Asymptotics of the sequence defined by $x_{n+1} = x_n + \frac{2}{x_n}$, $x_0=1$

Question

Let $$x_{n+1} = x_n + \dfrac{2}{x_n}, x_0=1$$ Compute the limit $$\lim_{n \to \infty} \dfrac{x_n}{\sqrt{n}}$$

If we had there a $1/2$ we would have the sequence which aproximates the square root of two. I speculated that the limit of $x_n$ is $2\sqrt{2}$, but it's of course false. Another ideea was to use Cesaro-Stolz lemma, but it didn't get me too far either. Rearranging the terms we get $$x_n (x_{n+1}-x_n) = 2$$ If $x_n$ diverges to infinity (which I am strongly inclined to believe) then it's terms should get infinetly closer to one another. Sth akin to the harmonic series $\sum \dfrac{1}{n}$. If this direction is ok, How could I proceed from here?

Answer 1

The sequence is strictly increasing, and not bounded (otherwise it would have a limit $x$ satisfying $x = x + \frac 2x$). Therefore $x_n \to \infty$.

Then $$ x_{n+1}^2 - x_n^2 = \left( x_n + \frac{2}{x_n}\right)^2 - x_n^2 = 4 + \frac{4}{x_n^2} \to 4, $$ and Cesàro-Stolz implies that $$ \frac{x_n^2}{n} \to 4 \Longrightarrow \frac{x_n}{\sqrt n} \to 2 $$

Answer 2

This sequence problem is similar to others. Let $\;y := 1/n,\; z := \ln(y).\;$ If $\;x_{n+1} = x_n + 2/x_n\;$ with initial value $\;x_0=1,\;$ then the constant $\;c= .538934322...\;$ depends on $\;x_0\;$ and yields the series $$\;x_n = 2\sqrt{n}\;(1 + (c-z/8)y + (-(4c-1)^2/32 +z(4c-1)/32 -z^2/128)y^2 + O(y^3))$$ which is the asymptotic expansion of $\;x_n.$ Similar questions on MSE include question 2675217, question 843014.

By the way, the equation $\;x_n (x_{n+1}-x_n) = 2\;$ is the discrete analog of the equation $\;yy'=2\;$ which immediately implies $\;y^2 = 4x+C \;$ which translats back to $\;x_n = 2\sqrt{n}(1+c/n)\;$ which gives the initial asymptotic expansion. There is a simple recursion for getting further terms of the expansion.