Question

Hey, I need help with the method of solving

the series 2^n/(3^(n-1)) from n=1 to infinity

Answer 1

Okay, so you have:
\[\sum_{n=1}^{\infty}\frac{2^n}{3^{n-1}}\]?

Answer 2

yup, ps nice pic haha

Answer 3

easy

Answer 4

= 3 times sum of (2/3)^n

Answer 5

then S= a/(1-r)

Answer 6

everyone wants a piece of this one

Answer 7

= (3 x (2/3) ) / (1/3)

Answer 8

= 6

Answer 9

Haha, thank you. So what you need to do is get in geometric series form. You know that geometric series always start with a 1 but if you plug in n=1, you notice you get 2? So factor out a 2 to rewrite it. Giving:
\[\sum_{n=1}^{\infty}2(\frac{2}{3})^{n-1}\].
So when you plug in n=1 you get a 1.
So the series converges to:
\[\frac{2}{1-\frac{2}{3}}\].
Since |2/3|<1.
Which is 6.

Answer 10

Well a 2(1) :P Since the two can be pulled out of the entire summation.

Answer 11

thanks, didn't see it as a geometric series, thanks everyone