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!
 A: If $n\geq 1$,
$$
\frac{{a_n }}{{a_{n + 1} }} = \sqrt {\frac{{n^5  + 2024n^4  + 8508n^3  + 4096228n^2  + 5787803n + 1026060}}{{n^5  + 2023n^4  + 4464n^3  + 4090926n^2  + 5787384n + 1025640}}}  > 1,
$$
since every term in the numerator is larger than the corresponding one in the denominator.
A: Alternative approach:
Since each element of $a_1, a_2, \cdots $ is positive,
$a_1, a_2, \cdots $ is decreasing if and only if $(a_1)^2, (a_2)^2, \cdots$ is
decreasing.
Let $b_n$ represent the numerator of $(a_n)^2$, and let $c_n$ represent the denominator of
$(a_n)^2$.  Then it is sufficient to show that beyond a finite number of leading terms,
the fraction $\frac{c_n}{b_n}$ is increasing.
Personally, I favor polynomial long division here.
$$b_n(n - 2021) = c_n + D_1n + D_2$$
where $D_1, D_2$ are fixed constants.
Therefore, $\dfrac{c_n}{b_n}$ has the form
$$\left(n - 2021 - \frac{D_1n + D_2}{b_n}\right)\tag1$$
Clearly, there exists $N \in \Bbb{Z^+},$ such that for all $n\geq N, ~~\left|\dfrac{D_1n + D_2}{b_n}\right| < 1.$
Therefore, for $n \geq N,$ as $n \to (n+1)$, $(n - 2021)$ has increased by $1$ to $([n+1] - 2021)$, which
must overshadow, any effect of the corresponding fraction, 
$~\dfrac{D_1(n+1) + D_2}{b_{n+1}}.$

Edit
Note that the question requires that it be shown that the sequence $\langle a_n\rangle$ is decreasing, rather than showing that the corresponding function is (beyond a certain point) strictly decreasing.  This distinction allows my analysis.
A: $$a_n=\frac{\sqrt{n^{2}+2021n+420}}{\sqrt{n^{3}+2022n+420}} \implies a_n^2=\frac {n^{2}+2021n+420}{n^{3}+2022n+420 }$$
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
$$\frac{a_{n+1}^2 } {a_n^2}=1-\frac{1}{n}-\frac{2020}{n^2}+O\left(\frac{1}{n^3}\right)= 1-\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$
$$\frac{a_{n+1} } {a_n}=1-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)$$
A: Thank you for your answers, as there were no magic shortcuts, I thought i'd share my original way.
I will say though that for my case it was enough to show $a_n$ is decreasing at some point.
We define $f\left(x\right)=\sqrt{\frac{x^{2}+2021x+420}{x^{3}+2022x+420}},\;\;\forall x\ge 1$. by deriving we get:
$$f'\left(x\right)=\frac{\left(2x+2021\right)\left(x^{3}+2022x+420\right)-\left(x^{2}+2021x+420\right)\left(3x^{2}+2022\right)}{2f\left(x\right)\left(x^{3}+2022x+420\right)^{2}}$$
While its clear that both $f$ and the denominator are always positive, we check the limit of the nominator:
$$\lim_{x\to\infty}x^{4}\left[\left(2+\frac{2021}{x}\right)\left(1+\frac{2022}{x^{2}}+\frac{420}{x^{3}}\right)-\left(1+\frac{2021}{x}+\frac{420}{x^{2}}\right)\left(3+\frac{2022}{x^{2}}\right)\right]=\infty\left[2-3\right]=-\infty$$
Hence there exists $ N\in\mathbb{N}$ such that $f'(x)<0\;\;\forall x\ge N$, thus $\left(a_{n}\right)_{n=N}^{\infty}$ is decreasing.
