I am having trouble finding the sum of

enter image description here

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.

  • $\begingroup$ 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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.