Show that $x_{n}-\sqrt{2\ln(n)}$ converges to zero

Question

We have the sequence $x_{n+1}=x_{n}+\frac{1}{nx_{n}}$ with $x_{1}>0$. Show that the limit of the sequence $(x_{n}-\sqrt{2\ln(n)})=0$.
I showed that $x_{n}$ has limit $\infty$. Then I showed that the limit of the sequence $\frac{x_{n}}{\sqrt{2\ln(n)}}$ is 1. I tried calculating the limit that we want to show is 0, but I think I got something wrong and I dont know why. \begin{align} \lim_{n\rightarrow\infty}\left(x_{n}-\sqrt{2\ln(n)}\right) &=\lim_{n\rightarrow\infty}\sqrt{2}\frac{\frac{x_{n}}{\sqrt{2\ln(n)}}-1}{\frac{1}{\sqrt{\ln(n)}}}\\ &=\lim_{n\rightarrow\infty}\sqrt{2}\frac{\frac{x_{n+1}}{\sqrt{2\ln(n+1)}}-\frac{x_{n}}{\sqrt{2\ln(n)}}}{\frac{\sqrt{\ln(n)}-\sqrt{\ln(n+1)}}{\sqrt{\ln(n)\ln(n+1)}}}\\ &=\lim_{n\rightarrow\infty}\frac{\sqrt{\ln(n)}x_{n+1}-\sqrt{\ln(n+1)}x_{n}}{\sqrt{\ln(n)}-\sqrt{\ln(n+1)}} \end{align} that is infinity. What I did wrong and how I can do this problem?

Answer 1

I'm not sure how you got that last limit to be infinity, but here's an outline of one way to solve this problem.

First, notice that $$x_{n+1}^2 = \left(x_n + \dfrac{1}{nx_n}\right)^2 = x_n^2 + \dfrac{2}{n} + \dfrac{1}{n^2x_n^2}.$$

It is easy to check that $\{x_n\}$ is increasing. So you can bound $$x_n^2 + \dfrac{2}{n} \le x_{n+1}^2 \le x_n^2 + \dfrac{2}{n} + \dfrac{1}{n^2x_1^2}$$ $$\dfrac{2}{n} \le x_{n+1}^2 - x_n^2 \le \dfrac{2}{n} + \dfrac{1}{n^2x_1^2}$$

Now, sum this from $n = 1$ to $N - 1$ and see if you can get that $x_N^2 - 2\ln N$ is bounded by constants as $N \to \infty$. From there, it will be easy to show that $\lim_{N \to \infty}(x_N - \sqrt{2\ln N}) = 0$.