What is the radius of convergence of the Taylor series of $\frac{2\tan x}{1+4x^2}$ around $x=0?$

Question

What is the radius of convergence of the Taylor series of $\frac{2\tan x}{1+4x^2}$ around $x=0?$

I have to find out $a_n=\frac {f^{(n)}(0)}{n!}$, but the $n-$th derivative of the given function is not easy to handle. Please help. Thanks,

Answer 1

Hint. 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 $$b_{2n+1}=2[x^{2n+1}]\sum_{k=0}^{\infty}a_{2k+1} x^{2k+1}\cdot \sum_{k=0}^{\infty}(-4x^2)^{k}\\=2(a_{2n+1}-4a_{2n-1}+4^2a_{2n-3}+\dots +(-4)^na_1)$$ where $\sum_{k=0}^{\infty}a_{2k+1} x^{2k+1}$ is the Taylor's series at $0$ of $\tan(x)$.

Notice that $a_{2n+1}\sim \frac{C}{(\pi/2)^{2n+1}}$ (see for example Calculate and justify why one has this asymptotic for the nth odd coefficient of the tangent function). Show that $$\lim_{n\to +\infty}\sqrt[2n+1]{|b_{2n+1}|}=2$$ and therefore the radius of convergence is $1/2$.

Answer 2

If Complex Analysis is allowed: Let $P:=\{(2k+1) \frac{\pi}{2}:k \in \mathbb Z\} \cup \{\frac{i}{2},-\frac{i}{2}\}$ and $f(z):=\frac{2 \tan(z)}{1+4z^2}$.

Then $f$ is holomorphic on $ \mathbb C \setminus P$.

Hence the radius of convergence in question is given by

$$ \min \{|0-p|: p \in P\}=\frac{1}{2}.$$