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Where does the $u$ go?Use the substitution $u=z-1$ .
So the $R = 2$ then correct?That would be one method. Perhaps simpler is to use the fact that u+2= 2((u/2)+ 1)= 2(1- (-u/2)) and $\frac{1}{1- z}= \sum z^n$ (sum of a geometric series). So $\frac{1}{u+ 2}= \frac{1}{2}\frac{1}{1- (-u/2)}= \frac{1}{2}\sum (-u/2)^n$
First series expansionSo the $R = 2$ then correct?
$\displaystyle \lim_{n\to\infty}\left|\frac{(2z+3)\left(-\frac{u}{2}\right)^{n+1}}{(2z+3)\left(-\frac{u}{2}\right)^n}\right|$No, that is not correct. You should not have that "1/2" in front of the limit.
Right.The radius of convergence is 2 right?