Prove that $\lim ( \sqrt{n^2+n}-n) = \frac{1}{2}$ Here's what I have so far:
Given $\epsilon > 0$, we want to find N such that $\sqrt{n^2+n}-n < \epsilon$ for all $n>N$. 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.
 A: 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}
A: For $n>0$ we have 
(1). $ \sqrt {n^2+n}-n=\frac {n}{\sqrt {n^2+n}+n}.$
(2). $ n=\sqrt {n^2}<\sqrt {n^2+n}<\sqrt {n^2+n+\frac {1}{4}}=n+\frac {1}{2}.$ 
(3). Therefore $ \frac {1}{2}-\frac {1}{8n+2}=\frac {n}{(   n+\frac {1}{2})   +n}<\frac {n}{\sqrt {n^2+n} +n}<\frac {n}{n+n}=\frac {1}{2}.$
A: For each $\varepsilon>0$ you want to find $N$ such that, for $n>N$,
$$
\left|\sqrt{n^2+n}-n-\frac{1}{2}\right|<\varepsilon
$$
that is,
$$
-\varepsilon+\frac{1}{2}<\sqrt{n^2+n}-n<\varepsilon+\frac{1}{2}
$$
It is not restrictive to assume $0<\varepsilon<1/2$.
The inequality $\sqrt{n^2+n}-n>A$, for $A<1/2$, is the same as $\sqrt{n^2+n}>n+A$, and squaring gives
$$
n>2An+A^2
$$
that is $n>A^2/(1-2A)$. In the case of $A=(1-2\varepsilon)/2$ we have
$$
n>\frac{(1-2\varepsilon)^2}{8\varepsilon}
$$
The inequality $\sqrt{n^2+n}-n<B$, with $B>1/2$, is the same as $\sqrt{n^2+n}<n+B$ and, squaring,
$$
n<2Bn+B^2
$$
which is true for every integer $n$.
A: $$\sqrt{n^2+n}-n = n(\sqrt{1+1/n}-1)$$
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...
$$\left|\sum_{k=2}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^{k-1}}\right| = \left|\sum_{k=2}^\infty\frac{(-1)^{k-1}\cdot\frac{1}{2}\cdot\prod_{l=1}^{k-1}\left(\frac{2l-1}{2}\right)}{k! n^{k-1}}\right| < \sum_{k=2}^\infty\frac{\frac{1}{2}\prod_{l=1}^{k-1}\left(\frac{2l-1}{2}\right)}{k! n^{k-1}} < \sum_{k=2}^\infty\frac{(k-1)!}{2\cdot k!n^{k-1}} = \sum_{k=2}^\infty\frac{1}{2kn^{k-1}}$$
And I should hope you can take it from there.
