Prove $\frac{n}{\sqrt{1+n^2}}$ converges as $n \to \infty $

You have

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

RHS inequality is clear and LHS follows from

$$\sqrt{1+n^2} \le 1+n.$$

You can then apply squeeze theorem.

Observe that we can rewrite the given expression as \begin{align*} \frac{n}{\sqrt{1+n^{2}}} = \frac{1}{\sqrt{1 + \displaystyle\frac{1}{n^{2}}}} \end{align*}

Since the square root is a continuous function and the argument tends to one, it results the given limit converges to the same value as well. More precisely, we have \begin{align*} \lim_{n\rightarrow\infty}\frac{1}{\sqrt{1+\displaystyle\frac{1}{n^{2}}}} = \frac{1}{\sqrt{1 + \displaystyle\lim_{n\rightarrow\infty}\frac{1}{n^{2}}}} = \frac{1}{\sqrt{1 + 0 }} = 1 \end{align*}

As to its convergence, we may argument as follows. Consider the series \begin{align*} \sum_{n=1}^{\infty}\frac{1}{n^{2}} = \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \ldots \end{align*}

According to the integral test, such series converges. Therefore its general term converges to zero, from whence the result follows.

Let $n>N\in\Bbb N,\varepsilon>0$. Then$$|1-a_n|=\frac{\sqrt{1+n^2}-n}{\sqrt{1+n^2}}\\=\frac1{(n+\sqrt{1+n^2})\sqrt{1+n^2}}\\<\frac1{\sqrt{1+n^2}}~~~(*)<\frac1n<\frac1N$$$(*)$ since $\sqrt{1+n^2}+n>1$.

Now take $N>1/\varepsilon$.