Proving specific properties of $x_{n+1} = x_n + \frac{1}{nx_n}$ where $x_1 = \frac{1}{2}$

Question

First, I am asked to prove that $(x_n)_{n \geq 1}$ is divergent.

Secondly, I have to find the limit of $\left( \frac{x_n}{\sqrt{2\ln(n)}} \right)_{n \geq 1}$ (if it exists).

What I managed to show:

• $(x_n)_{n \geq 1}$ is monotonically increasing:

$$x_{n+1} - x_n = \frac{1}{nx_n} > 0$$

• $x_{n+1} \geq \frac{2}{\sqrt{n}}$:

$$x_{n+1} = x_n + \frac{1}{nx_n} \geq 2\sqrt{\frac{x_n}{nx_n}} = \frac{2}{\sqrt{n}} \textbf{ (AM-GM) }$$

• $x_{n+1} \leq \frac{1}{2} \left(H_{n}+1 \right)$:

$$ x_{n+1} = x_{n} + \frac{1}{nx_n} \leq x_{n} + \frac{1}{2n} \leq x_{n-1} + \frac{1}{2n} + \frac{1}{2(n-1)} \leq \dots \leq x_1 + \frac{1}{2}H_n = \frac{1}{2} \left( H_n + 1 \right)$$

• $(x_n)_{n \geq 1}$ is divergent:

Suppose the sequence converges with limit $l \in (0, \infty)$. From the last inequality we have: $$ x_{n+1} \leq \frac{1}{2} \left( H_n + 1 \right) \Leftrightarrow x_{n+1} - \frac{1}{2} \ln(n) \leq \frac{1}{2} \left( H_n - \ln(n) + 1 \right)$$

As $n \to \infty$ the LHS approaches becomes $l - \frac{1}{2} \lim_{n \to \infty} \ln (n)$ which diverges to $- \infty$, but the RHS converges to $\frac{1}{2}(\gamma + 1)$ where $\gamma$ is the Euler-Mascheroni Constant, which is a contradiction.

Edit: Thank you everyone for your comments, I now understand how to finish this assingment.

Answer 1

One can apply the Stolz–Cesàro theorem: $$ \lim_{n \to \infty} \frac{x_n^2}{2 \ln(n)} = \lim_{n \to \infty} \frac{x_{n+1}^2-x_n^2}{2 (\ln(n+1)-\ln(n))} $$ if the limit on the right exists. But $$ \frac{x_{n+1}^2-x_n^2}{2 (\ln(n+1)-\ln(n))} = \frac{2/n+ 1/(n^2x_n^2)}{2 \ln(1+1/n)} = \frac{1/n}{\ln(1+1/n)} \left( 1 + \frac{1}{2n x_n^2} \right) $$ converges to $1$, since $x_n \to +\infty$ is already known.

Answer 2

Clearly $$x_{n+1}^2 = x_n^2+\frac2{n} + \frac{1}{n^2x_n^2}\ge x_n^2+\frac2{n}$$ and hence $$ x_n^2\ge x_1^2+2\sum_{k=1}^{n-1}\frac1k=x_1^2+2\ln n+2\gamma+o(1). \tag1$$ On the other hand, it is easy to see $$ x_n\ge 1, \forall n\ge 2 $$ and hence for $n\ge 2$, $$x_{n+1}^2 = x_n^2+\frac2{n} + \frac{1}{n^2x_n^2}\le x_n^2+\frac{2}{n}+\frac1{n^2}. $$ From (1) one has, for $\forall n>2$ $$ x_n^2\le x_2^2+2\sum_{k=2}^{n-1}\frac{1}{k}+\sum_{k=2}^{n-1}\frac1{n^2}=x_2^2+2\ln (n-1)-2+\sum_{k=2}^{n-1}\frac1{n^2}+2\gamma+o(1). \tag2$$ From (1) and (2), one has, for $n>N$, $$ x_1^2+2\ln n+\gamma+o(1) \le x_n^2\le x_2^2+2\ln (n-1)-2+\sum_{k=2}^{n-1}\frac1{n^2}+2\gamma+o(1). $$ This implies $$ \lim_{n\to\infty}\frac{x_n^2}{2\ln n}=1.$$

Answer 3

Replacing the recursion by this simple version of a corresponding ODE

$$\frac{d}{dn} x(n) = \frac{1}{n x(n)}\tag{1}$$

gives the solution

$$x(n) = \pm \sqrt{C+2\log(n)}\tag{2}$$

which coincides with the result striktly derived by others. This is not likely to be an accident.