Why the sequence $\frac{2+a_n}{\sqrt{2+a_n^2}}$ is bounded

Question

Given $\left\{ a_{n}\::\:n=1,\:2,\:3,\:\cdots\right\} $ is an infinite sequence in $\mathbb{R}$, and every term is positive. How to prove that the set

\begin{equation} \left\{ \frac{2+a_{n}}{\sqrt{2+a_{n}^{2}}}\::\:n=1,\:2,\:3,\:\cdots\right\} \end{equation} has a limit point?

Bolzano-Weierstrass theorem says that every bounded infinite subset of $\mathbb{R}$ has a limit point. But how to prove it is bounded? I tried this \begin{equation} \left|\frac{2+a_{n}}{\sqrt{2+a_{n}^{2}}}\right|\leq\left|\frac{2+a_{n}}{\sqrt{a_{n}^{2}}}\right|=\left|\frac{2+a_{n}}{a_{n}}\right| \end{equation} but does not seem work.

Or these is another way to show it has a limit point, without using Bolzano-Weierstrass theorem

Answer 1

Hint: You can let intuition guide you: When $a_n$ is very small, it is negligible compared to $2$, so the fraction is close to $\sqrt{2}$. Similarly, when $a_n$ is very large, the fraction is very close to $a_n/\sqrt{a_n^2}=1$. In between these extremes, it is continuous, hence bounded. You could build that into a formal proof, but that is clearly overkill! But this quick reasoning gives you the answer very easily, it is for sure bounded.

Instead, note that you need to prove the existence of some $M$ so that $$ 2+a_n \le M\sqrt{2+a_n^2} .$$ Now square that inequality (note that both sides are positive), rearrange it a bit, say by collecting equal powers of $a_n$, and try find some value of $M$ that lets you prove it.

Answer 2

Alt. hint:   let $a = \sqrt{2} \tan t$, then:

$$ \frac{2+a}{\sqrt{2+a^2}} = \frac{\sqrt{2}}{\sqrt{2}}\cdot \frac{\sqrt{2} + \tan t}{\sqrt{1+\tan^2 t}} = (\sqrt{2}+\tan t) \cdot |\cos t| \;\le\; \sqrt{2}|\cos t| + |\sin t| $$