Calculate and justify why one has this asymptotic for the nth odd coefficient of the tangent function I would like to see a justification that why the nth (odd) coefficient of the tangent function in its Taylor series $c_n:=\frac{f^{(2n-1)}(0)}{(2n-1)!}$ is $$\sim \frac{2}{(\pi/2)^{2n}}.$$
The calculation was in this course notes (in spanish) from Universidad Autónoma de Madrid. 
The last identity in page 10 is stated thus using Cauchy formula. My calculations, following page 11, are that using the Residue theorem for a circle centered in the origin with radius $\pi/2<R<3\pi/2$ then $$c_n=\frac{1}{2\pi i}\cdot 2\pi i\cdot\left[ \text{Res} \left( \frac{\tan z}{z^{2n}} ,z=\frac{\pi}{2}\right)+\text{Res} \left( \frac{\tan z}{z^{2n}} ,z=-\frac{\pi}{2}\right)  \right].$$
I know how calculate the residues, for example $$\text{Res} \left( \frac{\tan z}{z^{2n}} ,z=\frac{\pi}{2}\right)=\left( \lim_{z\to\frac{\pi}{2}} \frac{z-\frac{\pi}{2}}{\cos z}\right) \cdot \left( \lim_{z\to\frac{\pi}{2}} \frac{\sin z}{z^{2n}} \right) =-1\cdot\frac{1}{ \left(  \frac{\pi}{2} \right)^{2n}}.$$
Also one can do a comparison with Wolfram Alpha for the other, Residue[(tan z)/z^(2n),{z,-pi/2}].
Thus I wrote $$c_n=-1\cdot\frac{1}{ \left(  \frac{\pi}{2} \right)^{2n}}-1\cdot\frac{1}{ (-1)^{2n}\left(  \frac{\pi}{2} \right)^{2n}},$$
and notice what is the difference between my calculations and those in page 11, because I have a sign in my coefficients and not the same identity (I have no the summand corcerning the integral), in the text is claimed (I say my interpretation and in my words) that for the first identity in page 11 one need to show that the size of the integral is little, in comparison with the first summand (I understand that it is required compute an upper bound of $|\int_{C_R}|$.) Also since the coeffients in the Taylor expansion of the tangent function are positive I know that in my calculations are mistakes.

Question. Justify rigurously that the odd coefficients in the Taylor series of the tangent function $$f(z)=\tan z$$ satisfy that $$\frac{f^{(2n-1)}(0)}{(2n-1)!}\sim \frac{2}{(\pi/2)^{2n}},$$
  using Cauchy and the Residues theorems for a the circle centered in the origin with radius $\pi/2<R<3\pi/2$. Thanks in advance.

 A: I propose a solution that uses already found results (in fact, that use complex functions theory): it suffices to use the expression of these coefficients using Bernoulli numbers (https://www.quora.com/Why-do-the-Bernouilli-numbers-appear-in-the-coefficients-of-the-Taylor-expansion-of-the-tan-function) and take into account the asymptotic approximation of these numbers that you will find in (https://en.wikipedia.org/wiki/Bernoulli_number#Asymptotic_approximation)
A: The statement of Cauchy's Integral Formula is the following
$$\int_{\gamma} f(z) dz = 2 \pi i \sum_{z_k \in \{Res(f)\}} Res(z_k, f(z))$$
For $f(z) = \tan(z) / z^{2n}$, there is a residue of 
$$z = 0 \implies Res(0, f(z)) = c_n = \frac{f^{(2n)}(0)}{(2n - 1)!}$$
As you have noted. There are also residues of 
$$Res(\frac{\pi}{2}, f(z)) = - \frac{1}{\left(\pi/2\right)^{2n}} \quad Res(\frac{\pi}{2}, f(z)) = - \frac{1}{(-1)^{2n}\left(\pi/2\right)^{2n}} = - \frac{1}{\left(\pi/2\right)^{2n}}$$
where I have noted that $(-1)^{2n} = 1$. This yields
$$\frac{1}{2\pi i }\int_{C_R} f(z) dz = c_n + \left(- \frac{1}{(\pi/2)^{2n}}\right) + \left(- \frac{1}{(\pi/2)^{2n}}\right) = c_n - \frac{2}{(\pi/2)^{2n}}$$
$$\implies c_n = \frac{2}{(\pi/2)^{2n}} + \frac{1}{2\pi i }\int_{C_R} f(z) dz$$
verifying what is written in the paper. 
I think your mistake was equating the residue corresponding to $c_n$ with the other two residues, as this is not the case. Specifically, we cannot say from the start that
$$c_n = \frac{1}{2\pi i} \int_{C_R} \frac{\tan(z)}{z^{2n}} dz$$
because this is not true when $\pi/2 < R < 3\pi / 2$. The reason is because Cauchy's integral formula requires $f$ to be holomorphic inside the region outlined by $C_R$, but this is not the case for $\tan(z)$, because holomorphicity fails at $z = \pm \pi/2$. Therefore, we have to use the residue formula as above.
From here, I think the proof provided in the paper holds up (though I admit my Spanish is a bit rusty). Hopefully this helps.
A: $\tan(z) =  \frac{-1}{z - \pi/2} + \frac{1}{z +\pi/2}$ +  (meromorphic function with poles at distance $\geq \frac{3\pi}{2}$ from $0$).  The dominant term in the asymptotics is from the poles at $\pm \pi/2$, with an exponentially smaller (in $n$) error term.  
For $|a|<1$, $$\frac{1}{z- a^{-1}} = \frac{-a}{1-az} = - \sum a^{n+1} z^n $$ from which the contribution of the two smallest poles to the coefficient of $z^{2n-1}$ in the Taylor series of $\tan(z)$ is $2a^{2n} = 2(2/\pi)^{2n}$.
