Converegence of $x_n = (\sqrt{n^2 + \sqrt{n}} -n)(\sqrt{n+1} + \sqrt{2n})$

Question

Let $x_n = (\sqrt{n^2 + \sqrt{n}} -n)(\sqrt{n+1} + \sqrt{2n})$ for each $n\geq 1$. I am looking for easy ways to show that $(x_n)$ is convergent. I have tried to show that is decreasing and bounded, but I got stucked. Also, I tried to find a formula $x_n =f(x_{n-1})$ but I failed. Besides, by using a computer, I can tell that it is convergent with limit $\lim x_n = (1+\sqrt{2})/2\approx 1.207106781$. Clearly, this is a root of $4x^2-4x-1$, maybe this suggest a way of solve this problem, but I can't relate the sequence with that second order polynomial.

What it is the best way to attack this problem of convergence of sequence? Are there any 'algebra trick' that make the problem easier? and, once the convergence is proved, how can I calculate $\lim x_n$ without the help of a computer? Thanks in advance.

Answer 1

We have that

$$\sqrt{n^2 + \sqrt{n}} -n=\frac{\sqrt{n}}{\sqrt{n^2 + \sqrt{n}} +n}$$

and then

$$x_n=\frac{\sqrt{n}(\sqrt{n+1} + \sqrt{2n})}{\sqrt{n^2 + \sqrt{n}} +n}$$

which can be easily solved factoring out $n$ from numerator and denominator.

Answer 2

Yes, you are right: $$x_n=\frac{\sqrt{1+\frac{1}{n}}+\sqrt2}{1+\sqrt{1+\frac{1}{n\sqrt{n}}}}\rightarrow\frac{1+\sqrt2}{2}$$

Answer 3

Use formula $$\boxed{\sqrt{a}-\sqrt{b} = {a-b \over \sqrt{a}+\sqrt{b}}}$$

\begin{align}x_n &= \color{red}{(\sqrt{n^2 + \sqrt{n}} -n)}(\sqrt{n+1} + \sqrt{2n})\\ &= \color{red}{{n^2 + \sqrt{n} -n^2\over \sqrt{n^2 + \sqrt{n}} +n }}\cdot \sqrt{n}\Big( \sqrt{1+{1\over n}} + \sqrt{2}\Big)\\ &= { n\over n\cdot (\sqrt{1+ {\sqrt{n}\over n^2}} +1) }\cdot \Big( \sqrt{1+{1\over n}} + \sqrt{2}\Big)\\ &\to {1+\sqrt{2}\over 2} \end{align}