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".


  • 5
    $\begingroup$ If $a,b,c,d\geq$ then the mediant $m=\frac{a+c}{b+d}$ is between $\frac{a}{b}$ and $\frac{c}{d}$. So, if $m\leq \frac{c}{d}$ then $\frac{a}{b}\leq \frac{c}{d}$. This gives you a method that involves less complicated computations. Let $P(n)$ be the numerator inside the root and $Q(n)$ the denominator inside the root. Then compute $\frac{\Delta P(n)}{\Delta Q(n)}$, where $\Delta P(n)=P(n+1)-P(n)$ and similarly for $\Delta Q(n)$. Note that $\frac{P(n+1)}{Q(n+1)}$ is the mediant of $\frac{P(n)}{Q(n)}$ and $\frac{\Delta P(n)}{\Delta Q(n)}$. $\endgroup$
    – plop
    May 25, 2021 at 17:19
  • $\begingroup$ @plop Thank you for the interesting, unusual way of thought, highly appreciated. $\endgroup$
    – GuyPago
    May 25, 2021 at 17:32

5 Answers 5


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.

  • $\begingroup$ That was the "messy" part I was taking about, I didn't manage to get to this form as we can't use calculators, but seeing it as is, it's clear. Thanks $\endgroup$
    – GuyPago
    May 25, 2021 at 19:02

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,
$~\dfrac{D_1(n+1) + D_2}{b_{n+1}}.$

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.

  • $\begingroup$ This will take me a while to understand, I will definitely look deeper into this as its an interesting way of thought. Thank you for sharing! $\endgroup$
    – GuyPago
    May 25, 2021 at 18:59

$$a_n=\frac{\sqrt{n^{2}+2021n+420}}{\sqrt{n^{3}+2022n+420}} \implies a_n^2=\frac {n^{2}+2021n+420}{n^{3}+2022n+420 }$$

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 $$\frac{a_{n+1}^2 } {a_n^2}=1-\frac{1}{n}-\frac{2020}{n^2}+O\left(\frac{1}{n^3}\right)= 1-\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$ $$\frac{a_{n+1} } {a_n}=1-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)$$


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:


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.


"I tried showing an+1≤an but it was too messy. "

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

  • $\begingroup$ Why is it enough to show? How can I say that the last inequality holds? thank you for answering. $\endgroup$
    – GuyPago
    May 25, 2021 at 18:54
  • $\begingroup$ I'm sorry I read "enough to show" rather than "easy enough to show", I literally thought the last inequality was the final step, my apologies. But if that's not a mess, what is? :) I appreciate your answer, it has been helpful. $\endgroup$
    – GuyPago
    May 25, 2021 at 19:12
  • 2
    $\begingroup$ It's hardly a messy. It expands out and the the corresponding terms of the cube clearly outweigh the terms of the square. $\endgroup$
    – fleablood
    May 25, 2021 at 19:31

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