Define a sequence $ (x_{n})_{n \in \mathbb{N}} $ of positive real numbers by $$ x_{1} := 1 \quad \text{and} \quad \forall n \in \mathbb{N}: \quad x_{n + 1} := x_{n} + \frac{n}{(x_{1} \times \cdots \times x_{n})^{1/n}}. $$ Define an associated sequence $ (y_{n})_{n \in \mathbb{N}_{\geq 2}} $ of positive real numbers by $$ \forall n \in \mathbb{N}_{\geq 2}: \quad y_{n} := \frac{x_{n}}{\ln(n)}. $$
I believe that the Cauchy-Schwarz Inequality yields $ \displaystyle \lim_{n \to \infty} y_{n} = \infty $. Is my assertion correct? A proof of it may be found in another post, but I haven’t gotten down to checking it yet. Thanks for your help!
So $x_{n+1} = x_n + \frac{n}{(\prod_{k=1}^n x_k)^{1/n}}$ and $x_1 = 1$.
You claim that $\frac{x_n}{\ln n} \to \infty$ and would like a proof.
btw, what is the "CS theorem"?
Obviously the $x_n$ are increasing.
Therefore, $x_1 < (\prod_{k=1}^n x_k)^{1/n} < x_n$ so that $x_{n+1} > x_n + \frac{n}{x_n} $ and $x_{n+1} < x_n + n $.
From the second inequality, $x_n < \frac{n(n+1)}{2} $. Putting this in the first inequality, $x_{n+1} > x_n + \frac{n}{\frac{n(n+1)}{2}} = x_n + \frac{2}{n+1} $.
From this, $x_n > 2\ln n+c$ for some real $c$. This isn't enough, but it's a start.
Suppose that $ x_n \sim r n$. Then, $$\prod_{k=1}^n x_k \sim r^n n! \sim r^n \sqrt{2\pi n}\frac{n^n}{e^n} = (nr/e)^n \sqrt{2\pi n} $$ so $$(\prod_{k=1}^n x_k)^{1/n} \sim (nr/e) (2\pi n)^{1/(2n)} $$ so $$x_{n+1} \sim x_n + \frac{n}{(nr/e) (2\pi n)^{1/(2n)}} = x_n + \frac{e}{r (2\pi n)^{1/(2n)}} .$$
Since $(2\pi n)^{1/(2n)} \to 1$, if $r =\frac{e}{r}$ (or $r = e^{1/2}$), $x_{n+1} \sim x_n + r$.
I think this shows that $x_n \sim e^{1/2} n$, which certainly implies $\frac{x_n}{\ln n} \to \infty$.
I'll check this computationally tomorrow. Now, time for sleep.