Prove sequence $x_n = \frac{3n+5}{\sqrt{4n^2 - 1}}$ is bounded.

Question

Let $n \in \mathbb N$ and: $$x_n = \frac{3n+5}{\sqrt{4n^2 - 1}}$$ Prove $x_n$ is a bounded sequence.

How can I show that the sequence is bounded? I was thinking about the monotonicity and ways to use it. Looking at the graph the sequence is monotonically decreasing $\forall n \ge1$.

If I could prove that fact then it would become easier to show that the sequence has an upper bound at $n = 1$ and some lower bound.

To prove monotonicity I've tried dividing and subtracting $x_n$ and $x_{n+1}$ but the result of those actions is not that easy to handle for me.

I'm not allowed to use methods from calculus.

What else could I try?

Answer 1

$$x_n=\frac{3n+5}{\sqrt{4n^2-1}}$$ Since $n^2<4n^2-1$ for any $n\ge1$. Then, $$x_n<\frac{3x+5}{\sqrt{n^2}}=\frac{3n+5}{n}=3+\frac{5}{n}\le8$$ For any $n>0$. On the other hand $x_n$ is strictly positive, thus $$x_n\in]0,8[$$ hence it is bounded.

Answer 2

$$x_n = \frac{3n+5}{\sqrt{4n^2 - 1}}=\frac{n(3+5/n)}{n\sqrt{4-1/n^2}}$$

Cancel out $n$.

Now note that $$|3+5/n|\leq 8$$ and $$1\leq|\sqrt{4-1/n^2}|$$

so $$|\frac{1}{\sqrt{4-1/n^2}}|\leq 1$$

That means $|\frac{3+5/n}{\sqrt{4-1/n^2}}|\leq 8$

This is not a sharp bound but it serves the purpose of showing that sequence is bounded.

Answer 3

$$ \lim_{n\to \infty} \frac {3n+5}{\sqrt{4n^2 - 1}}$$

$$=\lim_{n\to \infty} \frac{3n+5}{n \sqrt{4 - 1/{n^2}}} = 3/2$$

A convergent sequence is bounded.