Science:Math Exam Resources/Courses/MATH101/April 2012/Question 04 (b)/Solution 1

We want to use the Ratio test, so we look at the limit:

$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{a_{n+1}}{a_n}\right|& =\underset{n\to \mathrm{\infty }}{\lim }\frac{\frac{(2n+2)!}{((n+1{)}^2+1)((n+1)!{)}^2}}{\frac{(2n)!}{(n^2+1)(n!{)}^2}}\\ & =\underset{n\to \mathrm{\infty }}{\lim }\frac{(2n+2)!(n^2+1)(n!{)}^2}{(2n)!((n+1{)}^2+1)((n+1)!{)}^2}\\ & =\underset{n\to \mathrm{\infty }}{\lim }\frac{(2n+2)!}{(2n)!}\cdot \frac{(n^2+1)}{(n+1{)}^2+1}\cdot {\left(\frac{n!}{(n+1)!}\right)}^2\\ & =\underset{n\to \mathrm{\infty }}{\lim }(2n+2)(2n+1)\cdot \frac{n^2+1}{n^2+2n+2}\cdot \frac{1}{(n+1{)}^2}\\ & =\underset{n\to \mathrm{\infty }}{\lim }\frac{2n+2}{n+1}\cdot \frac{2n+1}{n+1}\cdot \frac{n^2+1}{n^2+2n+2}=2\cdot 2\cdot 1=4\end{array}$

Since this limit is greater than $1$, by the Ratio test, the series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges.