what is the radius of convergence of $\sum_{k=0}^\infty \frac{(−1)^k}{k!(n+k)!}\frac{z^{n+2k}}{2^{n+2k}}$

Question

let be z a complex number. I wanted to find the radius of convergence of this power serie $\sum_{k=0}^\infty \frac{(−1)^k}{k!(n+k)!}\frac{z^{n+2k}}{2^{n+2k}}$

Answer 1

You have $$ \sum_{k=0}^\infty \frac{(−1)^k}{k!(n+k)!}\frac{z^{n+2k}}{2^{n+2k}} =\frac{z^n}{2^n}\sum_{k=0}^\infty \frac{(−1)^k}{k!(n+k)!}\frac{z^{2k}}{2^{2k}}. $$ The radius of convergence is $$ \lim_{k\to\infty} \frac{|a_k|}{|a_{k+1}|} =\lim_{k\to\infty} \frac{(k+1)!(n+k+1)!4^k}{k!(n+k)!4^{k+1}} =\lim_{k\to\infty}\frac{(k+1)(n+k+1)}4=\infty $$ This gives you the radius of convergence for the series of the variable $z^2$. But it does not change things because the radius is infinite. If it were finite, we would have to take the square root.