Proof verification hat $x_n = \{ \sqrt{n^4 + n^3 +1} - \sqrt{n^4 - n^3 + 1} \}$ is an unbounded sequence

Question

This is the first time i'm proving a sequence is unbounded and have some troubles choosing the right approach.

Let $n \in \mathbb N$ and: $$ x_n = \{ \sqrt{n^4 + n^3 +1} - \sqrt{n^4 - n^3 + 1} \} $$ Prove $x_n$ is unbounded

Start with $x_n$:

$$ \begin{align} x_n &= \sqrt{n^4 + n^3 +1} - \sqrt{n^4 - n^3 + 1} = \\ &= \frac{(\sqrt{n^4 + n^3 +1} - \sqrt{n^4 - n^3 + 1})(\sqrt{n^4 + n^3 +1} + \sqrt{n^4 - n^3 + 1})}{\sqrt{n^4 + n^3 +1} + \sqrt{n^4 - n^3 + 1}} = \\ &=\frac{(\sqrt{n^4 + n^3 +1})^2 - (\sqrt{n^4 - n^3 + 1})^2}{\sqrt{n^4 + n^3 +1} + \sqrt{n^4 - n^3 + 1}} \\ &= \frac{2n^3}{\sqrt{n^4 + n^3 +1} + \sqrt{n^4 - n^3 + 1}} = \\ &= \frac{2n}{\sqrt{1 + {1\over n} + {1\over n^4}} + \sqrt{1 - {1\over n} + {1\over n^4}}} \end{align} $$

Now consider the denominator:

$$ d_1 =\sqrt{1 + {1\over n} + {1\over n^4}} \\ d_2 = \sqrt{1 - {1\over n} + {1\over n^4}} $$

$d_1$ is bounded with: $$ 1 < d_1 < \sqrt3 $$

But $d_2$ is also bounded: $$ 0 < d_2 \le 1 $$

Therefore their sum is bounded:

$$ 1 < d_1 + d_2 \le 1 + \sqrt3 $$

So finally we have $2n$ is unbounded and $d_1 + d_2$ is bounded. Hence:

$$ 2n \over d_1 + d_2 $$

is unbounded.

I'm wondering how is even unboundness proven? Should I use some constant say $M$ and prove that there exist $N > n$ such that $x_N > M$? If so then how could I do it for the given problem. Also i'm interested whether my approach is valid at least in some way because it feels like it's not.

Please note that this is a precalculus question. I know this could be easily solved by showing that the limit does not exist, but I'm not allowed to use calculus.

Answer 1

Your approach is entirely correct; by writing $x_n$ as a fraction and showing that the denominator is bounded whereas the numerator is unbounded, you can indeed conclude that the sequence is unbounded.

In general, to show that a sequence is unbounded it suffices to show that for every constant $M$ there exists an integer $N$, depending on $M$, such that $x_N>M$. In this case, to find an $N$ such that $$x_N=\sqrt{N^4+N^3+1}-\sqrt{N^4-N^3-1}=\frac{2N}{d_1+d_2}>M.$$ You already noted that $d_1+d_2<1+\sqrt{3}$ so you can take any $N$ satisfying $$N>\frac{1+\sqrt{3}}{2}M.$$

Answer 2

Your approach is good so far and in your last paragraph you nearly stated what is left to prove:

The thing is that for every $M\in\mathbb{R}$ you need to find an index $N\in\mathbb{N}$ such that $x_n\geq M$ for all $n\geq N$.

Use your lower bound of $x_n$ to find such an index for any given $M$.

Answer 3

Your approach is absolutely valid.

You have proved that $$x_n=\frac{2n}{d_1+d_2} \ge \frac{2n}{1+\sqrt{3}} \gt \frac{n}{2} $$

So for any given $M \gt 0$, you can choose $N = 2M$, s.t. $x_N \gt M$.

Answer 4

Alternative approach: assume $\alpha>\beta>0$ and consider that $$ \sqrt{\alpha}-\sqrt{\beta} = \int_{\beta}^{\alpha}\frac{dx}{2\sqrt{x}}=\frac{\alpha-\beta}{2\sqrt{\gamma}},\qquad \gamma\in(\beta,\alpha). $$ If $\alpha=n^4+n^3+1$ and $\beta=n^4-n^3+1$ we get $$\sqrt{n^4+n^3+1}-\sqrt{n^4-n^3+1} = \frac{n^3}{\sqrt{n^4+O(n^3)}} = n+O(1) $$ hence the given sequence is clearly divergent.