Proof verification of $x_n = \sqrt[3]{n^3 + 1} - \sqrt{n^2 - 1}$ is bounded

Question

Let $n \in \mathbb N$ and: $$ x_n = \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1} $$ Prove $x_n$ is bounded sequence.

Start with $x_n$:

$$ \begin{align} x_n &= \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1} = \\ &= n \left(\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}}\right) \end{align} $$

From here:

$$ \sqrt[^3]{1 + {1\over n^3}} \gt 1 \\ \sqrt{1 - {1\over n^2}} \lt 1 $$

Therefore:

$$ \sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}} \gt 0 $$

Which means $x_n \gt 0$.

Consider the following inequality:

$$ \sqrt[^3]{n^3 + 1} \le \sqrt{n^2 + 1} \implies \\ \implies x_n < \sqrt{n^2 + 1} - \sqrt{n^2 - 1} $$ Or: $$ x_n < \frac{(\sqrt{n^2 + 1} - \sqrt{n^2 - 1})(\sqrt{n^2 + 1} + \sqrt{n^2 - 1})}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} = \\ = \frac{2}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} <2 $$

Also $x_n \gt0$ so finally:

$$ 0 < x_n <2 $$

Have i done it the right way?

Answer 1

It seems right and it is bounded, as a minor observation we can improve the upper bound

$$x_n < \frac{2}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} \le\sqrt 2$$

Answer 2

$$x_n=\frac{(n^3+1)^2-(n^2-1)^3}{\sqrt[3]{(n^3+1)^6}+\sqrt[3]{(n^3+1)^5}\sqrt{n^2-1}+...}\rightarrow0,$$ which says that $\{x_n\}$ is bounded.

Answer 3

This may be weaker than the other ways (that are nice, including OP's), but seemed like an easy way to see it; note that $n \le \sqrt[3]{n^3+1} \le n+1$, and $n-1 \le \sqrt{n^2-1} \le n$ [can be verified by cubing & squaring, respectively]

So $\sqrt[3]{n^3+1} - \sqrt{n^2-1} \ge n - n = 0$ and $\sqrt[3]{n^3+1} - \sqrt{n^2-1} \le (n+1) - (n-1) = 2$.

Answer 4

Your prove is fine but a lot more work than necessary.

As $n \ge 1$ we have

$n = \sqrt[3]{n^3} < \sqrt[3]{n^3 + 1} < \sqrt[3]{n^3 + 3n^2 + 3n + 1} = \sqrt[3]{(n+1)^3} = n+1$

and

$n = \sqrt{n^2} > \sqrt{n^2 -1 } = \sqrt{n^2 - 2 + 1} \ge \sqrt{n^2 - 2n + 1} = \sqrt{(n-1)^2} = n-1$.

So $0 = n - n < \sqrt[3]{n^3 + 1} - \sqrt{n^2 -1} < (n+1) - (n-1) = 2$.

Answer 5

$$\sqrt[3]{n^3+1}-\sqrt{n^2-1}=\sqrt[3]{n^3+1}-n-\sqrt{n^2-1}+n \\=\frac1{(n^3+1)^{2/3}+(n^3+1)^{1/3}n+n^2}+\frac1{\sqrt{n^2-1}+n}$$

and both terms are decreasing. The supremum is with $n=1$,

$$x_1=\sqrt[3]2.$$