for which value of $a$ that$\sum_{n=0}^{\infty} \frac{1}{u_{n}^{a}}$ converges? We are given an arbitrary real positive $u_0$. The sequence $\{u_n\}_{n\ge 0}$ is defined by $u_{n+1}=u_ne^{-u_n}$ for $n\ge 0$. Find the values of $a\in\mathbb{R}$ for which the sequence $\sum_{n=0}^{\infty} \frac{1}{u_{n}^{a}}$ is convergent.
Thank you for your replies

Hints :


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*First, using the fact that $xe^{-x}\le x$ and that the equation $x=xe^{-x}$ has only one solution (zero), show that $(u_n)$ converges to $0$.

*Using the definition of $u_n$, show that $\frac{1}{u_{n+1}}-\frac{1}{u_n}\sim 1$.

*By sum, conclude that $u_n\sim \frac{1}{n}$.

