1. ## Squeeze Theorem

How would I use the squeeze theorem to evaluate for:
$\displaystyle lim_{n\to\infty}\{\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2 + n}} \}$

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

3. 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.

4. Originally Posted by xxlvh
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.
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...