# How do you use the limit comparison test to determine if Sigma (5n-3)/(n^2-2n+5) from [1,oo) is convergent or divergent?

Nov 30, 2016

${\sum}_{n = 1}^{\infty} \frac{5 n - 3}{{n}^{2} - 2 n + 5}$ is divergent

#### Explanation:

$\frac{5 n - 3}{{n}^{2} - 2 n + 5 + 7 n + \frac{5}{4}} < \frac{5 n - 3}{{n}^{2} - 2 n + 5}$ but

$\frac{5 n - 3}{{n}^{2} - 2 n + 5 + 7 n + \frac{5}{4}} = \frac{5 n - 3}{n + \frac{5}{2}} ^ 2 = \frac{5}{n + \frac{5}{2}} - \frac{\frac{31}{2}}{n + \frac{5}{2}} ^ 2$

We know that ${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2$ is convergent and $\frac{1}{n + \frac{5}{2}} ^ 2 < \frac{1}{n} ^ 2$ so

${\sum}_{n = 1}^{\infty} \frac{\frac{31}{2}}{n + \frac{5}{2}} ^ 2$ is convergent. Now examining

${\sum}_{n = 1}^{\infty} \frac{5}{n + \frac{5}{2}} = {\sum}_{n = 3}^{\infty} \left(\frac{5}{n + \frac{1}{2}}\right) > 5 {\sum}_{n = 4}^{\infty} \frac{1}{n}$

and we know that ${\sum}_{k = 4}^{\infty} \frac{1}{n}$ is divergent.

Concluding

${\sum}_{n = 1}^{\infty} \frac{5 n - 3}{{n}^{2} - 2 n + 5}$ is divergent