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The Question:

Find the supremum of the set $${\{\sqrt[4]{n^4+n^3}-n:n\in \mathbb{N}\}}$$ And then it tells us to plug large values of n to determine a suitable guess, show that is an upper bound and then prove it is the smallest upper bound.

I followed the question, finding a suitable guess for s is 1/4, and showed that this is an upper bound just fine. My issue lies with proving that there is no smaller upper bound. At this point, my working looks like this as I try to prove by contradiction:

Assume h is some other upper bound, such that h < 1/4.
$${\sqrt[4]{n^4+n^3}-n < h}$$ $$n^4+n^3 < (h+n)^4$$

But after expansion, all I can cancel is $n^4$ which leaves me with a lot of unknowns to various powers and a really complicated solution to do by hand

$$n^3 < h^4 + 4h^3n + 6h^2n^2 + 4hn^3$$

Which means I know I've gone down the wrong route but I'm not sure which way I should go about proving this. I adapted an answer from a different book example, but that only went up to power 2, so simplifying this way was much easier.

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    $\begingroup$ Please don't vandalize your question text at any time, but especially after you have already received several answers. $\endgroup$ Nov 28, 2021 at 2:18

4 Answers 4

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As yourself said we have: $$n^3 < h^4 + 4h^3n + 6h^2n^2 + 4hn^3 \Longrightarrow 0 < h^4 + 4h^3n + 6h^2n^2 + (4h-1)n^3$$ But we know for enough large $n$ the sign of polynomial is agree with its leader term, so: $$0<(4h-1)n^3 \overset{n>0}{=\Longrightarrow} 0<4h-1 \Longrightarrow \frac{1}{4}<h$$ It's that contradiction you look it up.

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  • $\begingroup$ Thank you, this is exactly what I was looking for. Knew I was over-complicating it in my head. $\endgroup$ Nov 2, 2019 at 15:15
  • $\begingroup$ @Ali Ashja':Very nice +1 $\endgroup$
    – Khosrotash
    Jan 27, 2021 at 8:15
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$\sqrt [4] {n^{4}+n^{3}} -n=n[(1+\frac 1n)^{1/4}-1]=n(1+\frac 1 {4n}+o(\frac 1 {n})-1)$ so the limit is $\frac 1 4$.

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In order to prove that $\require{cancel}\sup\left\{\sqrt[4]{n^4+n^3}-n\,\middle|\,n\in\mathbb N\right\}=\frac14$, it is enough, since $\frac14$ is an upper bound of $\left\{\sqrt[4]{n^4+n^3}-n\,\middle|\,n\in\mathbb N\right\}$, to prove that$$\lim_{n\to\infty}\sqrt[4]{n^4+n^3}-n=\frac14,$$which follows from the fact that\begin{align}\sqrt[4]{n^4+n^3}-n&=\sqrt[4]{n^4+n^3}-\sqrt[4]{n^4}\\&=\frac{\cancel{n^4}+n^3-\cancel{n^4}}{\sqrt[4]{n^4+n^3}^3+\sqrt[4]{n^4+n^3}^2\times n+\sqrt[4]{n^4+n^3}\times n^2+n^3}\end{align}

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Let $$ f(n) = \sqrt[4]{n^4+n^3}-n = n ((1+\frac{1}{n})^{1/4}-1) $$ Now by Bernoulli's inequality, $(1+\frac{1}{n})^{1/4} \le 1+\frac{1}{4n}$, so $$ f(n) \le n ((1+\frac{1}{4n})-1) = \frac{1}{4} $$ Now we need to show that this is the smallest upper bound. To do so, let $x = \frac{1}{n}$ and re-consider Bernoulli's inequality, $(1+x)^{1/4} \le 1+\frac{x}{4}$ or $(1+x) \le (1+\frac{x}{4})^4 = 1 + x + (3 x^2)/8 + x^3/16 + x^4/256$ or $0 \le (3 x^2)/8 + x^3/16 + x^4/256$, and the equality limit actually is attained for $x \to 0$ which is $n \to \infty$. So there is no smaller limit and the supremum is $1/4$.

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