Let $a_n = \frac{9^n}{n + 5^n}$.

At large $n$ value, $a_n$ is expected to behave like $\frac{9^n}{5^n}$, therefore it diverges.

Using the direct comparison test, how can I find $b_n$ (has to be smaller than $a_n$ to prove that $a_n$ diverges)?

  • Hi! I've used MathJax to make your post a bit easier to read. I encourage you to do the same for future questions (see this link for some tips about it). And since you are a new user, you might be interested in this other link about the question format (not that there is anything particularly bad about your post). – Arnaud D. Aug 10 at 10:17

By Binomial Theorem $5^{n}=(1+4)^{n}=1+4n+...+4^{n}>1+4n >n$ so $\frac {9^{n}} {n+5^{n}} > \frac {9^{n}} {2(5^{n})}$. Take $b_n=\frac {9^{n}} {2(5^{n})}$.

  • This is also known as Bernoulli's inequality. – Arnaud D. Aug 10 at 10:19
  • Hi! Thanks for that, what do I do after this? How do I evaluate if this is divergent or convergent? – Alicia Aug 10 at 10:21
  • @Alicia The geometric series $\sum (\frac 9 5)^{n}$ is divergent because the common ratio $ \frac 9 5$ exceeds $1$. – Kavi Rama Murthy Aug 10 at 10:23
  • Yes, but since this is bigger than the original series, it won't be valid to compare no? – Alicia Aug 10 at 10:26
  • @Alicia Read my answer carefully. The original series is bigger than the new series $\sum b_n$, not the other way. – Kavi Rama Murthy Aug 10 at 10:28

We have $a_n \ge \frac{1}{n}$ for all $n$.

  • Hi! How did you get that? Thanks! – Alicia Aug 10 at 10:22
  • I think the question is about the divergence of the sequence, not the associated series. – Arnaud D. Aug 10 at 10:23
  • Yes it seems about sequences! – gimusi Aug 10 at 11:00

We have that eventually $6^n \ge n+5^n$ therefore

$$a_n = \frac{9^n}{n + 5^n}\ge \frac{9^n}{6^n}=\left(\frac32\right)^n\to \infty$$

indeed by induction

  • $n=1\implies 6\ge 1+5$

  • assuming $6^n \ge n+5^n$ true we have

$$6^{n+1}=6\cdot 6^n\ge 6n+6\cdot 5^n\ge (n+1)+5^{n+1}$$

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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