Prove a sequence is not convergent using a formal definition My attempt: (By contradiction)
Let's assume $x_{n}$ is convergent. If $x_{n}$ is convergent, then is a cauchy sequence. Thus, there exists an N such that $$\left | x_{N+1} - x_{N} \right |<\epsilon.$$
Then, replacing:
$$\left | N-\sqrt{N} - N-1+\sqrt{N+1} \right |<
N-\sqrt{N} - N-1+\sqrt{N+1} <\epsilon ...(1)$$
but $$\sqrt{N+1}>(\sqrt{N}+1)$$ is always true.
So (1) is only posible for certain values of <\epsilon. Then $x_{n}$ is not convergent.
 A: Herein, we present three distinct ways to formally show that the sequence of interest is unbounded.

APPROACH $1$

Note that for $n> 4$, $n-\sqrt{n}>\sqrt{n}$.  
Therefore, given any number $B>0$, 
$$|n-\sqrt{n}|>B$$
whenever $n>\max(4,B^2)$.


APPROACH $2$

Alternatively, note that for $n>1$
$$\begin{align}
n-\sqrt{n}&=\frac{n^2-n}{n+\sqrt{n}}\\\\
&>\frac{n-1}{2}
\end{align}$$
Therefore, given any number $B>0$, 
$$|n-\sqrt{n}|>B$$
whenever $n>2B+1$.


APPROACH $3$

And here is one more way forward that is direct.  We proceed with Brute force and write
$$\begin{align}
|n-\sqrt{n}|&=\left(\sqrt{n}-\frac12\right)^2-\frac14\\\\
&>B\\\\
\end{align}$$
whenever $n>B+\frac12 +\sqrt{B+\frac14}$.
A: Note that for $n > 4$, we have $\sqrt{n} > 2$ which implies $n > 2\sqrt{n}$ (by multiplying both sides by $\sqrt{n}$).
Thus for $n > 4$, subtracting $\sqrt{n}$ from both sides of $n > 2\sqrt{n}$ yields
$$n - \sqrt{n} > \sqrt{n}$$
Therefore, for $n > 4$ we have $x_n > \sqrt{n}$ which shows that it is unbounded since $\sqrt{n}$ is unbounded.
A: I would say that an indirect proof would not help in solving such a problem. Try the following "two-stage estimation" method, which is frequently seen in an epsilon-argument:
If $n \geq 1$, then $n - \sqrt{n} = \sqrt{n}(\sqrt{n}-1)$; let $M >0 $. Since $n > 4$ implies $\sqrt{n}(\sqrt{n} -1) > \sqrt{n}$ (to estimate "out" the term $\sqrt{n}-1$) and $n > M^{2}$ implies $\sqrt{n} > M$, we have got
$$
n > \max \{ 4, M^{2} \} \Rightarrow n - \sqrt{n} > M.
$$
This just shows that the sequence $(n - \sqrt{n})$ grows indefinitely as $n$ exceeds every bound; hence the sequence simply diverges and it cannot converge by definition.
You can see such a type of argument frequently when it comes to problems such as proving continuity.
