frac{1}{n!z^n}+...\right)$$

Well, it seems fairly easy to see what products are going to give us the coefficient of $\,z^{-1}\,$:

(first term left) times (second term right), (second left) times (third right),...,(n-th left) times ((n+1)-th right),..., so:$$\frac{1}{z}+\frac{1}{2z}+\frac{1}{6z}+...+\frac{1}{n!z}+...=\frac{1}{z}\sum_{n=1}^\infty\frac{1}{n!}=\frac{1}{z}(e-1)$$

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