# Direct Comparison Test (Divergence)

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 '18 at 10:17
• @Alicia Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… – gimusi Sep 6 '18 at 23:52

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 '18 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 '18 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 '18 at 10:23
• Yes, but since this is bigger than the original series, it won't be valid to compare no? – Alicia Aug 10 '18 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 '18 at 10:28

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

• Hi! How did you get that? Thanks! – Alicia Aug 10 '18 at 10:22
• I think the question is about the divergence of the sequence, not the associated series. – Arnaud D. Aug 10 '18 at 10:23
• Yes it seems about sequences! – gimusi Aug 10 '18 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

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