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

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

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