anonymous 5 years ago How do you find the sum of the series of 3^(k-1)/4^(3k+1) ; k = 0 to infinity? (infinite series)

1. anonymous

You just have to do some manipulation on the series to see that it's actually geometric, and then all you need to do is use the sum of a geometric series formula. So...

2. anonymous

$\sum_{k=0}^{\infty}\frac{3^{k-1}}{4^{3k+1}}=\sum_{k=0}^{\infty}\frac{3^k.3^{-1}}{(4^3)^k.4}=\frac{1}{12}\sum_{k=0}^{k=\infty}(\frac{3}{64})^k$

3. anonymous

The geometric series has first term a=1 and common ratio r=3/64. The sum is then,$\frac{a}{1-r}=\frac{1}{1-3/64}=\frac{64}{61}$

4. anonymous

You have to now multiply this answer by 1/12 to get your sum:$\sum_{k=0}^{\infty}\frac{3^{k-1}}{4^{3k+1}}=\frac{16}{183}$

5. anonymous

Please fan me :) I punch out answers and no one bothers to reward! Cheers.

6. anonymous

Wow thank you and how do I fan you? I am new to this site

7. anonymous

There should be a [thumbs up] icon next to my name and the words, "Become a fan" underlined as a link. You just click on that...it's blue...

8. anonymous

oh lol I had refresh it to see the 'become a fan' button but anyway thank you for your help!

9. anonymous

cheers...no worries.