Finding radius of convergence for series of $\tan(z)$ I want to find radius of convergence for Maclaurin series of $\tan(z)$ without finding the series itself. Is it possible to do? If so, how to derive it.
 A: We will prove that the radius of convergence for Maclaurin series of $\tan$ is $\rho(\tan)=\frac{\pi}{2}$.
First we show that for all $n\in\mathbb{N}$, there exists $P_n\in\mathbb{N}[X]$ such that
$$ \tan^{(n)}=P_n\circ\tan $$
The case $n=0$ is trivial (take $P_0=X$), if $P_0,\ldots,P_n$ are well defined, we use the identity $\tan'=1+\tan^2$. Thus, by Leibniz formula, we have
$$ \tan^{(n+1)}=(1+\tan^2)^{(n)}=(\tan^2)^{(n)}=\sum_{k=0}^n{\binom{n}{k}\tan^{(k)}\tan^{(n-k)}} $$
We then define $$ P_{n+1}=\sum_{k=0}^n{\binom{n}{k}P_kP_{n-k}}\in\mathbb{N}[X] $$
because $P_k\in\mathbb{N}[X]$ for all $0\leqslant k\leqslant n$ by hypothesis.
Let $x\in[0,\frac{\pi}{2}[$, we have $\tan^{(n)}(x)=P_n(\underbrace{\tan x}_{\geqslant 0})\geqslant 0$ for all $n\in\mathbb{N}$ so that
$$ \forall n\in\mathbb{N},\,0\leqslant \sum_{k=0}^{n}\frac{\tan^{(k)}(0)}{k!}x^k=\tan(x)-\underbrace{\int_0^x\frac{(x-t)^n}{n!}\tan^{n+1}(t)dt}_{\geqslant 0}\leqslant \tan(x) $$
Thus the power series $\varphi(x):=\sum_{k=0}^{+\infty}\frac{\tan^{(k)}(0)}{k!}x^k$ converges for all $x\in[0,\frac{\pi}{2}[$ and $\rho(\tan)\geqslant \frac{\pi}{2}$.
We now show that $\varphi(x)=\tan(x)$ for all $x\in]-\frac{\pi}{2},\frac{\pi}{2}[$. Let $a_n=\frac{\tan^{(n)}(0)}{n!}$, using the formula verified by $(P_n)_{n\in\mathbb{N}}$, dividing by $n!$ and evaluating at $X=0$ gives
$$ \forall n\in\mathbb{N}^*,\,(n+1)a_{n+1}=\sum_{k=0}^n a_k a_{n-k} $$
and $a_0=0$, $a_1=1$. Thus
$$ \varphi'(x)=\sum_{n=0}^{+\infty}(n+1)a_{n+1}x^n=1+\sum_{n=1}^{+\infty}\left(\sum_{k=0}^n a_k a_{n-k}\right)x^n=1+\varphi(x)^2 $$
It follows that $(\arctan\circ\varphi)'=1$ and, using $\varphi(0)=a_0=0$, we have for all $x\in]-\frac{\pi}{2},\frac{\pi}{2}[$
$$ \varphi(x)=\tan(x) $$
Thus $\lim\limits_{x\rightarrow \frac{\pi}{2}^-}\varphi(x)=\lim\limits_{x\rightarrow \frac{\pi}{2}^-}\tan x=+\infty$ so that $\rho(\tan)=\frac{\pi}{2}$.
A: Hint:
It is the quotient of the power series for $\sin z$ by the power series for $\cos z$, which is not $0$ at $z=0$, and the radius of convergence of both is $\infty$. The series for $\tan z$ is  obtained dividing the power series for $\sin z$ by the power series for $\cos z$ by increasing power order, and you just have to add the constraint that $\cos z$ can't be $0$.
