# I'm having trouble finding the sum of an infinite series.

I am having trouble finding the sum of

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.

• 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? – Brian M. Scott Feb 17 '15 at 4:07
• Would (4^n)/(14^n)+(9^n)/(14^n) work? – JMartinez Feb 17 '15 at 4:11
• It would indeed. Then you can use the fact that $\frac{a^n}{b^n}=\left(\frac{a}b\right)^n$. – Brian M. Scott Feb 17 '15 at 4:18

Hint: $$\sum_{n=1}^{\infty}{\Big(\frac{4^n + 9^n}{14^n}}\Big) = \sum_{n=1}^{\infty}\Big[ \Big(\frac{4}{14}\Big)^n + \Big(\frac{9}{14} \Big)^n \Big]$$