Let $a_{n} = \sqrt{n^{2}+n} - n$, for $n\in\textbf{N}$. Is the sequence $(a_{n})_{n=1}^{\infty}$ monotonic?

Question

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?

Answer 1

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.

Answer 2

As you wrote, for $n\ge 1,$

$$a_n=\frac{1}{b_n}$$

with

$$b_n=\sqrt{1+\frac 1n}+1$$

It is clear that $\forall n\ge 1\;\; b_n\ne 0$ and that

$$(b_n) \;\text{ is decreasing}$$ because $$b_n-b_{n+1}=\sqrt{1+\frac 1n}-\sqrt{1+\frac{1}{n+1}}$$ $$=\frac{ 1 }{n(n+1)(\sqrt{1+\frac 1n}+\sqrt{1+\frac{1}{n+1}})}>0$$

So, $ (a_n) $ is increasing.

Answer 3

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 $a_n$