Squeeze Theorem

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September 19th 2009, 08:07 AM
xxlvh

Squeeze Theorem

How would I use the squeeze theorem to evaluate for:
$lim_{n\to\infty}\{\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2 + n}} \}$
September 19th 2009, 08:31 AM
Plato

$\frac{n}<br /> {{\sqrt {n^2 + n} }} \leqslant \sum\limits_{k = 1}^n {\frac{1}<br /> {{\sqrt {n^2 + k} }}} \leqslant \frac{n}<br /> {{\sqrt {n^2 + 1} }}$
September 19th 2009, 08:40 PM
xxlvh

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.
September 19th 2009, 11:21 PM
bkarpuz

Quote:

Originally Posted by xxlvh

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...