How do I show that the following series converges in $\mathbb{C}$? Problem Statement

Does the series
  $$S := \sum_{n=0}^{\infty}{\frac{1}{2n^2 - i^n}}$$
  converge in $\mathbb{C}$?

Attempt
Since
$$i^n = \begin{cases}
1, n \equiv 0 \pmod 4 \\
i, n \equiv 1 \pmod 4 \\
-1, n \equiv 2 \pmod 4 \\
-i, n \equiv 3 \pmod 4 \\
\end{cases}
$$
we can write $S$ as
$$S = S_0 + S_1 + S_2 + S_3$$
where
$$S_0 := \sum_{n=0 \\ n \equiv 0 \pmod 4}^{\infty}{\frac{1}{2n^2 - 1}}$$
$$S_1 := \sum_{n=1 \\ n \equiv 1 \pmod 4}^{\infty}{\frac{1}{2n^2 - i}}$$
$$S_2 := \sum_{n=2 \\ n \equiv 2 \pmod 4}^{\infty}{\frac{1}{2n^2 + 1}}$$
$$S_3 := \sum_{n=3 \\ n \equiv 3 \pmod 4}^{\infty}{\frac{1}{2n^2 + i}}.$$
Divergence Test does not apply, as each of the following are true:
$$\left|\frac{1}{2n^2 - 1}\right| \to 0, n \to \infty$$
$$\left|\frac{1}{2n^2 - i}\right| \to 0, n \to \infty$$
$$\left|\frac{1}{2n^2 + 1}\right| \to 0, n \to \infty$$
$$\left|\frac{1}{2n^2 + i}\right| \to 0, n \to \infty$$
I think Ratio Test also does not apply, as the resulting 
$$\left|\frac{s_{n+1}}{s_n}\right|$$
for each of $S_1, S_2, S_3, S_4$ approaches $L=1$ as a limit.
Finally, I think I need to use the Comparison Test.  However, I am not sure how to apply the test in this case.  Any helpful hints will be appreciated.
 A: The series converges absolutely, that is the following series is convergent
$$\sum_{n=0}^{\infty}{\frac{1}{|2n^2 - i^n|}}\leq \sum_{n=0}^{\infty}\frac{1}{2n^2 - 1}<+\infty$$
(note that by triangle inequality $|2n^2 - i^n|\geq |2n^2| - |i^n|=2n^2-1$).
Then use the fact that absolute convergence implies convergence.
A: Your series is absolutely convergent since $\sum_{n\geq 1}\frac{1}{2n^2-1}$ is convergent.
What about computing a closed form for it? We are dealing with
$$ \sum_{m\geq 0}\left(\frac{1}{2(4m+1)^2-i}+\frac{1}{2(4m+2)^2+1}+\frac{1}{2(4m+3)^2+i}+\frac{1}{2(4m+4)^2-1}\right)$$
and in general
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b} $$
for any $a\neq b$ with positive real part, with $\psi(x)=\frac{d}{dx}\log\Gamma(x)$. By exploiting the reflection formula for the $\Gamma$ function, $\Gamma(x)\,\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$, we get that the original series equals
$$ \frac{1}{2}+\frac{\pi}{8\sqrt{2}}\left(\tanh\frac{\pi}{4\sqrt{2}}-\cot\frac{\pi}{4\sqrt{2}}\right)+\frac{1+i}{16}\left[\psi\left(\frac{7-i}{8}\right)-\psi\left(\frac{5+i}{8}\right)\right]+\frac{1-i}{16}\left[\psi\left(\frac{3+i}{8}\right)-\psi\left(\frac{1-i}{8}\right)\right]$$
i.e.
$$ \approx 0.70926022705888734623 + 0.19724548429701715455\,i.$$
A: In $S_3$ and $S_1$ you can multiply both numerator and denominator with $2n^2-i$ and $2n^2+i$ respectivelt.
After you do that, prove that $S_3,S_1$ converge absolutely with respecte to the complex absolute value.
Then they converge besause $\mathbb{C}$ is complete and every absolutely convergent series converges.
Combine the convergence of the $4$ series and you have your conclusion.
