The sequence $a_n=\frac{2n+1}{(n-1)^2}$ decreases monotonically How could we show that the sequence $a_n=\frac{2n+1}{(n-1)^2}$ decreases monotonically? 
When we take the quotient $\frac{a_n}{a_{n+1}}$ we get $\frac{n^2(2n+1)}{(n-1)^2(2n+3)}$. Howcan we conclude that this quotient is $\geq 1$ ? 

Every term of the sequence is positive. So, the sequence is decreasing if, and only if, the ratio $a_{n+1}/a_{n} < 1$ for every integer $n \geq 2$.
\begin{equation*}
\frac{a_{n+1}}{a_{n}} = \frac{(2n+3)(n-1)^2}{(2n+1)n^{2}} = \frac{(2n+3)(n^{2} - 2n + 1)}{2n^{3} + n^{2}}
= \frac{2n^{3} - n^{2} - 4n + 3}{2n^{3} + n^{2}} .
\end{equation*}
For every integer $n \geq 2$,
\begin{equation*}
2n^{2} + 4n - 3 = \left(n + 2 - \sqrt{\frac{5}{2}}\right)\left(n + 2 + \sqrt{\frac{5}{2}}\right) > 0 .
\end{equation*}
So,
\begin{equation*}
2n^{3} - n^{2} - 4n + 3 < 2n^{3} + n^{2} ,
\end{equation*}
and $a_{n+1}/a_{n} < 1$.
