The monotony of an implicit sequence

Solution 1:

First of all, note that $u_n>1+\frac{1}{n}$ since $$ {f_n(1+\frac{1}{n})=(1+\frac{1}{n})^n-1-n-\frac{1}{n}<e-1-n<e-4<0 \\\text{and $f_n(x)$ is increasing} } $$

To prove why $u_n^2+(n-1)u_n-(n+1)$ is always positive for $n\ge 3$ and $u_n>1+\frac{1}{n}$, note that the roots of $x^2+(n-1)x-(n+1)$ are $$ {r_1=\frac{1-n-\sqrt {n^2+2n+5}}{2}<0<u_n\\ r_2=\frac{1-n+\sqrt {n^2+2n+5}}{2} \\=\frac{2n+2}{\sqrt {n^2+2n+5}+n-1} \\<\frac{2n+2}{n+1+n-1}=1+\frac{1}{n}<u_n } $$ hence, both of the roots of $x^2+(n-1)x-(n+1)$ are less than $u_n$ and since the coefficient of $x^2$ is positive, then $u_n^2+(n-1)u_n-(n+1)>0$ and the proof is complete $\blacksquare$