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.