1. 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= $\sqrt{n^2+2n}-n$

I have the answer that the limit as n goes to infinity is 1. How would I calculate this limit for the sequence?

2. $\sqrt {n^2 + 2n} - n = \dfrac{{2n}}{{\sqrt {n^2 + 2n} + n}}$

3. Originally Posted by Plato
$\sqrt {n^2 + 2n} - n = \dfrac{{2n}}{{\sqrt {n^2 + 2n} + n}}$
I just want to add that Plato probably got this result by using the standard technique of multiplying the original expression by $\displaystyle \frac{\sqrt{n^2 + 2n} + n}{\sqrt{n^2 + 2n} + n}$.

4. 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?

5. 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.

6. Ok, I have divided the top and bottom of that limit by n, and I get this....

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

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!

7. $\dfrac{{2n}}{{\sqrt {n^2 + 2n} + n}} = \dfrac{2}{{\sqrt {1 + \frac{2}{n}~} + 1}}$

8. Got it Thanks!!!!

9. I corrected that typo.

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