# show the convergence of $\sum_{k=1}^{\infty}{\frac{(2i)^k}{5^k}}$ using the Leibniz criterion

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}

• Is $i$ supposed to be the complex number satifying $i^2=-1$? Then $a_k$ is not monotonously decreasing, and $i^{k+1} \ne (-1)^k$. Feb 3, 2018 at 16:54

The complex-version geometric series goes through: $\displaystyle\sum_{n=0}^{\infty}z^{n}=\dfrac{1}{1-z}$, where the convergence interval is $|z|<1$.
You can't order complex numers, so you can't use Leibniz criterion.But the proof is possible without using Leibniz criterion. If we set $x=2i/5$ we have $|x|<1$ and the geometric series gives us the desired result
$\sum_{n=1}^{\infty}x^n=\frac{1}{1-x}-1=\frac{1}{1-2i/5}=-\frac{4}{29} + \frac{10}{29}i$
• You are using the same letter $i$ for two different things ... Feb 3, 2018 at 16:55