Prove that $\lim ( \sqrt{n^2+n}-n) = \frac{1}{2}$ Here's what I have so far:

Given $\epsilon &gt; 0$, we want to find N such that $\sqrt{n^2+n}-n &lt; \epsilon$ for all $n&gt;N$. And so:

$( \sqrt{n^2+n}-n-\frac{1}{2}) \cdot \frac{\sqrt{n^2+n}-(n+\frac{1}{2})}{\sqrt{n^2+n}-(n+\frac{1}{2})}$ 
$= \frac{(n^2+n)-(n+\frac{1}{2})}{\sqrt{n^2+n} + (n+\frac{1}{2})}$

And I'm not sure how to go on from here. Help would be appreciated.
Hint:
\begin{align}
\sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n} 
\end{align}
and
\begin{align}
\left|\frac{n}{\sqrt{n^2+n}+n} -\frac{1}{2}\right| = \left|\frac{n-\sqrt{n^2+n}}{2\sqrt{n^2+n}+2n}\right| = \frac{n}{2(\sqrt{n^2+n}+n)^2} \leq \frac{n}{8n^2}
\end{align}
$$\sqrt{n^2+n}-n = n(\sqrt{1+1/n}-1)$$

Which by the binomial theorem is equal to

$$n\left(-1 + \sum_{k=0}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^k}\right) = n\left(\sum_{k=1}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^k}\right)$$

(With the product equal to 1 for $k = 0$.)  So now move the n inside the sum to get

$$\sum_{k=1}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^{k-1}} = \frac{1}{2} + \sum_{k=2}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^{k-1}}$$

So $\epsilon$ would be the absolute value of $\sum_{k=2}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^{k-1}}$, and all that remains is to show this approaches zero for sufficiently large n.  Well, we'll start by saying...

$$\left|\sum_{k=2}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^{k-1}}\right| = \left|\sum_{k=2}^\infty\frac{(-1)^{k-1}\cdot\frac{1}{2}\cdot\prod_{l=1}^{k-1}\left(\frac{2l-1}{2}\right)}{k! n^{k-1}}\right| &lt; \sum_{k=2}^\infty\frac{\frac{1}{2}\prod_{l=1}^{k-1}\left(\frac{2l-1}{2}\right)}{k! n^{k-1}} &lt; \sum_{k=2}^\infty\frac{(k-1)!}{2\cdot k!n^{k-1}} = \sum_{k=2}^\infty\frac{1}{2kn^{k-1}}$$

And I should hope you can take it from there.