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Math Help - Squeeze Theorem

  1. #1
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    Red face Squeeze Theorem

    How would I use the squeeze theorem to evaluate for:
    lim_{n\to\infty}\{\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2 + n}} \}
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  2. #2
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    \frac{n}<br />
{{\sqrt {n^2  + n} }} \leqslant \sum\limits_{k = 1}^n {\frac{1}<br />
{{\sqrt {n^2  + k} }}}  \leqslant \frac{n}<br />
{{\sqrt {n^2  + 1} }}
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  3. #3
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    So after that step I would solve for the limits of those outer functions,
    how would I start for:
    lim_{n\to\infty}\frac{n}{\sqrt{n^2+n}}
    I was going to factor out the highest power of "n" but don't know how to apply that in the denominator since it's under a square root sign.
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  4. #4
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by xxlvh View Post
    So after that step I would solve for the limits of those outer functions,
    how would I start for:
    lim_{n\to\infty}\frac{n}{\sqrt{n^2+n}}
    I was going to factor out the highest power of "n" but don't know how to apply that in the denominator since it's under a square root sign.
    Plato already completed the proof, but there is somethings missing with you.
    I hope the following will fill it.
    \lim_{n\to\infty}\frac{n}{\sqrt{n^{2}+n}}=\lim_{n\  to\infty}\frac{1}{\frac{\sqrt{n^{2}+n}}{n}}
    =\lim_{n\to\infty}\frac{1}{\sqrt{\frac{n^{2}+n}{n^  {2}}}} (as n\to\infty, we may take n>0)
    =\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{1}{n}}}
    =\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{1}{n}}}=1 (since 1/\sqrt{1+\cdot} is continuous in a neighbourhood of 0)
    =\frac{1}{\sqrt{1+\lim\limits_{n\to\infty}\frac{1}  {n}}}=\frac{1}{\sqrt{1+0}}=1
    I guess you can show now the limit on the right-hand side...
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