How to write the general solution for the recurrence relation: $x_{n+1} = x_n e ^ {\frac{1}{x_n}}$?

Question

Consider the sequence $(x_n)_{n\geq0}$, with $x_0>0$ and satisfying the recurrence relation: $$x_{n+1} = x_n e ^ {\frac{1}{x_n}},$$ how you go about writing $x_n$ in terms of $x_0$, and the difference $x_{n+1} - x_n$?

Answer 1

I doubt one can find a closed formula giving $x_n$ in terms of $x_0$, however for every $x\in {\mathbb R}$, one has \begin{equation} e^x \ge 1 + x \end{equation} It follows that if $x_n > 0$, \begin{equation} x_{n+1}\ge x_{n}(1 + \frac{1}{x_{n}}) = x_{n} + 1 \end{equation} hence $x_n \ge x_0 + n$ when $x_0 > 0$, and $x_n\to \infty$.

When $x_0 < 0$, we see that $x_n< 0$ for all $n$, and $x_n$ increases. It is easy to conclude that $x_n\to 0$.