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Math Help - Decide if sequence converges/diverges

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    Junior Member utopiaNow's Avatar
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    Decide if sequence converges/diverges

    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.
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    Quote Originally Posted by utopiaNow View Post
    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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