Convergence of the sequence $ \sqrt {n-2\sqrt n} - \sqrt n $

Question

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?

Answer 1

There's a small mistake. You seem to think that to study the convergence of the sequence$$\left(\frac{\sqrt{n-2\sqrt n}}{\sqrt n}\right)_{n\in\mathbb N}$$is equivalent to the convergence of its square. It is true in this case because it is a sequence of real numbers greater than $0$; but it is false in general.

Answer 2

It's perfect except for the justification that $2\sqrt {n}/n $ converges to $0 $. What you have written only proves that it is bounded above by $1 $.

Answer 3

It is true, but formally I think it would be more elegant to say

$\frac{\sqrt{n-2\sqrt{n}}}{\sqrt{n}}=\sqrt{\frac{n-2\sqrt{n}}{n}}=\sqrt{1-\frac{2}{\sqrt{n}}}\longrightarrow1$

Answer 4

$$ \sqrt{n-2\sqrt{n}}-\sqrt{n}=\sqrt{n\left(1-\frac{2}{\sqrt{n}}\right)}-\sqrt{n}=\sqrt{n}\left(\sqrt{1-\frac{2}{\sqrt{n}}}-1\right) $$ and $$ \sqrt{1-\frac{2}{\sqrt{n}}}\underset{(+\infty)}{=}1-\frac{1}{\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right) $$ Finally $$ \sqrt{n-2\sqrt{n}}-\sqrt{n}\underset{(+\infty)}{=}1+o\left(1\right) $$ Meaning that

$$ \sqrt{n-2\sqrt{n}}-\sqrt{n}\underset{n \rightarrow +\infty}{\rightarrow}1 $$

Answer 5

It can be much shorter with asymptotic analysis: \begin{align} \sqrt {n-2\sqrt n\,} - \sqrt n&=\sqrt n\biggl(\sqrt{1-\frac2{\smash[b]{\sqrt n}\,}}-1\biggr)=\sqrt n\biggl(1-\frac1{\sqrt n}+o\biggl(\frac1{\sqrt n}\biggr)-1\biggr)\\[1ex] &=-1+o(1) \to -1. \end{align}