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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?

Ivalin
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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 $2$.

Ninad Munshi
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  • Is that a theorem or is it the general definition of the radius of convergence? Because the definition I have doesn't say that – Ivalin Jan 20 '21 at 23:09
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    @Ivalin it is a consequence of the analyticity of holomorphic functions https://math.stackexchange.com/questions/886731/proof-that-radius-of-convergence-extend-to-nearest-singularity – Ninad Munshi Jan 20 '21 at 23:19
  • This is very helpful, thanks! – Ivalin Jan 20 '21 at 23:29