<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] Maybe the answer that gives the direct clue about what is going on is going through a proof that
$$\lim_{n \to +\infty} (\sqrt{n^2+n+c_2}-n)-(\sqrt{n^2+n+c_1}-n)=0$$
or equally
$$\lim_{n \to +\infty} \sqrt{n^2+n+c_2}-\sqrt{n^2+n+c_1}=0$$
Now write this to make it obvious as
$$\lim_{n \to +\infty} \frac{1}{\sqrt{n^2+n+c_1}}(\frac{\sqrt{1+\frac{1}{n}+\frac{c_2}{n^2}}}{\sqrt{1+\frac{1}{n}+\frac{c_1}{n^2}}}-1)=0$$
So we can pick whatever constant we want. We pick $\frac{1}{4}$ and have
$$\lim_{n \to +\infty} \sqrt{n^2+n}-n=\lim_{n \to +\infty} (\sqrt{n^2+n+\frac{1}{4}}-n)=$$ $$\lim_{n \to +\infty} (\sqrt{(n+\frac{1}{2})^2}-n)=\lim_{n \to +\infty} n+\frac{1}{2}-n=\frac{1}{2}$$
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