Find the maximal ring where the following series converges: $$\sum_{n=1}^\infty\frac{3^n+2^n}{(z-5)^n}+\sum_{n=0}^{\infty}\frac{n^2}{20^n}(z-5)^{2n}$$
I think that taking the minimum between the two separate radiuss' will be enough. The right series is taylor series which is entire and analytic in $\mathbb C$ and $$\displaystyle{\overline{\lim}_{n\to\infty}\frac {\sqrt[n]{n^2}}{20}}=\frac 1 {20}$$ so $|z-5|<0.05$ but for the second series (the laurent series which has main singularity in z=5), I cannot find how to calculate the radius of convergence. How can I do so? (and by the way, am I correct with the taylor series?)
Best Answer
For the first one, you can use comparison test* to realize this is just a question of
$$\lim\sup_{n\to\infty}\left|{3^n\over (z-5)^n}\right|^{1/n}<1\iff {3\over |z-5|}<1\iff |z-5|>3$$
Similarly the second one gives:
$$\limsup_{n\to\infty}\left|{n^2(z-5)^{2n}\over 20^n}\right|^{1/n}<1\iff |z-5|^2<20\iff |z-5|<\sqrt{20}$$
so you just get $3<|z-5|<2\sqrt 5$.
Note the two inequalities point in different directions, this is because you are getting an annulus, it's not about a minimum, because we're not talking about two Taylor series that converge on a disc with a common center, we're talking about it diverging inside one disc and outside the other.
*The comparison is that $3^n+2^n<2\cdot 3^n$ so that
$$\sum_{n=1}^\infty \left|{3^n+2^n\over (z-5)^n}\right|<2\sum_{n=1}^\infty \left|{3^n\over (z-5)^n}\right|$$
Note that it is justified because we ask about convergence at each point in the annulus, so we can assume $z$ is fixed, and each term still represents some number, just like with series without variables.