Prove Using Definition that the Sequence $\frac{n+1}{\sqrt{n}+2}$ Diverges to Infinity Formally, a sequence $x_n$ diverges to infinity whenever for all $M>0$ there exists $N(M)$ such that $n>N(M)$ implies $x_n>M$.
Prove formally, using the definition, that the following sequence diverges to infinity:
\begin{equation}
x_n=\frac{n+1}{\sqrt{n}+2}
\end{equation}
Proof. Let $M>0$. Take $N(M)=?$...
 A: Hint You want for all $n >N(M)$ to have:
$$\frac{n+1}{\sqrt{n}+2} >M $$
Solve this inequality for $n$.
Or easier, you can observe that
$$\frac{n+1}{\sqrt{n}+2}> \frac{n-4}{\sqrt{n}+2}=\frac{(\sqrt{n}-2)(\sqrt{n}+2)}{\sqrt{n}+2}$$
and if you make this last term greater than $M$ you are done.
A: Hints: 
$$x_n\ge \frac{n-4}{\sqrt{n}+2}=\sqrt{n}-2 > M$$
Can you proceed from here?
A: Let $M>0$ be given. We require
$$\frac{n+1}{\sqrt{n}+2} >M \ \forall n>N$$
We know that
$$\frac{n+1}{\sqrt{n}+2} > \frac{n}{\sqrt n+ \sqrt n} = \frac{\sqrt n}{2} \mbox{for } n > 4$$
So take
$$N=4M^2$$
and we have
$$\frac{n+1}{\sqrt{n}+2}>M \ \forall n>N=max\{4, 4M^2 \} $$
A: The standard thing to do is multiply top and bottom by (something like) the conjugate of the bottom. So:
$$
\frac{n+1}{\sqrt n+2}
=
\frac{(n+1)(\sqrt n - 2)}{n-4}\,.
$$
Now, as long as $n\ge9$, you get $(n+1)\big/(n-4)\le2$, while $\sqrt n-2$ gets big. That should do it for you, I think.
A: Hint:
$n=t^2 \implies \dfrac{\sqrt{n}+2}{n+1}=\dfrac{t+2}{t^2+1}=\dfrac{t}{t^2+1}+\dfrac{2}{t^2+1}$
$t\to \infty \implies \dfrac{t}{t^2+1}+\dfrac{2}{t^2+1} \to0$
A: \begin{equation}
x_n=\frac{n+1}{\sqrt{n}+2}
\end{equation}
term by term
$$\begin{align}\left[\sqrt{n+2} < \sqrt{n}+2\right]&\implies\left[\frac{n+1 \color{red}{+1-1}}{\sqrt{n+2}}>\frac{n+1}{\sqrt{n}+2}\right] \\
&\implies \sqrt{n+2}-\frac{1}{\sqrt{n+2}} > x_n \\
&\implies a_n > x_n
\end{align}$$
and 
$$\begin{align}
\left[n+1>n-4\right] &\implies \left[\frac{n+1}{\sqrt{n}+2} >\frac{n-4}{\sqrt{n}+2}\right] \\
&\implies x_n >\sqrt{n}-2 \\
&\implies x_n > b_n
\end{align}$$
by the squeeze theorem:
$\lim _{n \to \infty} x_n \to \infty$ if and only if $\lim _{n \to \infty} a_n \to \infty$ and $\lim _{n \to \infty} b_n \to \infty$ 
