## anonymous 5 years ago Find the sum of the infinite geometric series. 1/3^7 + 1/3^9 + 1/3^11 + 1/3^13...

1. anonymous

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

2. anonymous

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$\frac{1}{9}s_n=\frac{1}{9}+\frac{1}{9^2}+\frac{1}{9^3}+...+\frac{1}{9^n}$

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

ok let me try it

7. anonymous

Whats that fraction simplified?

8. anonymous

$\frac{9}{8.3^7}=\frac{1}{1944}$

9. anonymous

Correct! Thanks

10. anonymous

fan me! :P

11. anonymous

Done

12. anonymous

:)