I would like to show the convergence of $\sum_{k=1}^{\infty}{\frac{(2i)^k}{5^k}}$ using the Leibniz criterion.
Question: Is that proof correct?
$\begin{align} \sum_{k=1}^{\infty}{\frac{(2i)^k}{5^k}} &= \sum_{k=0}^{\infty}{\frac{(2i)^{k+1}}{5^{k+1}}} = \sum_{k=1}^{\infty}{\left( \frac{ 2i}{5} \right)^{k+1}} \\ \end{align}$
Let $a_k = \left( \frac{ 2i}{5} \right)^{k+1}$ and thus a monotonously decreasing sequence. And using the Leibniz criterion the sequence $\sum_{k=1}^{\infty}(-1)^ka_k$ converges. Thus $ \sum_{k=1}^{\infty}{\left( \frac{ 2i}{5} \right)^{k+1}}$
$\begin{align} \sum_{k=1}^{\infty}{\left( \frac{ 2i}{5} \right)^{k+1}} = \sum_{k=1}^{\infty}{i^{k+1}\left( \frac{ 2}{5} \right)^{k+1}} = \sum_{k=1}^{\infty}{(-1)^{k} \left( \frac{ 2}{5} \right)^{k+1}} = \sum_{k=1}^{\infty}{(-1)^{k}a_k} \\ \end{align}$