# Evaluating the series $\sum_{k=1}^\infty \frac{2\times 3^k}{4^{2k+1}}$

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

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!

• Next stop: getting the hang of titles! :) The key directive is that people should be able to know something about your question without having to open it. – Asaf Karagila Aug 5 '18 at 7:53
• Thank you Asaf, had no idea we could directly integrate problems in the title and have mathjax pick it up! – jackbenimbo Aug 5 '18 at 20:51

## 1 Answer

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 $$S_\infty=\dfrac{\dfrac12\dfrac{3}{16}}{1-\dfrac{3}{16}}=\dfrac{3}{26}$$

• so you reduced 2/4, how are you getting 3/16 though? (Thanks in advance) – jackbenimbo Aug 5 '18 at 6:31
• the power $4$ is $2k+1$, $(4^2)^k*4$. – Nosrati Aug 5 '18 at 6:32
• I see it clearly now thank you so much for your time on a Saturday evening! Cheers – jackbenimbo Aug 5 '18 at 6:41
• you are welcome. – Nosrati Aug 5 '18 at 6:41