Rozwiązanie 2.1

$a_{n+1}-a_n=\frac{-n^2-2n+2}{{\left(n+1\right)}^2}-\frac{-n^2+3}{n^2}=\frac{n^2\left(-n^2-2n+2\right)}{n^2{\left(n+1\right)}^2}-\frac{\left(-n^2+3\right){\left(n+1\right)}^2}{n^2{\left(n+1\right)}^2}=$

$=\frac{-n^4-2n^3+2n^2}{n^2{\left(n+1\right)}^2}-\frac{\left(-n^2+3\right)\left(n^2+2n+1\right)}{n^2{\left(n+1\right)}^2}=$

$=\frac{-n^4-2n^3+2n^2}{n^2{\left(n+1\right)}^2}-\frac{-n^4-2n^3-n^2+3n^2+6n+3}{n^2{\left(n+1\right)}^2}=$

$=\frac{-n^4-2n^3+2n^2-\left(-n^4-2n^3-n^2+3n^2+6n+3\right)}{n^2{\left(n+1\right)}^2}=$

$=\frac{-n^4-2n^3+2n^2+n^4+2n^3+n^2-3n^2-6n-3}{n^2{\left(n+1\right)}^2}=\frac{-6n-3}{n^2{\left(n+1\right)}^2}$

Skoro $n\ge 1$, to $-6n\le -6$, czyli $-6n-3\le -9$. Zatem $-6n-3<0$.

Skoro $n>0$, to $n^2>0$ oraz ${\left(n+1\right)}^2>0$. Zatem $n^2{\left(n+1\right)}^2>0$.

Wobec tego $\frac{-6n-3}{n^2{\left(n+1\right)}^2}<0$, czyli ciąg $\left(a_n\right)$ jest malejący, co należało wykazać.