$$\sum_{k=1}^\infty \frac{2\times 3^k}{4^{2k+1}}$$
Hi all, I finally am getting the hang of MathJax (sort of) thank goodness! I was hoping for some help on a problem involving series. I am stuck trying to re-write this problem to make it easier to solve. For example, I was hoping to use the fact that a Sum = $$S\infty= \frac{a_1}{1-r}$$
to solve but unlike a problem with simply k+1 in the denominator, this has a constant infront of it which is throwing me off.. any tips on how approach solving or a trick that I am unaware of?
Thank you!
This is a geometric series $$\sum_{k=1}^\infty \frac{2*3^k}{4^{2k+1}}=\sum_{k=1}^\infty \dfrac12\left(\frac{3}{16}\right)^{k}$$ with $a_1=\dfrac12\dfrac{3}{16}$ and $q=\dfrac{3}{16}$, then $$S_\infty=\dfrac{\dfrac12\dfrac{3}{16}}{1-\dfrac{3}{16}}=\dfrac{3}{26}$$