anonymous
  • anonymous
Find the sum of the infinite geometric series. 1/3^7 + 1/3^9 + 1/3^11 + 1/3^13...
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
The form of the nth term is\[a_n=\frac{1}{9^{n}}\]for the series\[\frac{1}{3^7}\sum_{n=0}^{\infty}\frac{1}{3^{2n}}=\frac{1}{3^7}\sum_{n=0}^{\infty}\frac{1}{9^{n}}\]
anonymous
  • 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\[\frac{1}{9}s_n=\frac{1}{9}+\frac{1}{9^2}+\frac{1}{9^3}+...+\frac{1}{9^n}\]
anonymous
  • anonymous
Subtract the second from the first:\[(1-\frac{1}{9})s_n=1-\frac{1}{9^n}\rightarrow s_n=\frac{1-\frac{1}{9^n}}{8/9}=\frac{9}{8}(1-\frac{1}{9^n})\]

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anonymous
  • anonymous
The series will be the limit of the following sequence of partial sums:\[\lim_{n \rightarrow \infty}\frac{1}{3^7}s_n=\lim_{n \rightarrow \infty}\frac{1}{3^7}.\frac{9}{8}.(1-\frac{1}{9^n})=\frac{9}{8.3^7}\]
anonymous
  • anonymous
\[\frac{1}{3^7}+\frac{1}{3^9}+...=\frac{9}{8.3^7}\]
anonymous
  • anonymous
ok let me try it
anonymous
  • anonymous
Whats that fraction simplified?
anonymous
  • anonymous
\[\frac{9}{8.3^7}=\frac{1}{1944}\]
anonymous
  • anonymous
Correct! Thanks
anonymous
  • anonymous
fan me! :P
anonymous
  • anonymous
Done
anonymous
  • anonymous
:)

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