Calculate $\lim_{n\to\infty}\sqrt{n^2+n} - n$. Calculate $\lim_{n\to\infty}\hspace{2 pt}a_n$, where $a_n = \sqrt{n^2+n} - n$ is a sequence of complex numbers.
I performed the ratio test:
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty}\frac{\sqrt{(n+1)^2 + (n+1)} - (n+1)}{\sqrt{n^2+n} - n} = \lim_{n\to\infty}\frac{\sqrt{n^2+3n+2} - (n+1)}{\sqrt{n^2+n} - n}$$
which ultimately leads to, when multiplied by $\frac{\frac{1}{n}}{\frac{1}{n}}$,
$$\lim_{n\to\infty}\frac{\sqrt{1+\frac{3}{n}+\frac{2}{n^2}} - (1+\frac{1}{n})}{\sqrt{1+\frac{1}{n}} - 1}$$
Should I multiply by the complex conjugate? It leads to a rather messy expression. I am stuck.
How can one perform the root test here? I attempted: $$\lim_{n\to\infty}\sqrt[n]{\sqrt{n^2+n} - n} \Longleftrightarrow \lim_{n\to\infty}((n^2 + n)^{\frac{1}{2}} - n)^{\frac{1}{n}}$$
I have also no idea how to proceed.

$$a_n =\sqrt{n^2+n}-n = \frac{(n^2+n)-n^2}{\sqrt{n^2+n}+n}=\frac{n}{\sqrt{n^2+n}+n}=\frac{1}{\sqrt{1+\frac{1}{n}}+1} \Rightarrow \lim_{n \to \infty} a_n = \frac{1}{2}$$
