Proving the convergence of $a_n = \frac{n}{n+\sqrt n}$

Question

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.

Answer 1

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 $n \le n + \sqrt{n}$.

Answer 2

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

$|a_m-a_n| = a_m-a_n < \frac{1}{\sqrt{n}} \leq \frac{1}{\sqrt{N}} \leq \varepsilon$.

Answer 3

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 $\lim\left(\dfrac{n}{n+\sqrt{n}}\right)= 1$