I want to compute the radius of convergence of $\sum_{n=0}^{\infty} \sin(in) \left( \frac{z}{i+1}\right)^{2n}$. I used two methods but those methods give different answers, so I want to see where my mistake is. I should only use the Hadamard Formula: The radius of convergence of $\sum_{n\geq0} a_nz^n$ is given by $R = \liminf_{n \to \infty} 1/\sqrt[n]{|a_n|}$
First method: Consider $w=z^2$, then we want to compute the radius of convergencec of $\sum_{n\geq0}^{\infty} \frac{\sin(in)}{(1+i)^{2n}}w^n.$ Applying the formula, the liminf is easy to compute and we get that the radius of this new series satisfy $|w|<2/e$ so the radius of convergence of the original series is $\sqrt{2/e}$ (because it satisfies $|z|^2 < \frac{2}{e}$.
Second method: We can describe the series as $\sum_{n=0}^{\infty} a_nz^n$ given by the formula $a_n=0$ if $n$ is odd, $a_n = \frac{\sin(in)}{(1+i)^{2n}}$ if $n$ is even. Here, is where I think I'm wrong computing the liminf, because I get the same expression of the first method for the series in function of $w$. I'm pretty sure that when computing the liminf, I have to use the fact that $n$ is even, but I don't know how. If I proceed without this fact in consideration, we get that the radius of convergence is $2/e$ (same limit as the first method).
Where is my mistake?
By the way, the limit that I reached in the first and second method is $\lim_{m \to \infty} \frac{2}{\sinh(m)^{1/m}}$, maybe I'm wrong here too.
Consider $\sum_{k=0}^\infty a_kz^k$.
For even power terms subseries $(k=2n)$ i.e. $\sum_{n=0}^\infty \frac{i \sinh(k/2)}{(1+i)^k}z^k$;
$$\frac{1}{R_e}=\lim_{k\rightarrow\infty}\begin{vmatrix} \frac{i \sinh(k/2)}{(1+i)^k}\end{vmatrix}^{1/k}=\frac{\sqrt e}{\sqrt2}$$
For odd power subseries $(k=2n+1)$ i.e. $\sum_{n=0}^\infty b_kz^k$ s.t. $b_k=0 ~\forall k$;
$\frac{1}{R_o}=\lim_{k\rightarrow\infty}\begin{vmatrix}0\end{vmatrix}^{1/k}=0$
$\therefore \frac{1}{R}=\sup\{1/R_e, 1/R_o\}=\sqrt e/\sqrt2\implies R=\frac{\sqrt2}{\sqrt e}$