What is the radius of convergence of the taylor series of$ \frac{2\tan x}{1+4x^2}$ around $x = 0$ ?
my attempsts:The function $f(x) := \frac{2\tan x}{1+4x^2}$ is odd and the coefficient of $x^{2n+1}$ of its Taylor expansion at $0$ is equal to $$\frac{2\tan x}{1+4x^2}=2\sum_{n=0}^{\infty}(-1)^{2n+1}\frac{ x^{2n+1}}{2n+1}\cdot \sum_{k=0}^{\infty}(-4x^2)^{k}$$
Now i can not able proceed further ...pliz help as i have found this solution but i did not understand this...solution me...What is the radius of convergence of the Taylor series of $\frac{2\tan x}{1+4x^2}$ around $x=0?$