<s>[INST] Computing the series$\sum\limits_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} $ Show convergence of 
$\begin{align}
\sum_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} &= \sum_{k=1}^{\infty}{\frac{3^{k-1}}{5^k}}+ \sum_{k=1}^{\infty}{\frac{(2i)^k}{5^k}} \\
&= \sum_{k=1}^{\infty}{\frac{1}{3} \cdot \left( \frac{3}{5} \right) ^k} + \sum_{k=1}^{\infty}{ \left( \frac{2i}{5} \right )^k} \\
&= \frac{1}{3} \cdot \sum_{k=1}^{\infty}{ \left( \frac{3}{5} \right) ^k} + \sum_{k=1}^{\infty}{ \left( \frac{2i}{5} \right )^k} \\
\end{align}$
The first part of the sum converges because it is the geometric series with $q= \frac{3}{5}, 1> \left| \frac{3}{5} \right|$. 
$$\sum_{n=1}^{\infty}{\left(\frac{2i}{5} \right)^k}$$ 
Question: Why does that series diverge (WolframAlpha)?
I mean, if $q=\frac{2i}{5}, |q|<1$ then it should be the geometric series and thus converge?
 [/INST] by geometrics sum wfor any complex such that:   $|z|<1$ we have 

Therefore
$$\sum_{k=1}^{\infty} z^k= \lim_{n\to\infty}\sum_{k=1}^{n} z^k =\lim_{n\to\infty} \frac{z(1-z^{n+1})}{1-z}  = \frac{z}{1-z}  $$
since $\lim_{n\to\infty} z^n =0 $ since $|z|<1$.
Hence
 \begin{align}
\sum_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} &= \sum_{k=1}^{\infty}{\frac{3^{k-1}}{5^k}}+ \sum_{k=1}^{\infty}{\frac{(2i)^k}{5^k}} \\
&= \sum_{k=1}^{\infty}{\frac{1}{3} \cdot \left( \frac{3}{5} \right) ^k} + \sum_{k=1}^{\infty}{ \left( \frac{2i}{5} \right )^k} \\
&= \frac{1}{3} \cdot \sum_{k=1}^{\infty}{ \left( \frac{3}{5} \right) ^k} + \sum_{k=1}^{\infty}{ \left( \frac{2i}{5} \right )^k} \\&= \frac{1}{3}\frac{\frac{3}{5}}{1-\frac{3}{5}}+ \frac{\frac{2i}{5}}{1-\frac{2i}{5}}
\end{align}
</s>