Question

Find the sum of the infinite geometric series. 1/3^7 + 1/3^9 + 1/3^11 + 1/3^13...

Answer 1

The form of the nth term is\[a_n=\frac{1}{9^{n}}\]for the series\[\frac{1}{3^7}\sum_{n=0}^{\infty}\frac{1}{3^{2n}}=\frac{1}{3^7}\sum_{n=0}^{\infty}\frac{1}{9^{n}}\]

Answer 2

Let the partial sum of the first a_n terms be\[s_n=1+\frac{1}{9}+\frac{1}{9^2}+...+\frac{1}{9^{n-1}}\]Then\[\frac{1}{9}s_n=\frac{1}{9}+\frac{1}{9^2}+\frac{1}{9^3}+...+\frac{1}{9^n}\]

Answer 3

Subtract the second from the first:\[(1-\frac{1}{9})s_n=1-\frac{1}{9^n}\rightarrow s_n=\frac{1-\frac{1}{9^n}}{8/9}=\frac{9}{8}(1-\frac{1}{9^n})\]

Answer 4

The series will be the limit of the following sequence of partial sums:\[\lim_{n \rightarrow \infty}\frac{1}{3^7}s_n=\lim_{n \rightarrow \infty}\frac{1}{3^7}.\frac{9}{8}.(1-\frac{1}{9^n})=\frac{9}{8.3^7}\]

Answer 5

\[\frac{1}{3^7}+\frac{1}{3^9}+...=\frac{9}{8.3^7}\]

Answer 6

ok let me try it

Answer 7

Whats that fraction simplified?

Answer 8

\[\frac{9}{8.3^7}=\frac{1}{1944}\]

Answer 9

Correct! Thanks

Answer 10

fan me! :P

Answer 11

Done

Answer 12

:)