limitof the series $\sum_{k=2}^{\infty}{\frac{2^{k+1}+3^k}{4^{k-1}}}$

Question

how to compute the limit of the series

$\sum_{k=2}^{\infty}{\frac{2^{k+1}+3^k}{4^{k-1}}}$ ?

$\begin{align} \sum_{k=2}^{\infty}{\frac{2^{k+1}+3^k}{4^{k-1}}} &= \sum_{k=2}^{\infty}{\frac{2^{k+1}}{4^{k-1}}} + \sum_{k=2}^{\infty}{\frac{3^k}{4^{k-1}}} \\ &= 4 \sum_{k=2}^{\infty}{\frac{2^{k+1}}{4^k}} + 4 \sum_{k=2}^{\infty}{\frac{3^k}{4^k}} \\ &= 8 \sum_{k=2}^{\infty}{\left(\frac{2}{4} \right )^k} + 4 \sum_{k=2}^{\infty}{\left(\frac{3}{4} \right )^k} \\ &= 8 \cdot \left( \frac{1}{1-\frac{2}{4}}- \left(\frac{2}{4} \right )^0-\left(\frac{2}{4} \right )^1 \right) + 4 \left( \frac{1}{1-\frac{3}{4}} - \left(\frac{3}{4} \right )^0 - \left(\frac{3}{4} \right )^1 \right) \\ &= 8 \cdot \frac{1}{2} + 4 \cdot \frac{5}{3} = \frac{32}{2}\\ \end{align}$

Which is wrong. Wolfram Alpha returns $13$ as a limit.

Question: Where could be my mistake? Or how could it be done correctly?

Answer 1

There is a mistake in the penultimate step.

In fact the sum of the second series is $\frac 94$ because

$$\sum_{k=2}^{\infty}{\left(\frac{3}{4} \right )^k}= \left( \frac{1}{1-\frac{3}{4}} - \left(\frac{3}{4} \right )^0 - \left(\frac{3}{4} \right )^1 \right)=\left(4-1-\frac 34\right)=3-\frac 34=\frac{12-3}4=\frac 94$$ and then the total sum is 13

Answer 2

I think it's better $$\sum_{k=2}^{\infty}\left(8\left(\frac{1}{2}\right)^k+4\left(\frac{3}{4}\right)^k\right)=\frac{8\cdot\frac{1}{4}}{1-\frac{1}{2}}+\frac{4\cdot\frac{9}{16}}{1-\frac{3}{4}}=4+9=13$$

Answer 3

note that $$\frac{2^{k+1}}{4^{k-1}}+\frac{3^k}{4^{k-1}}=8\cdot\left(\frac{2}{4}\right)^k+4\cdot \left(\frac{3}{4}\right)^k$$