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I am having trouble finding the sum of

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Since I'm sure this is a geometric series and the first term would be (13/14) I used the equation (13/14)/(1-(13/14)) but that didn't seem to be the answer. Any help would be appreciated.

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  • $\begingroup$ It’s not a geometric series: $4^2+9^2\ne 13^2$, for instance. Can you see how to split it as a sum of two series that genuinely are geometric? $\endgroup$ – Brian M. Scott Feb 17 '15 at 4:07
  • $\begingroup$ Would (4^n)/(14^n)+(9^n)/(14^n) work? $\endgroup$ – JMartinez Feb 17 '15 at 4:11
  • $\begingroup$ It would indeed. Then you can use the fact that $\frac{a^n}{b^n}=\left(\frac{a}b\right)^n$. $\endgroup$ – Brian M. Scott Feb 17 '15 at 4:18
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Hint: $$\sum_{n=1}^{\infty}{\Big(\frac{4^n + 9^n}{14^n}}\Big) = \sum_{n=1}^{\infty}\Big[ \Big(\frac{4}{14}\Big)^n + \Big(\frac{9}{14} \Big)^n \Big]$$

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  • $\begingroup$ Maybe you would want to show a one-line proof of why the series at all converges. $\endgroup$ – Landon Carter Feb 17 '15 at 4:13
  • $\begingroup$ Since there are two fractions would I have to do [(4/14)/(1-(4/14))+(9/14)/(1-(9/14))]? $\endgroup$ – JMartinez Feb 17 '15 at 4:15

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