(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

[itex]\sqrt{n^{2}+2n} - n[/itex]

2. Relevant equations

Triangle Inequality, etc...

3. 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]

Now, [itex]\sqrt{n^{2}+2n} - n < |\sqrt{n^{2}+2n}-(n+1)|[/itex]

Multiply the RHS by 1 in the following manner:

[itex]|\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}}[/itex]

[itex]= \frac{n^{2}+2n+1-n^{2}-2n}{\sqrt{n^{2}+2n}+(n+1)}[/itex]

[itex]= \frac{1}{\sqrt{n^{2}+2n}+(n+1)}[/itex]

Now, by observation,

[itex]\frac{1}{\sqrt{n^{2}+2n}+(n+1)} < \frac{1}{n}[/itex]

we know [itex]\forall\epsilon>0 \exists N\in N \ni n \geq N => \frac{1}{n} < \epsilon[/itex]

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

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

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