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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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Of course limit is 1. I was just trying to show that it is convergent without using the limit value, like ratio test or showing that is is a Cauchy Sequence. But how do I do it? –  Euclidean Sep 19 '12 at 13:18
11  
Why not rearrange? $$ \frac{n}{n+\sqrt{n}} = \frac{1}{1+\frac{1}{\sqrt{n}}}. $$ –  Antonio Vargas Sep 19 '12 at 13:29
    
Your edit doesn't work. You have to show it's bounded before you can apply the least upper bound property. –  anonymous Sep 19 '12 at 14:44
    
@AntonioVargas : Your comment is better than the answers posted below. Why not make it an answer? –  Michael Hardy Sep 19 '12 at 15:40
    
@anonymous is it correct now? –  Euclidean Sep 20 '12 at 13:22

2 Answers 2

up vote 7 down vote accepted

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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4  
That seems like it will be a pain, since it will involve computations of $$\frac m{m+\sqrt m} - \frac n{n+\sqrt n}$$ which look messy to me, with not much hope of simplification. –  MJD Sep 19 '12 at 13:36
    
@Ales how will you show that the sequence is a Cauchy sequence. –  Euclidean Sep 19 '12 at 14:45
3  
@MJD I added the proof. It is not that complicated. –  Aleš Bizjak Sep 19 '12 at 16:58
    
You're right, it wasn't as bad as I feared. Thanks. –  MJD Sep 19 '12 at 17:58

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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Hmm, the least upper bound property for bounded sequence. Thanks :) –  Euclidean Sep 19 '12 at 13:39
    
"Show that it is increasing and bounded" seems more complicated than what is needed here. In the expresion $\dfrac{1}{1+\frac{1}{\sqrt{n}}}$, the fact that $1/\sqrt{n}\to0$ as $x\to\infty$, and that $w\mapsto1/(1+w)$ is continuous, shows that the original sequence converges. –  Michael Hardy Sep 19 '12 at 16:17
    
I know but the OP does not want to use the limit argument. –  vanna Sep 19 '12 at 16:29

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