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.
Best Answer
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.