Convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(2i)^{n}\cdot n!}{n^{n}}$ I have to show the series
$$\sum_{n=1}^{\infty}\frac{(2i)^{n}\cdot n!}{n^{n}}$$
converges. I know it does and I tried to use the ratio test, but in the final limit, I got
$$\lim_{n\to\infty}2i\left[\left(1+\frac{1}{n}\right)\right]^{-1}$$
which results at
$$\frac{2i}{e}$$
and I don't know if I can't say it's smaller than 1 because of the imaginary unity.
 A: By CS and Stirling we have $$\left|\sum_{n=1}^{\infty}\frac{(2i)^{n}\cdot n!}{n^{n}}\right| \leq \sum_{n=1}^{\infty}\frac{2^{n}\cdot n!}{n^{n}} \leq \sum_{n=1}^{\infty}\frac{2^{n}\cdot \sqrt{2\pi n} \, n^n \, e^{\frac{1}{12n}}}{n^{n} \, e^n} \leq \, e^{\frac{1}{12}} \sum_{n=1}^{\infty}\frac{2^{n}\cdot \sqrt{2\pi n} }{e^n} \, .$$
Regarding your actual question you can do the same again, but using CS first before you apply the ratio test.
A: \begin{align*}
\sum_{n=1}^{\infty}\frac{(2i)^{n}{n!}}{n^n}
&=\sum_{n=1}^{\infty}\frac{(2e^{i\frac{\pi}{2}})^{n}{n!}}{n^n}\\
& =\sum_{n=1}^{\infty}\frac{2^n{n!}\cos(n\frac{\pi}{2})}{n^n}+i\sum_{n=1}^{\infty}\frac{2^n{n!}\sin(n\frac{\pi}{2})}{n^n}\\
&=\sum_{n=1}^{\infty}(a)_{n}+i\sum_{n=1}^{\infty}(b)_{n}
\end{align*}
1)we have
$\sum_{n=1}^{\infty}((a)_{n}+i(b)_{n})$ is convergent $\implies\sum_{n=1}^{\infty}a_{n}$ and $\sum_{n=1}^{\infty}b_{n}$convergens \
So
\begin{align*}
\sum_{n=1}^{\infty}a_{n}&=\sum_{n=1}^{\infty}\frac{2^n{n!}\cos(n\frac{\pi}{2})}{n^n}\\
&=(\sum_{n\in 2N^*}^{}\frac{2^n(n!)\cos(n\frac{\pi}{2})}{n^n}+\sum_{n\in 2N+1}^{}\frac{2^n(n!)\cos(n\frac{\pi}{2})}{n^n})\\
&=\sum_{n\in 2N^*}^{}\frac{2^n(n!)\cos(n\frac{\pi}{2})}{n^n}\
=\sum_{p=1}^{\infty}\frac{2^{2p}((2p)!)\cos(p\pi)}{(2p)^{2p}}\
\end{align*}
$|\frac{a_{n+1}}{a_{n}}|=|\frac{2^(2p+2)(2p+2)!\cos(\pi(p+1))(2p)^{2p}}{(2p+2)^(2p+2)\cos(\pi(p))2^(2p)(2p)!}|=4|\frac{2p+1}{2p+2}||(1-\frac{1}{p+1})^{2p}|$\
$\lim_{n\mapsto\infty}|\frac{a_{n+1}}{a_{n}}|=\frac{4}{e^2}$
After calculating the limit, we find$\frac{4}{e^2}$ so this series is converjent .and the second part in the same way
$\sum_{n=0}^{\infty}a_{n}$is convergent and $\sum_{n=1}^{\infty}b_{n}$is convergent $\implies \sum_{n=1}^{\infty}\frac{(2i)^n(n!)}{n^n}$ is convergent
