Convergence of Sequence

• Oct 4th 2010, 04:44 PM
RU2010
Convergence of Sequence
I need help figuring out how the convergence of a sequence.

I have the answer, but I'm not understanding how it is calculated.

It is the sequence,

an=$\displaystyle \sqrt{n^2+2n}-n$

I have the answer that the limit as n goes to infinity is 1. How would I calculate this limit for the sequence?
• Oct 4th 2010, 04:57 PM
Plato
$\displaystyle \sqrt {n^2 + 2n} - n = \dfrac{{2n}}{{\sqrt {n^2 + 2n} + n}}$
• Oct 4th 2010, 05:07 PM
mr fantastic
Quote:

Originally Posted by Plato
$\displaystyle \sqrt {n^2 + 2n} - n = \dfrac{{2n}}{{\sqrt {n^2 + 2n} + n}}$

I just want to add that Plato probably got this result by using the standard technique of multiplying the original expression by $\displaystyle \displaystyle \frac{\sqrt{n^2 + 2n} + n}{\sqrt{n^2 + 2n} + n}$.
• Oct 4th 2010, 08:37 PM
RU2010
I still don't understand where to go from here. I've tried using L'Hopital's rule here, but that doesn't seem to work either. Can you show how the limit as n goes to infinity is 1?
• Oct 4th 2010, 09:10 PM
mr fantastic
Quote:

Originally Posted by RU2010
I still don't understand where to go from here. I've tried using L'Hopital's rule here, but that doesn't seem to work either. Can you show how the limit as n goes to infinity is 1?

Divide by n the numerator and denominator of the expression in post #2 (and which you ought to derive using the suggestion in post #3). Note that 1/n --> 0 etc.
• Oct 5th 2010, 01:18 PM
RU2010
Ok, I have divided the top and bottom of that limit by n, and I get this....

$\displaystyle \lim _{n\rightarrow \infty }2\, \left( {\frac {\sqrt {{n}^{2}+2\,n}}{n }}+1 \right) ^{-1}$

How do I simplify the radical part on the bottom? I should end up with 2/2 so that the entire limit is 1. But I'm stuck on simplifying the radical.

Thanks!
• Oct 5th 2010, 01:26 PM
Plato
$\displaystyle \dfrac{{2n}}{{\sqrt {n^2 + 2n} + n}} = \dfrac{2}{{\sqrt {1 + \frac{2}{n}~} + 1}}$
• Oct 5th 2010, 01:32 PM
RU2010
Got it Thanks!!!!
• Oct 5th 2010, 01:35 PM
Plato
I corrected that typo.

That is simple first year algebra. Divide through by $\displaystyle n$.