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

Question

anonymous · Answer

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...

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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...

anonymous · Answer

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

anonymous · Answer

cheers...no worries.