Decide if sequence converges/diverges

**utopiaNow** · June 10th 2009, 03:38 PM

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

I rearranged terms to get $a_n = \sqrt{n^2 + n} - n$ . 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 $a_n \to 0.5$ . So now I'm trying to prove it converges by showing that $a_n$ 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 $a_n$ is divergent? I'm not sure at this point, any suggestions would be appreciated.

Thanks in advance.

**TheEmptySet** · June 10th 2009, 03:48 PM

Originally Posted by utopiaNow

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

I rearranged terms to get $a_n = \sqrt{n^2 + n} - n$ . 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 $a_n \to 0.5$ . So now I'm trying to prove it converges by showing that $a_n$ 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 $a_n$ is divergent? I'm not sure at this point, any suggestions would be appreciated.

Thanks in advance.

Multiply the numerator and denominator by the conjugate to get

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

$a_n=\frac{1}{\sqrt{1+\frac{1}{n}}+1}$

Now as n goes to infintiy we get

$\lim_{n \to \infty}\frac{1}{\sqrt{1+\frac{1}{n}}+1}=\frac{1}{2 }$

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