Prove $\frac{n}{\sqrt{1+n^2}}$ converges as $n \to \infty $ I have been trying to prove this for the past several hours but I am stuck. 
My hypothesis is it converges to $1$.
I know a few ways to prove this:
Let $a_n=\frac{n}{\sqrt{1+n^2}}$


*

*Given $\epsilon > 0$ find an $N \in \mathbb{N}$ s.t. if $n > N$, then $|a_n-1| < \epsilon$.

*Show $a_n$ is monotonically increasing and it is bounded.

*Use comparison lemma and find another series $b_n \to b$, and show $|a_n-1| \leq C|b_n-b|$, where $C \in \mathbb{R^+}$.


For #1 I have tried to manipulate $\frac{n}{\sqrt{1+n^2}}$ and come up with something that resembles $\frac{something}{n}$ or any other power of $n$ so that I can then let $N \gt \frac{something}{\epsilon}$ or something similar. But I always end up with $n$ in the numerator.
For #2 I am having problem proving $a_n < a_{n+1}$ directly. I don't know how else to approach this part. Showing boundedness is trivial.
For #3 I tried manipulating the original inequality but it got ugly pretty fast.
Can anyone please give me some hints?
 A: 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.
A: 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.
A: 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 $N>1/\varepsilon$.
A: To show the sequence is monotonically increasing, note that
$$
 (n+1)^2(n^2+1)  - n^2[(n+1)^2 + 1]   = 2n+1 > 0
$$
from which it follows
$$
\frac{n}{\sqrt{n^2+1}} < \frac{n+1}{\sqrt{(n+1)^2+1}}.
$$
Since you can already prove boundedness, the sequence converges.
A: Hint: what happens if you analyze the following expression? $$\left( \frac{n}{\sqrt{1+n^2}} \right )^{-2}$$
A: Binomial series:
$$\lim\limits_{n\to\infty}\dfrac n {\sqrt{n^2+1}}=\lim\limits_{n\to\infty}\left(1+\dfrac 1 {n^2}\right)^{-\frac12}=\lim\limits_{n\to\infty}\left(1+O\left(\dfrac1 {n^2}\right)\right)=1.$$
