# Analysis: What does Sqrt(n^2+2n)-n converge to?Prove it.

1. Oct 3, 2011

### Chinnu

1. The problem statement, all variables and given/known data

Formulate a conjecture about the convergence or divergence of the sequence:

$\sqrt{n^{2}+2n} - n$

2. Relevant equations

Triangle Inequality, etc...

3. The attempt at a solution

Start by noticing that $(n+1)^{2} = (n^{2}+2n+1), so, (n+1) = \sqrt{n^{2}+2n+1}$

Now, $\sqrt{n^{2}+2n} - n < |\sqrt{n^{2}+2n}-(n+1)|$

Multiply the RHS by 1 in the following manner:

$|\sqrt{n^{2}+2n}-(n+1)| * \frac{\sqrt{n^{2}+2n+1}+\sqrt{n^{2}+2n}}{\sqrt{n^{2}+2n+1}+\sqrt{n^{2}+2n}}$

$= \frac{n^{2}+2n+1-n^{2}-2n}{\sqrt{n^{2}+2n}+(n+1)}$

$= \frac{1}{\sqrt{n^{2}+2n}+(n+1)}$

Now, by observation,

$\frac{1}{\sqrt{n^{2}+2n}+(n+1)} < \frac{1}{n}$

we know $\forall\epsilon>0 \exists N\in N \ni n \geq N => \frac{1}{n} < \epsilon$

I'm not sure what to do now...

Last edited: Oct 4, 2011
2. Oct 3, 2011

### Chinnu

Also, could someone tell me why the latex formating didn't work in the above post...

3. Oct 3, 2011

### SammyS

Staff Emeritus
Yes,

You terminated them with a backslash [\itex] rather than a slash [/itex] .

4. Oct 4, 2011

### Dick

And I think you want to multiply numerator and denominator by the 'conjugate' sqrt(n^2+2n)+n. Then think about the limit.

5. Oct 4, 2011

### Chinnu

lol, stupid mistake,

Thank you

6. Oct 4, 2011

### SammyS

Staff Emeritus
A mistake... NOT a stupid mistake.

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