<s>[INST] 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?
 [/INST] Let $f(x)=\sqrt{x^2+x}-x$.
By AM-GM we have $\sqrt{x(x+1)}< \dfrac{x+(x+1)}{2}\ $ thus $\ f(x)< \dfrac 12\ $ and $\ a_n< \dfrac 12$
This also means $A=2x+1-2\sqrt{x(x+1)}>0$
Since $f'(x)=\dfrac{A}{2\sqrt{x(x+1)}}$ then $f$ is $\nearrow$ and so is $a_n$
</s>