Prove that $\lim_{n\to\infty} (\sqrt{n^2+n}-n) = \frac{1}{2}$ Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon$
$\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| < \epsilon$
$\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} < \epsilon$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} > \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \sqrt{n} > \frac{1-2 \epsilon}{2}$
$\Rightarrow n > \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$

Using the fractional binomial theorem we have
$$
\Bigl(1+\frac{1}{n}\Bigr)^{1/2}=1+\frac{1}{2n}+o\Bigl(\frac{1}{n}\Bigr)
$$
where $o(1/n)$ denotes a quantity that grows asymptotically slower than $1/n$ as $n\to\infty$. Multiplying through by $n$ yields
$$
\sqrt{n^2+n}=n+\frac{1}{2}+o(1),
$$
which is equivalent to the limit
$$
\lim_{n\to\infty}\sqrt{n^2+n}-n=\frac{1}{2}.
$$
