Convergence of the sequence $ \sqrt {n-2\sqrt n} - \sqrt n $ Here's my attempt at proving it:
Given the sequence $$ a_n =\left( \sqrt {n-2\sqrt n} - \sqrt n\right)_{n\geq1} $$
To get rid of the square root in the numerator:
\begin{align}
\frac {\sqrt {n-2\sqrt n} - \sqrt n} 1 \cdot \frac {\sqrt {n-2\sqrt n} + \sqrt n}{\sqrt {n-2\sqrt n} + \sqrt n} &= \frac { {n-2\sqrt n} - \ n}{\sqrt {n-2\sqrt n} + \sqrt n} = \frac { {-2\sqrt n}}{\sqrt {n-2\sqrt n} + \sqrt n} \\&= \frac { {-2}}{\frac {\sqrt {n-2\sqrt n}} {\sqrt n} + 1}
\end{align}
By using the limit laws it should converge against:
$$
\frac { \lim_{x \to \infty} -2 } { \lim_{x \to \infty} \frac {\sqrt {n-2\sqrt n}}{\sqrt n}  ~~+~~\lim_{x \to \infty} 1}
$$
So now we have to figure out what $\frac {\sqrt {n-2\sqrt n}}{\sqrt n}$ converges against:
$$
\frac {\sqrt {n-2\sqrt n}}{\sqrt n} \leftrightarrow \frac { {n-2\sqrt n}}{ n} = \frac {1-\frac{2\sqrt n}{n}}{1}
$$
${\frac{2\sqrt n}{n}}$ converges to $0$ since:
$$
2\sqrt n = \sqrt n + \sqrt n \leq \sqrt n ~\cdot ~  \sqrt n = n
$$
Therefore $~\lim_{n\to \infty} a_n = -1$
Is this correct and sufficient enough? 

It is true, but formally I think it would be more elegant to say
$\frac{\sqrt{n-2\sqrt{n}}}{\sqrt{n}}=\sqrt{\frac{n-2\sqrt{n}}{n}}=\sqrt{1-\frac{2}{\sqrt{n}}}\longrightarrow1$
