How would I use the squeeze theorem to evaluate for:

$\displaystyle 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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- Sep 19th 2009, 08:07 AMxxlvhSqueeze Theorem
How would I use the squeeze theorem to evaluate for:

$\displaystyle lim_{n\to\infty}\{\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2 + n}} \} $ - Sep 19th 2009, 08:31 AMPlato
$\displaystyle \frac{n}

{{\sqrt {n^2 + n} }} \leqslant \sum\limits_{k = 1}^n {\frac{1}

{{\sqrt {n^2 + k} }}} \leqslant \frac{n}

{{\sqrt {n^2 + 1} }}$ - Sep 19th 2009, 08:40 PMxxlvh
So after that step I would solve for the limits of those outer functions,

how would I start for:

$\displaystyle lim_{n\to\infty}\frac{n}{\sqrt{n^2+n}}$

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. - Sep 19th 2009, 11:21 PMbkarpuz
**Plato**already completed the proof, but there is somethings missing with you.

I hope the following will fill it.

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

$\displaystyle =\lim_{n\to\infty}\frac{1}{\sqrt{\frac{n^{2}+n}{n^ {2}}}}$ (as $\displaystyle n\to\infty$, we may take $\displaystyle n>0$)

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

$\displaystyle =\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{1}{n}}}=1$ (since $\displaystyle 1/\sqrt{1+\cdot}$ is continuous in a neighbourhood of $\displaystyle 0$)

$\displaystyle =\frac{1}{\sqrt{1+\lim\limits_{n\to\infty}\frac{1} {n}}}=\frac{1}{\sqrt{1+0}}=1$

I guess you can show now the limit on the right-hand side...