Computing the series$\sum\limits_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} $

Question

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?

Answer 1

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}

Answer 2

If $i$ satisfies $i^2=-1$ then $$\sum_{k=1}^\infty\left(\frac{2i}5\right)^k$$ converges in $\Bbb C$ to $$\frac{2i/5}{1-2i/5}=\frac{2i}{5-2i}=\frac{-4+10i}{29}$$ (geometric series) as $|2i/5|<1$.

Answer 3

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.