Proving the convergence of $a_n = \frac{n}{n+\sqrt n}$ How do I show that the sequence $\{x_n\}$ where $x_n=\frac{n}{n+\sqrt n}$ is convergent, without using the value to which $\{x_n\}$ converges?
EDIT
Ok, so this is how I proceeded to prove this convergence based on vanna's answer. Please tell me if I am wrong anywhere.
First I show that the sequence is increasing 
$$x_n -x_m = \frac{\sqrt n -\sqrt m}{\sqrt{mn} +\sqrt m + \sqrt n + 1} \gt 0 \qquad\forall n\gt m$$
I will now show that it is bounded
$$n\lt n+\sqrt n \implies x_n \lt 1 \quad\forall n \in \mathbb{N} $$
By the Least Upper Bound Property there exists a Least Upper Bound $L$ for the sequence $\{x_n\}$ since is it bounded. 
Since L is the least upper bound,  
$$\exists x_k \in \{x_n\} \mid x_k \in (L-\epsilon, L]  \; \;\forall \epsilon \gt 0\;$$
or else there will exist another number 
$$L-\frac{\epsilon }{2}$$ 
such that 
$$\forall k \in \mathbb{N} \quad x_k\le L-\frac {\epsilon }{2}$$
but this is contradiction as $L$ is the least upper bound.
Hence 
$$\quad \forall \epsilon \gt 0 \quad\exists k \in \mathbb{N}\; \; \mid  x_k \in (L-\epsilon, L]$$
Now since $\{ x_n\} $ is monotonically increasing sequence and bounded by $L$ ,
$$\;\forall i \ge k \;,x_i \in (L-\epsilon, L]$$ 
Hence 
$$\forall \epsilon\ge 0, \; \exists k\in \mathbb{N} \; \mid \; |x_i-L| \lt \epsilon$$ 
And thus we proved that the series converges and it converges to it's Least Upper Bound. 
RE-EDIT
Is there some way I can prove it is convergent by first proving it is Cauchy sequence.
 A: If you really don't want to show that $1$ is the limit, but just that it exists, then you can show that it is a Cauchy sequence.
Here it goes. Assume $m,n\in\mathbb{N}$ and WLOG $m \geq n$. Then 
$a_m-a_n = \frac{m}{m+\sqrt{m}}-\frac{n}{n+\sqrt{n}} = \frac{n m + m\sqrt{n} - m n - n \sqrt{m}}{(m+\sqrt{m})(n+\sqrt{n})}=\frac{\sqrt{m n}(\sqrt{m}-\sqrt{n})}{(m+\sqrt{m})(n+\sqrt{n})} = \frac{\sqrt{m n}(\sqrt{m}-\sqrt{n})}{\sqrt{mn}(\sqrt{m}+1)(\sqrt{n}+1)}=\frac{\sqrt{m}-\sqrt{n}}{(\sqrt{m}+1)(\sqrt{n}+1)}$
Thus $0 \leq a_m - a_n \leq \frac{\sqrt{m}}{\sqrt{m}+1} \frac{1}{\sqrt{n}+1}\leq \frac{1}{\sqrt{n}+1} < \frac{1}{\sqrt{n}}$.
Suppose now you are given an $\varepsilon > 0$. Let $N=\lceil \frac{1}{\varepsilon^2}\rceil$. If $n, m \geq N$ and WLOG $m \geq n$, then
$|a_m-a_n| = a_m-a_n < \frac{1}{\sqrt{n}} \leq \frac{1}{\sqrt{N}} \leq \varepsilon$.
A: Show that it is increasing and bounded.
To show that it is increasing, study the function
$$x \mapsto \frac{x}{x+\sqrt{x}} = \frac{1}{1+\frac{1}{\sqrt{x}}}$$
Boundedness is trivial since $n \le n + \sqrt{n}$.
A: I'm using the definition of the limit of the sequence to prove it. It's much easy to suspect that the limit will be $1$ (you can also check it by algebraic properties of limits). Now $n+\sqrt{n}> n\Rightarrow \dfrac{n}{n+\sqrt{n}}< 1$. Thus $1-\dfrac{n}{n+\sqrt{n}}= \left|\dfrac{n}{n+\sqrt{n}}-1\right|= \dfrac{\sqrt{n}}{n+\sqrt{n}}= \dfrac{1}{1+\sqrt{n}}<\dfrac{1}{\sqrt{n}}$.
Now for any given $\varepsilon> 0$, by Archimedean property $\exists k\in\mathbb{N}$ such that $\dfrac{1}{k}< \varepsilon^2$. Thus $\forall n\geq k\Rightarrow \dfrac{1}{n}\leq \dfrac{1}{k}< \varepsilon^2\Rightarrow \dfrac{1}{\sqrt{n}}\leq \dfrac{1}{\sqrt{k}}<\varepsilon$.
Hence $\left|\dfrac{n}{n+\sqrt{n}}-1\right|<\varepsilon\hspace{10pt}\forall n\geq k$
Since $\varepsilon> 0$ is choosen arbitrarily, we conclude $\lim\left(\dfrac{n}{n+\sqrt{n}}\right)= 1$
