Direct Comparison Test (Divergence)

Question

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)?

Answer 1

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 $b_n=\frac {9^{n}} {2(5^{n})}$.

Answer 2

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

Answer 3

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

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