Find Laurent series of rational function $f(z)={1 \over (z+1)^2(z+2)}$

Question

I need to find Laurent expansion with center in z=0 and on annulus that includes point 7/2 of function f(z)

$$f(z)={1 \over (z+1)^2(z+2)}$$

ok, so then I got this partial fractons

$${1 \over (z+1)^2}+{-1 \over (z+1)} +{1 \over (z+2)}$$

I integrated the first one and eventually it equals to the second one. Then I used sum of geometric series and got this

$$\sum_{i=1}^\infty {(-1)^n 2^{n-1}\over{(z-1)}^n} $$

Then i derived the sum and got this

$$\sum_{i=1}^\infty {(-1)^{n+1} (n+1) 2^{n-1}\over{(z-1)^{n+1} }}$$

Are those two sums correct expansion of the the first two fractions?

Answer 1

You have

$f(z) = -\frac {1}{z+1} + \frac {1}{(z+1)^2} + \frac {1}{z+2}$

$\frac {1}{(z+1)^2} = -\frac {d}{dx}\frac {1}{z+1}$

$\frac {7}{2}$ is outside the Taylor polynomials for all 3.

$\frac {1}{z+1} = \frac {z^{-1}}{1 + z^{-1}} = $$z^{-1}\sum_{n=0}^{\infty} (-1)^n z^{-n}\\\sum_{n=1}^{\infty} (-1)^{n+1} z^{-n}$

and that will converge at $z = \frac {7}{2}$

$-\frac {d}{dx} \sum_{n=1}^{\infty} (-1)^{n+1} z^{-n}=$$\sum_{n=1}^{\infty} (n)(-1)^{n+1} z^{-n-1}\\\sum_{n=1}^{\infty} (n-1)(-1)^{n} z^{-n}$

$\frac {1}{z+2} = \frac {z^{-1}}{1 + 2z^{-1}} = $$z^{-1}\sum_{n=0}^{\infty} (-1)^n 2^nz^{-n}\\\sum_{n=1}^{\infty} (-1)^{n+1}2^{n-1} z^{-n}$

And put it together $\sum_{n=1}^{\infty} (-1)^n(n-2^{n-1})z^{-n}$