# 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.

#### Solutions Collecting From Web of "Proving the convergence of $a_n = \frac{n}{n+\sqrt n}$"

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

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

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