Use the special comparison test to find whether the following series converge or diverge.

$\sum _{n=9}^{\infty }$$\frac{(2n+1)(3n-5)}{\sqrt{n^2-73}}$

1. If $\sum _{n=1}^{\infty }{b}_n$is a convergent series of positive terms and $a_{n}0$and $\frac{a_n}{b_n}$tends to a (finite) limit, then $\sum _{n=1}^{\infty }{a}_n$converges.

2. If $\sum _{n=1}^{\infty }{d}_n$is a divergent series of positive terms and $a_{n}0$and $\frac{a_n}{d_n}$tends to a limit greater than 0 ( or tends to +$\infty$), then $\sum _{n=1}^{\infty }{a}_n$ diverges.

The given series is $\sum _{n=9}^{\infty }\frac{(2n+1)(3n-5)}{\sqrt{n^2-73}}$

$\sum _{n=9}^{\infty }\frac{(2n+1)(3n-5)}{\sqrt{n^2-73}}$= $\sum _{n=9}^{\infty }\frac{6n^2-7n-5}{\sqrt{n^2-73}}$

Solve the series:

$\sum _{n-9}^{\infty }\frac{6n^2}{\sqrt{n^2}}$=$\sum _{n=9}^{\infty }\frac{6n^2}{n}$

= $\sum _{n=9}^{\infty }6n$

Now, the integral test of the series $\sum _{n=9}^{\infty }6n$is as follows:

role="math" localid="1653830107025" ${\int }_{}^{\infty }6ndn=6{\left[\frac{n^2}{2}\right]}^{\infty }$

= $\infty$

The value is infinite therefore the series diverges. So, the use test (b) to show that the given series is divergent.

Solve the limit as:

$\underset{n\to \infty }{\lim }$= $\underset{n\to \infty }{\lim }\frac{6n^2-7n-5}{\sqrt{n^2-73}}÷6n$

= $\underset{n\to \infty }{\lim }\frac{6n^2-7n-5}{\sqrt{n^2-73}}\times \frac{1}{6n}$

= $\underset{n\to \infty }{\lim }\frac{n^2\left[6-{\displaystyle \frac{7}{n}}-{\displaystyle \frac{5}{n^2}}\right]}{\sqrt[n]{1-{\displaystyle \frac{73}{n^2}}}}\times \frac{1}{6n}$

= $\frac{6-0-0}{6}$

= 1

Here $\frac{a_n}{d_n}>0$, therefore the series diverges.