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 $\lim_{n  \rightarrow \infty}x_n = 0$

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 $\lim_{n \to \infty} x_n = \dfrac12$.
