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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.

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    @AntonioVargas : Your comment is better than the answers posted below. Why not make it an answer?2012-09-19

2 Answers 2

7

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

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

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    You're right, it wasn't as bad as I feared. Thanks.2012-09-19
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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 $n \le n + \sqrt{n}$.

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    I know but the OP does not want to use the limit argument.2012-09-19