# Decide if sequence converges/diverges

• Jun 10th 2009, 03:38 PM
utopiaNow
Decide if sequence converges/diverges
The given sequence is: $\displaystyle a_n = n\left(\sqrt{1 + \frac{1}{n}} - 1\right)$.

I rearranged terms to get $\displaystyle a_n = \sqrt{n^2 + n} - n$. I couldn't tell by inspection if this is convergent or divergent, so I started plugging in large values of n and noticed that it seems to be that $\displaystyle a_n \to 0.5$. So now I'm trying to prove it converges by showing that $\displaystyle a_n$ has an upper bound and is non-decreasing. However I'm having trouble showing it has an upper bound, therefore leading me to think maybe that $\displaystyle a_n$ is divergent? I'm not sure at this point, any suggestions would be appreciated.

• Jun 10th 2009, 03:48 PM
TheEmptySet
Quote:

Originally Posted by utopiaNow
The given sequence is: $\displaystyle a_n = n\left(\sqrt{1 + \frac{1}{n}} - 1\right)$.

I rearranged terms to get $\displaystyle a_n = \sqrt{n^2 + n} - n$. I couldn't tell by inspection if this is convergent or divergent, so I started plugging in large values of n and noticed that it seems to be that $\displaystyle a_n \to 0.5$. So now I'm trying to prove it converges by showing that $\displaystyle a_n$ has an upper bound and is non-decreasing. However I'm having trouble showing it has an upper bound, therefore leading me to think maybe that $\displaystyle a_n$ is divergent? I'm not sure at this point, any suggestions would be appreciated.

$\displaystyle a_n=\frac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n}=\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}=\frac{n}{\sqrt{n^2+n}+n}=$
$\displaystyle a_n=\frac{1}{\sqrt{1+\frac{1}{n}}+1}$
$\displaystyle \lim_{n \to \infty}\frac{1}{\sqrt{1+\frac{1}{n}}+1}=\frac{1}{2 }$