# How do you determine whether the sequence a_n=sqrt(n^2+n)-n converges, if so how do you find the limit?

Jan 13, 2017

See below.

#### Explanation:

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

but

$\frac{n}{\sqrt{{n}^{2} + n} + n} = \frac{n}{n \left(\sqrt{1 + \frac{1}{n}} + 1\right)} = \frac{1}{\sqrt{1 + \frac{1}{n}} + 1}$

so

${\lim}_{n \to \infty} {a}_{n} = {\lim}_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n}} + 1} = \frac{1}{2}$

so the sequence is convergent and converges to $\frac{1}{2}$