<s>[INST] 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 "toolbox".
thanks!
 [/INST] $$a_n=\frac{\sqrt{n^{2}+2021n+420}}{\sqrt{n^{3}+2022n+420}} \implies a_n^2=\frac {n^{2}+2021n+420}{n^{3}+2022n+420 }$$
Long division or Taylor series
$$a_n^2 =\frac{1}{n}+\frac{2021}{n^2}-\frac{1602}{n^3}+O\left(\frac{1}{n^4}\right)$$ Doing the same
$$a_{n+1}^2=\frac{1}{n}+\frac{2020}{n^2}-\frac{5643}{n^3}+O\left(\frac{1}{n^4}\right)$$
$$\frac{a_{n+1}^2 } {a_n^2}=\frac{\frac{1}{n}+\frac{2020}{n^2}-\frac{5643}{n^3}+O\left(\frac{1}{n^4}\right) } {\frac{1}{n}+\frac{2021}{n^2}-\frac{1602}{n^3}+O\left(\frac{1}{n^4}\right) }$$ Long division again
$$\frac{a_{n+1}^2 } {a_n^2}=1-\frac{1}{n}-\frac{2020}{n^2}+O\left(\frac{1}{n^3}\right)= 1-\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$
$$\frac{a_{n+1} } {a_n}=1-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)$$
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