91Ó°ÊÓ

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

2. If $\underset{n=1}{\overset{∞}{∑}}{d}_n$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 +$∞$), then $\underset{n=1}{\overset{∞}{∑}}{a}_n$ diverges.

The given series is $\underset{n=9}{\overset{∞}{∑}}\frac{(2n+1)(3n-5)}{\sqrt{n^2-73}}$

$\underset{n=9}{\overset{∞}{∑}}\frac{(2n+1)(3n-5)}{\sqrt{n^2-73}}$= $\underset{n=9}{\overset{∞}{∑}}\frac{6n^2-7n-5}{\sqrt{n^2-73}}$

Solve the series:

$\underset{n-9}{\overset{∞}{∑}}\frac{6n^2}{\sqrt{n^2}}$=$\underset{n=9}{\overset{∞}{∑}}\frac{6n^2}{n}$

= $\underset{n=9}{\overset{∞}{∑}}6n$

Now, the integral test of the series $\underset{n=9}{\overset{∞}{∑}}6n$is as follows:

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

= $∞$

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:

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

= $\underset{n→∞}{\lim }\frac{6n^2-7n-5}{\sqrt{n^2-73}}×\frac{1}{6n}$

= $\underset{n→∞}{\lim }\frac{n^2\left[6-{\displaystyle \frac{7}{n}}-{\displaystyle \frac{5}{n^2}}\right]}{\sqrt[n]{1-{\displaystyle \frac{73}{n^2}}}}×\frac{1}{6n}$

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

= 1

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