Is $x_n =\sqrt{n} (\sqrt{n+1}− \sqrt{n})$ , $n \ge 1$ convergent? $$x_n =\sqrt{n} (\sqrt{n+1}− \sqrt{n}) ,\quad n \ge 1$$
My work is:
First i analyzed the convergence of this sequence in 2 parts:
a) $\lim \sqrt{n+1}− \sqrt{n} = \lim \frac{(\sqrt{n+1}− \sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}} = \lim \frac{n+1-n}{\sqrt{n+1}+\sqrt{n}} = \lim \frac{1}{ \infty} = 0$ when $n  \rightarrow \infty$.
But now how can I conclude that $\lim_{n  \rightarrow \infty}x_n = 0$
 A: A basic trick you should learn is that:
$$\sqrt{n+1}- \sqrt{n} = \frac{(\sqrt{n+1}- \sqrt{n}) \cdot (\sqrt{n+1}+ \sqrt{n})}{1 \cdot (\sqrt{n+1}+ \sqrt{n})} = \frac{1}{\sqrt{n+1} + \sqrt{n}} $$
Using that yields:
$$x_n=\frac{\sqrt n}{\sqrt{n+1}+\sqrt n}=\frac{\sqrt n}{\sqrt n\left(\sqrt{1+\frac{1}{n}}+1\right)} = \frac{1}{\sqrt{1+\frac{1}{n}} + 1}$$
Hence:
$$\lim_{x\to\infty} x_n = \frac{1}{2}$$
A: Simply
$$x_n=\sqrt n(\sqrt{n+1}-\sqrt n)=\frac{\sqrt n}{\sqrt{n+1}+\sqrt n}=\frac{\sqrt n}{\sqrt n\left(\sqrt{1+\frac{1}{n}}+1\right)}$$
now cancel and pass to the limit.
A: Is it correct this?
$\begin{align}
x_n
&=\sqrt{n} (\sqrt{n+1}− \sqrt{n})\\
&=\sqrt{n} (\sqrt{n+1}− \sqrt{n})\dfrac{\sqrt{n+1}+ \sqrt{n}}{\sqrt{n+1}+ \sqrt{n}}\\
&=\sqrt{n} \dfrac{(\sqrt{n+1}+ \sqrt{n})(\sqrt{n+1}− \sqrt{n})}{\sqrt{n+1}+ \sqrt{n}}\\
&=\sqrt{n} \dfrac{(n+1)-n}{\sqrt{n+1}+ \sqrt{n}}\\
&= \dfrac{\sqrt{n}}{\sqrt{n+1}+ \sqrt{n}}\\
\end{align}
$
so
$x_n
> \dfrac{\sqrt{n}}{2\sqrt{n+1}}
$
and
$x_n
< \dfrac{\sqrt{n}}{2\sqrt{n}}
=\dfrac12
$.
Doing the same thing to bound
the difference between these two bounds,
$\begin{align}
\dfrac12-\dfrac{\sqrt{n}}{2\sqrt{n+1}}
&=\dfrac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n+1}}\\
&=\dfrac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n+1}}\dfrac{\sqrt{n+1}+ \sqrt{n}}{\sqrt{n+1}+ \sqrt{n}}\\
&=\dfrac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+ \sqrt{n})}{2\sqrt{n+1}(\sqrt{n+1}+ \sqrt{n})}\\
&=\dfrac{1}{2\sqrt{n+1}(\sqrt{n+1}+ \sqrt{n})}\\
&<\dfrac{1}{4n}\\
\end{align}
$
so
$\dfrac12 -\dfrac1{{4n}}
< x_n < \dfrac12$
so $\lim_{n \to \infty} x_n = \dfrac12$.
A: This is a little sloppy, but I'm not sure how to clean it up at the moment, however its different so I thought of sharing.
$$x_{n+m} - x_n = \sqrt{n+m}\sqrt{n+m+1} - \sqrt{n}\sqrt{n+1} - m ≈(n+m)-n-m=0$$
Because for large enough $n$ there is little difference between $\sqrt{n}$ and $\sqrt{n+1}$ .  And every Cauchy Sequence is convergent.
A: What can you say about the limit of $\sqrt{n}/(\sqrt{n+1} + \sqrt{n})?$
A: Just to try something different:
$$\sqrt n(\sqrt{n+1}-\sqrt n) =
\sqrt n \sqrt n(\sqrt{1+\frac1n}-\sqrt 1)=
n(\sqrt{1+\frac1n}-1)=
\frac{\sqrt{1+\frac1n}-1}{\frac1n}$$
If we look at the derivative of $f(x)=\sqrt{x}$ at the point $1$ we see that
$$f'(1)=\lim_{t\to 0} \frac{f(1+t)-f(1)}t = \lim_{t\to 0} \frac{\sqrt{1+t}-1}t.$$
But we know that $f'(x)=(\sqrt{x})'=\frac1{2\sqrt{x}}$, hence $f'(1)=\frac12$
and 
$$\lim_{t\to 0} \frac{\sqrt{1+t}-1}t = \frac12.$$
Now if $n\to\infty$ then $\frac1n\to0^+$, so the above equation yields
$$\lim_{n\to\infty} \frac{\sqrt{1+\frac1n}-1}{\frac1n} = \frac12.$$

In fact this answer is "bad" in the sense that when you are being taught about limits of sequences you probably have not learned about derivatives yet. And even if you learned about derivatives, this is probably the limit that was used to derive the derivative of $\sqrt x$. 
But I thought that it is at least interesting to mention the connection between this question and derivatives. (Other useful answers have been given already. And you might be reminded of this later, when you learn about derivatives.) 
