<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] Wolfram Alpha says it diverges because you typed in $n=1$ to infinity instead of $k=1$ to infinity.  Correct the typo and it works.
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