<s>[INST] 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?
$\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon$
$\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| < \epsilon$
$\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} < \epsilon$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} > \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \sqrt{n} > \frac{1-2 \epsilon}{2}$
$\Rightarrow n > \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
 [/INST] (Yet another solution - sorry!)
$$
\left(1 + \frac1{2n}\right)^2 = 1 + \frac1n + \frac1{4n^2} > 1 + \frac1n,
$$
therefore
$$
\sqrt{1 + \frac1n} < 1 + \frac1{2n},
$$
therefore
$$
\sqrt{n^2 + n} - n = n\left(\sqrt{1 + \frac1n} - 1\right) < \frac12,
$$
but also
$$
\sqrt{n^2 + n} - n = \frac1{\sqrt{1 + \frac1n} + 1} > \frac1{2\left(1 + \frac1{4n}\right)}
> \frac{1 - \frac1{4n}}{2} = \frac12 - \frac1{8n},
$$
therefore
$$
\frac12 - \epsilon < \sqrt{n^2 + n} - n < \frac12
\quad \left(\epsilon > 0, \ n \geqslant \frac1{8\epsilon}\right).
$$
</s>