<s>[INST] Find the radius of convergence of the power series representation of $\frac{tan(\frac{z}{2})}{(z+2)(z+3)}$ We are given the following function:
$$\frac{tan(\frac{z}{2})}{(z+2)(z+3)}$$
How do we find the radius of convergence for the power series, representing that function around $z = 0$?
The power series representation for $tan(\frac{z}{2})$ is:
$$\frac{z}{2} + \frac{z^3}{24} + \frac{z^5}{240} + O(z^7)$$
The power series representation for $\frac{1}{(z+2)(z+3)}$ is:
$$\frac{1}{6} - \frac{5z}{36} + \frac{19z^2}{216} + O(z^3)$$
Then I multiply both to obtain:
$$\frac{z}{12} - \frac{5z^2}{72} + \frac{11z^3}{216} - \frac{5z^4}{162} + \frac{697z^5}{38800} + O(z^6)$$
I don't know how to find the radius of convergence from that. Is there a way to convert it to a representation of summation and the nth term in the power series, so that for example, we can use the ratio test to find the radius or is there another way to find it?
 [/INST] The closest singularities of the function are located at $z=-2,-3,\pm\pi$. The closest one to $0$ is $-2$, thus the radius of convergence of the series centered at $0$ would be $2$.
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