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 $\geq 1$ ? 
 A: Alternatively, one may write it as the sum of two decreasing sequences:
$$
\frac{2n+1}{(n-1)^2}=\frac{2(n-1)+3}{(n-1)^2}=\frac{2}{n-1}+\frac{3}{(n-1)^2},\qquad n\ge 2.
$$
A: or we compute $$a_{n+1}-a_n=\frac{2n+3}{n^2}-\frac{2n+1}{(n-1)^2}=-{\frac {2\,{n}^{2}+4\,n-3}{{n}^{2} \left( n-1 \right) ^{2}}}<0$$ since we have $n>0$
A: $\frac{n^2(2n+1)}{(n-1)^2(2n+3)}>1 \iff 2n^3+n^2>(n^2-2n+1)(2n+3) \iff 2n^3+n^2>2n^3+3n^2-4n^2-6n+2n+3 \iff 2n^2+4n-3>0 \iff (\sqrt{2}n+\sqrt{2})^2-5>0 \iff (\sqrt{2}n+\sqrt{2})^2>5$ 
Since the last inequality true and we have only biconditionals, it follows that the first inequality, which is precisely what we want, is correct. Therefore the method is basically using what we want to prove. However, keep in mind that you cannot use $\implies$ for once because then you cannot work backwards.
A: 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 $a_{n+1}/a_{n} < 1$.
