# Convergence of $x_{n+1} = x_n + \dfrac{(\vert x_n \vert)^{1/2}}{n^2}$

Let $(x_n)$ be the sequence defined by $x_{n+1} = x_n + \dfrac{(\vert x_n \vert)^{1/2}}{n^2}$ for $n \geq 2$ and $x_1$ be any real number. Then I want to prove that $x_n$ is convergent.

It is obvious that $x_n$ is monotonically increasing. I also wanted to show that $x_n$ is bounded above by using the inequality $x_{n+1} < x_n + \dfrac{\vert x_n \vert}{n^2}$ but I was not successful.

Suppose $x_m > 0$ for some $m$. Rearrange the recurrence to get $\dfrac{x_{n+1}-x_n}{\sqrt{|x_n|}} = \dfrac{1}{n^2}$.

Since $\dfrac{1}{\sqrt{|x|}}$ is decreasing for $x > 0$, a left endpoint Riemann sum will overestimate the integral.

Hence, $2\sqrt{x_n}-2\sqrt{x_m} = \displaystyle\int_{x_m}^{x_n}\dfrac{1}{\sqrt{|x|}}\,dx \le \sum_{k = m}^{n-1}\dfrac{x_{k+1}-x_k}{\sqrt{|x_k|}} = \sum_{k = m}^{n-1}\dfrac{1}{k^2} \le \sum_{k = 1}^{\infty}\dfrac{1}{k^2} = \dfrac{\pi^2}{6}$.

Therefore, $x_n \le \left(\sqrt{x_m}+\dfrac{\pi^2}{12}\right)^2$ for all $n \ge m$, and thus, $x_n$ is bounded above.

If $x_m \not > 0$ for any $m$, then $x_n$ is bounded above by $0$.

• This was beautiful. – user135520 Dec 21 '15 at 19:06
• It is not clear to me why $x_n<(x_m^{1/2}+C)^2$ for $n>m$ implies boundedness – Marco Disce Feb 28 '16 at 11:38
• Since $x_m$ is some finite value, $(\sqrt{x_m}+\tfrac{\pi^2}{12})^2$ is also some finite value. Hence, $x_n < (\sqrt{x_m}+\tfrac{\pi^2}{12})^2$ means that $x_n$ is bounded above for $n > m$, and thus, $x_n \le \max\{x_1,x_2,\ldots,x_m,(\sqrt{x_m}+\tfrac{\pi^2}{12})^2\}$ for all $n$. – JimmyK4542 Feb 28 '16 at 23:40