Study of a complex function 
Let $f(z)=\frac z {\sinh(z)}$. Determine its singularities in the extended complex plane and the residues in the poles; then show that $f$ can be written as a series of powers centered in $0$ with radius of convergence equal to $\pi$. Show also that this series is of the form $$1+\sum_{k=1}^\infty a_{2k} z^{2k},$$ and that the coefficients $a_{2k}$ satisfy this property: for every $\epsilon \gt 0$, there is a constant $C$ such that $|a_{2k}|\le C \bigl(\frac {1+\epsilon} \pi \bigl)^{2k}$.

Now, the singularities are in $\{i \pi t:t\in \mathbb N,\ t\ne 0\}$, and they are simple poles each one with residue exactly $i \pi t$. To show that $f$ can be written as a series of powers like above, it should be enough to notice that $f$ is holomorphic in the open disk $D_0(\pi)$, and is not holomorphic in any open disk $D_0(R)$, with $R\gt \pi$, right? Then, using Taylor on the hyperbolic sine, we obtain $$\frac 1 z\sum_{k=0}^\infty \frac {z^{2k+1}} {(2k+1)!}=1+\sum_{k=1}^\infty \frac 1 {(2k+1)!} z^{2k}.$$ My problem is that I don't know how to prove the final part, i.e. that $|a_{2k}|\le C \bigl(\frac {1+\epsilon} \pi \bigl)^{2k}$ for every positive $\epsilon$ and some $C$. I think that the criterion $\bigl(\frac 1 \pi\bigl)^{2k} = \lim \sup a_{2k}$ for the radius of convergence is involved, however I don't see how to continue.

Plus, the problem asks to prove that the series $\frac 1 z + \sum_{k=1}^\infty a_{2k} \frac 1 {2k +z}$ converges to a holomorphic function $F$ in $A=\mathbb C\setminus \{-2t:t\in \mathbb N\}$.

Since $a_{2k}=\frac 1 {(2k+1)!}$, we can say that the series is $\frac 1 z + \sum_{k=1}^\infty \frac 1 {(2k+1)!} \frac 1 {2k +z}$, that is defined exactly in $A$. In order to prove that $F$ is holomorphic, we must prove pointwise convergence and the uniform convergence over any compact set of $A$. Since as $k$ goes to infinity the terms of the series are asymptotic to $\frac 1 {(2k+1)!} \frac 1 {2k}$, pointwise convergence is immediate. Moreover, for any compact set $B\subset A$, we have $\sup_B |z|\lt M$, so definitively  $\frac 1 {(2k+1)!} \frac 1 {2k +z} \lt \frac 1 {(2k+1)!} \frac 1 {2k - M}$. Applying Weierstrass criterion, it follows the uniform convergence.
I would like to know if there is any mistake in what I wrote, and above all I would like to know how to prove that inequality, $|a_{2k}|\le C \bigl(\frac {1+\epsilon} \pi \bigl)^{2k}$. Thanks in advance
 A: You obtained the power series of $\sinh z /z$ and not $z/\sinh z$. Since $z/\sinh z$ is an even function and is holomoprhic in the open disc with radius $\pi$ centered at $0$, it admits a power series expansion of the given form (odd powers are missing because the function is even). The constant term is $1$ because the limit of the function at the origin is $1$. By the Cauchy-Hadamard formula
$$
\frac{1}{\pi } = \mathop {\lim \sup }\limits\limits_{n \to  + \infty }  \left| {a_{2n} } \right|^{1/2n} .
$$
Thus, for any $\varepsilon>0$ there is an $n_0>0$, such that for all $n>n_0$
$$
\left| {a_{2n} } \right|^{1/2n}  \le \frac{{1 + \varepsilon }}{\pi } \Leftrightarrow \left| {a_{2n} } \right| \le \left( {\frac{{1 + \varepsilon }}{\pi }} \right)^{2n} .
$$
Now, for $a_2,a_4,\ldots,a_{2n_0}$, there exists a $K>0$ depending on $n_0$ (i.e., depending on $\varepsilon$) such that
$$
\left| {a_{2n} } \right| \le K\left( {\frac{{1 + \varepsilon }}{\pi }} \right)^{2n} 
$$
(you may take $K = \mathop {\max }\limits_{1 \le n \le n_0 } \left( {\frac{{1 + \varepsilon }}{\pi }} \right)^{ - 2n} \left| {a_{2n} } \right|$). Now with $C=\max(1,K)$, we have
$$
\left| {a_{2n} } \right| \le C\left( {\frac{{1 + \varepsilon }}{\pi }} \right)^{2n} 
$$
for all $n\geq 1$. The $C$ depends only on $\varepsilon$.
Just for your interest,
$$
a_{2n}  = \frac{{2(1 - 2^{2n - 1} )}}{{(2n)!}}B_{2n} \sim ( - 1)^n \frac{2}{{\pi ^{2n} }},
$$
where $B_k$ denotes the $k$th Bernoulli number.
