Let $a_{n} = \sqrt{n^{2}+n} - n$, for $n\in\textbf{N}$. Is the sequence $(a_{n})_{n=1}^{\infty}$ monotonic? Let $a_{n} = \sqrt{n^{2}+n} - n$, for $n\in\textbf{N}$. Show that $a_{n}$ converges as $n\to\infty$. What is the limit? Is the sequence $(a_{n})_{n=1}^{\infty}$ monotonic?
MY ATTEMPT
The answer to the first question is $a_{n}\to 1/2$ as $n\to\infty$. Indeed,
\begin{align*}
\lim_{n\to\infty}\sqrt{n^{2}+n} - n = \lim_{n\to\infty}\frac{n}{\sqrt{n^{2} + n} + n} = \lim_{n\to\infty}\frac{1}{\sqrt{1 + 1/n} + 1} = \frac{1}{2}
\end{align*}
To test for monotonicity, we can try to study the behavior of the quotient:
\begin{align*}
\frac{a_{n+1}}{a_{n}} & = \frac{\sqrt{(n+1)^{2} + n + 1} - n - 1}{\sqrt{n^{2} + n} - n}\\\\
& = \frac{(n+1)^{2} + n + 1 - (n+1)^{2}}{n^{2} + n - n^{2}}\times\frac{\sqrt{n^{2} + n} + n}{\sqrt{(n+1)^{2} + n + 1} + n + 1}\\\\
& = \frac{n+1}{n}\times\frac{\sqrt{n^{2} + n} + n}{\sqrt{(n+1)^{2} + n + 1} + n + 1}
\end{align*}
But then I get stuck, because the first factor is greater than one and the second is smaller than one.
Can someone please finish my attempt or provide an alternative approach?

Since the sequence$$\left(\sqrt{1+\frac1n}\right)_{n\in\Bbb N}$$is decreasing and since you proved that$$\sqrt{n^2+n}-n=\frac1{\sqrt{1+\frac1n}+1},$$your sequence is increasing.
