Proving taylor coefficients of $\tan {\pi z \over 2}$ follow $\lim \limits_{n\to\infty} a_{2n+1}={4\over\pi}$. I've stumbled upon the following question while studying for a test in complex analysis:
Given the following Taylor series: $\tan {\pi z \over 2} = \sum \limits _{n=0}^{\infty} a_{2n+1} z ^ {2n+1}$
Prove that: $\lim \limits_{n\to\infty} a_{2n+1}={4\over\pi}$.
I've tried using Cauchy's integral formula for the $n^{th}$ derivative of $\tan {\pi z \over 2}$, but didn't get much progress.
If it helps, this is the third part of the question. The two others are:


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*Find all the singularity points of $\tan {\pi z \over 2}$, classify them and find the residues. (There are singularities at $\{1 + 2k; k \in \Bbb Z\}$, all are simple poles with  residue $-{2\over\pi}$)

*What is the radius of convergence of the Taylor series: $\tan {\pi z \over 2} = \sum \limits _{n=0}^{\infty} a_{2n+1} z ^ {2n+1}$? (It's 1 because $\tan {\pi z \over 2}$ has singularities in -1, 1)
I'm struggling with this question for several hours, so any help would be appreciated.
 A: We note that
$$\mathrm{Res}(\tan(\frac{\pi z}{2}),1) = \frac{\sin(\frac{\pi}{2})}{-\frac{\pi}{2}\sin(\frac{\pi}{2})} = -\frac{2}{\pi} = \frac{\sin(-\frac{\pi}{2})}{-\frac{\pi}{2}\sin(-\frac{\pi}{2})} = \mathrm{Res}(\tan(\frac{\pi z}{2}),-1)$$
This implies that $f(z):=\tan(\frac{\pi z}{2})+\frac{2}{\pi(z-1)}+\frac{2}{\pi(z+1)}$ is holomorphic in $D_2(0)$. Note
$$\frac{2}{\pi(z-1)}+\frac{2}{\pi(z+1)} = \frac{4z}{\pi(z^2-1)},$$
so we know that that for $z\in\Bbb D$ we have
$$f(z)=\tan(\frac{\pi z}{2})-\frac{4z}{\pi(1-z^2)} = \sum_{n=0}^\infty a_{2n+1}z^{2n+1} - \frac{4z}{\pi}\sum_{n=0}^\infty z^{2n} = \sum_{n=0}^\infty \left(a_{2n+1}-\frac{4}{\pi}\right)z^{2n+1}.$$
This is $f$'s Taylor series in $\Bbb D$, but since $f$ has no singularities in $D_2(0)$ it converges there as well, and in particular at $z=1$. Since the coefficients of a convergent series tend to zero, we're done.
A: The answer of Jonathan Y. is probably something you are looking for. However, say you knew the Taylor series about the origin for the tangent function
$$tan({\pi \over 2}z) = \sum^{\infty}_{i=0}{(-1)^{n+1} 2^{2n}(2^{2n}-1)B_{2n} \over (2n)!} ({\pi \over 2}z)^{2n-1}$$
After substitution of the Bernoulli number 
$$B_n = (-1)^{n+1}{2(2n)! \over (2\pi)^{2n}}\zeta(2n)$$
to an $n$th coefficient one easily sees
$${4 \over \pi}{2^{2n}-1 \over 2^{2n}}\zeta(2n) \xrightarrow{n \to \infty} {4 \over \pi}$$
($\zeta$ is the Riemann zeta function)
