Show that the sequence $x_n=\sum\limits_{k=1}^n\frac1{\sqrt{k+1}+\sqrt{k}}$ is unbounded. Consider the sequence $\{x_n\}_{n\ge1}$ defined by $$x_n=\sum_{k=1}^n\frac{1}{\sqrt{k+1}+\sqrt{k}}, \forall n\in\mathbb{N}.$$ Is $\{x_n\}_{n\ge 1}$ bounded or unbounded. 
I solved the problem as stated in the answer posted by me. Is it possible to solve the problem in a more better and rigorous manner?
 A: Here goes the solution I found out. Let $a_k$ be defined as $$a_k=\frac{1}{\sqrt{k}+\sqrt{k+1}}, \forall k\in\mathbb{N}.$$ Then, $$x_n=\sum_{k=1}^n a_k, \forall n\in\mathbb{N}.$$ Now $$a_k=\frac{1}{\sqrt{k}+\sqrt{k+1}}=\frac{\sqrt{k+1}-\sqrt{k}}{(k+1)-k}=\sqrt{k+1}-\sqrt{k}, \forall k\in\mathbb{N}.$$
This implies that $$x_n=\sqrt{n+1}-1, \forall n\in\mathbb{N}.$$ Now intuition says that $\{x_n\}_{n\ge 1}$ is an unbounded sequence. But let us try to prove it rigorously. 
Let us assume that $\{x_n\}_{n\ge 1}$ is bounded above. This implies that, we can find $M\in\mathbb{R}$ such that $x_n\le M, \forall n\in\mathbb{N}$. Now let $\lceil M\rceil=n_1\implies M\le n_1.$ 
Now $x_{n_1^2+4n_1+3}=\sqrt{n_1^2+4n_1+3+1}-1=\sqrt{(n_1+2)^2}-1=n_1+1.$ 
This implies that $x_{n_1^2+4n_1+3}=n_1+1>n_1\ge M\implies x_{n_1^2+4n_1+3}>M.$ But we have assumed that $x_n\le M, \forall n\in\mathbb{N}.$ Contradiction. 
This implies that $\{x_n\}_{n\ge 1}$ is not bounded above, which in turn implies that $\{x_n\}_{n\ge 1}$ is not bounded.  
A: $\sqrt {k+1}+\sqrt k \leq \sqrt {2k}+\sqrt k<3\sqrt k$. Hence the given sum is at least $\sum\limits_{k=1}^{n} \frac 1 {3\sqrt k}$ Now use the fact that the series $\sum\limits_{k=1}^{n} \frac 1 {3\sqrt k}$ is divergent.
