Question

Whats does the series from n=0 to infinity 3^(1-n) converge to?

Answer 1

\[3^{1-n}=\frac{1}{3^{n-1}}=3\frac{1}{3^n}\]

Answer 2

this is a geometric sum

Answer 3

Right, and I know the formula for the sum of a geometric series is a/1-r. But I cant get the series into the right form to use that.

Answer 4

\[\sum_{n=0}^{\infty}3\frac{1}{3^n}=3\sum_{n=0}^{\infty}\left(\frac{1}{3}\right)^n\]

Answer 5

But dont I need it in the form ar^(n-1) where n greater than or equal to 0?

Answer 6

that is if the sum starts from n=1

Answer 7

yours starts at n=0

Answer 8

you can modify what you have


\[\sum_{n=0}^{\infty}3\frac{1}{3^n}=3\sum_{n=0}^{\infty}\left(\frac{1}{3}\right)^n\]

\[=3\sum_{n=1}^{\infty}\left(\frac{1}{3}\right)^{n-1}\]

Answer 9

Ahh I see, so my sum should come to 9/2?

Answer 10

yes

Answer 11

Awesome! thanks for the help!

Answer 12

np