- #1
Chinnu
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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...
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