Show that $a_n=\frac{\sqrt{n^{2}+2021n+420}}{\sqrt{n^{3}+2022n+420}}$ is decreasing I need to show that the sequence $a_n=\frac{\sqrt{n^{2}+2021n+420}}{\sqrt{n^{3}+2022n+420}}$ is decreasing.
I tried showing $a_{n+1}\le a_n$ but it was too messy. 
I did manage to do it by showing that $f'(x)$ will be negative at some point by taking the limit of the nominator of the derivative to infinity,  but this method was pretty exhausting. I wonder if there a simpler method I just didn't see, and i'd be happy to have it in my &quot;toolbox&quot;.
thanks!
If $n\geq 1$,
$$
\frac{{a_n }}{{a_{n + 1} }} = \sqrt {\frac{{n^5  + 2024n^4  + 8508n^3  + 4096228n^2  + 5787803n + 1026060}}{{n^5  + 2023n^4  + 4464n^3  + 4090926n^2  + 5787384n + 1025640}}}  &gt; 1,
$$
since every term in the numerator is larger than the corresponding one in the denominator.
&quot;I tried showing an+1≤an but it was too messy. &quot;
No it's not.
$a_{n+1} \le a_n \iff $
$\frac{\sqrt{(n+1)^{2}+2021(n+1)+420}}{\sqrt{(n+1)^{3}+2022(n+1)+420}} \le \frac{\sqrt{n^{2}+2021n+420}}{\sqrt{n^{3}+2022n+420}}\iff$
$\sqrt{n^{3}+2022n+420}\sqrt{(n+1)^{2}+2021(n+1)+420}\le \sqrt{n^{2}+2021n+420}\sqrt{(n+1)^{3}+2022(n+1)+420}\iff$
$(n^{3}+2022n+420)((n+1)^{2}+2021(n+1)+420)\le (n^{2}+2021n+420)((n+1)^{3}+2022(n+1)+420)$
$(n^3 + 2022n + 420)(n^2 + 2023n + 2422) \le (n^2 + 2021n + 420)(n^3 +3n^2 + 2025n +2443) \iff$
$n^5 + 2023n^4+ (2022+2422)n^3 + (420 + 2022*2023)n^2 +(2022*2422+2023*420)n + 420*2422 \le n^5 + (3+2021)n^4 + (420+3*2021+2025)n^3 + (420*3 + 2021*2025+ 2443)n^2 + (2021*2443 + 420*2025)n + 420*2443\iff$
$n^5 + \color{red}{2023}n^4+ \color{green}{(2022+2422)}n^3 +  \color{blue}{(420 + 2022*2023)}n^2 +\color{purple}{(2022*2422+2023*420)}n + 420*2422 \le n^5  \color{red}{(3+2021)}n^4 +  \color{green}{(420+3*2021+2025)}n^3 + \color{blue}{(420*3 + 2021*2025+ 2443)}n^2 + \color{purple}{(2021*2443 + 420*2025)}n + 420*2443$
ANd as all the terms on the LHS are distinctly less than or equal to the corresponding terms on the RHS that is clearly true