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

[Chinnu]

Homework Statement

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

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

Homework Equations

Triangle Inequality, etc...

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...

 

[Chinnu]

Homework Statement

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

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

Homework Equations

Triangle Inequality, etc...

The Attempt at a Solution

Start by noticing that [itex](n+1)^{2} = (n^{2}+2n+1), so, (n+1) = \sqrt{n^{2}+2n+1}[\itex]<br /> <br /> Now, [itex]\sqrt{n^{2}+2n} - n < |\sqrt{n^{2}+2n}-(n+1)|[\itex]<br /> <br /> Multiply the RHS by 1 in the following manner:<br /> <br /> [itex]|\sqrt{n^{2}+2n}-(n+1)| x \frac{\sqrt{n^{2}+2n+1}+\sqrt{n^{2}+2n}}{\sqrt{n^{2}+2n+1}+\sqrt{n^{2}+2n}}[\itex]<br /> <br /> [itex]= \frac{n^{2}+2n+1-n^{2}-2n}{\sqrt{n^{2}+2n}+(n+1)}[\itex]<br /> <br /> [itex]= \frac{1}{\sqrt{n^{2}+2n}+(n+1)}[\itex]<br /> <br /> Now, by observation, <br /> <br /> [itex]\frac{1}{\sqrt{n^{2}+2n}+(n+1)} < \frac{1}{n}[\itex]<br /> <br /> we know $\forall\epsilon>0 \existsN\inN \ni n \geq N => \frac{1}{n} < \epsilon[\itex]<br /> <br /> I'm not sure what to do now...$[/itex][/itex][/itex][/itex][/itex][/itex]

[itex][itex][itex][itex][itex][itex]$<br /> Also, could someone tell me why the latex formating didn't work in the above post...$[/itex][/itex][/itex][/itex][/itex][/itex]

 

[SammyS]

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

Yes,

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

 

[Dick]

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

 

[Chinnu]

Yes,

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

You may want to Edit you Original Post.

lol, stupid mistake,

Thank you

 

[SammyS]

lol, stupid mistake,

Thank you

A mistake... NOT a stupid mistake.