Let the sequence $(x_n)_{n\geq1}$ be defined recursively as follows: $x_1 = \sqrt{2}, \quad x_{n+1} = x_n + \frac{1}{\sqrt{n} \cdot x_n} \quad \text{for } n \ge 1$
This is a problem from the 2024 UTCN entrance exam.
I’ve tried majorization, or induction, but I can’t prove that the series isn’t bounded.
Assume to the contrary that $(x_n)$ is bounded by some $M > 0$. Choose $m \in \mathbb N$ such $M \le \sqrt{m}$. Then $x_n \le M \le \sqrt{m} \le \sqrt n$ for all $n \ge m$. We therefore get for $n \ge m$ $$x_{n+1} - x_n = \frac{1}{\sqrt n x_n} \ge \frac{1}{\sqrt n \sqrt n} = \frac 1 n . \tag{1}$$ By induction on $k$ we get $$x_{m+k} \ge \sum_{r=m}^{m+k-1} \frac{1}{r} . \tag{2}$$ The base case $k=1$ follows from $(1)$ because $x_m > 0$. In the induction step we use again $(1)$: $$x_{m+k+1} = x_{m+k} + (x_{m+k+1} - x_{m+k}) \ge \sum_{r=m}^{m+k-1} \frac{1}{r} + \frac{1}{m+k} = \sum_{r=m}^{m+(k+1)-1} \frac{1}{r} $$
It is well-known that the harmonic series $\sum_{r=1}^\infty \frac{1}{r}$ diverges. Thus $\sum_{r=m}^{m+k-1} \frac{1}{r} \to \infty$ as $k \to \infty$. This shows that $(x_n)$ is not bounded, a contradiction.
Squaring the recurrence immediately leads to $$x^2_{k+1}=x^2_k+\frac2{\sqrt{k}}+\frac1{k x^2_k}\tag1,$$ i.e. $$x^2_{k+1}>x^2_k+\frac2{\sqrt{k}}>x^2_k+\frac4{\sqrt{k}+\sqrt{k+1}}=x^2_k+4(\sqrt{k+1}-\sqrt{k}).$$ By induction, $$x^2_n\ge4\sqrt{n}-2,\tag2$$ with equality only for $n=1$. This is sufficient for divergence. Of course, plugging (2) into (1) gives an upper bound, too, and since $\displaystyle\sum^\infty_{k=1}\frac1{k\sqrt{k}}<\infty$, we obtain $$x^2_n=4\sqrt{n}+O(1).$$
EDIT: the inequality $$2 \sqrt[4]{n} - 1 < x_n < 2 \sqrt[4]{n}$$ mentioned in a comment is not true. The lower bound is more accurately covered by (2). $x^2_n-4\sqrt{n}$ converges, naturally (monotone and bounded), but to a positive limit, as it seems. Convergence is slow, so it might be interesting to calculate a decent approximation of that limit.