<s>[INST] $u_{n+1}=u_n^{u_n}$ 
*

*$u_0>1$

*$u_{n+1}=u_n^{u_n}$
We want to show that :

*

*the sequence $u_n$ diverges

*$ \sum \dfrac{1}{u_n}$ converges

*$ \exists N \in \mathbb{N} , \forall k : u_{N+k} \geq k+2$

*$\exists C >0,  \forall n \geq N  : \sum_{k=n+1}^{ \infty} \dfrac{1}{u_k} \leq \dfrac{C}{u_{n+1}}  $

*$\forall n \geq N : u_n \sum_{k=n+1}^{ \infty} \dfrac{1}{u_k}  \leq \dfrac{C} {u_n^{u_n-1}  }$

My attempt :
$
\begin{align*}
\phi(x) &=x^x \\
\phi'(x)&= ( \ln x +1 ) x^x \\
\end{align*}
$
$\phi$ is decreasing on  $]0, \dfrac{1} {e}]$ then increasing on  $[  \dfrac{1} {e}, +\infty[$
1.
$
\begin{align*}
u_1&= u_0^{u_0} \\
u_2&=u_0^{ u_0 \times u_1} \\
u_n &> u_0^{u_0 ^n} \\
u_n &> \exp ( u_0^n \ln (u_0) ) \\
u_0 &>1 \\
u_n &\to \infty
\end{align*}
$
2.
$
\begin{align*}
w_n&= u_0^{  u_0^n } \\
u_n &> w_n \\
\dfrac{1}{u_n} &<   \dfrac{1}{w_n} \\
\dfrac{w_{n+1}  }{w_n} &= u_0^{  u_0^n (1-u_0) } \\
\dfrac{w_{n+1}  }{w_n} &< 1 \\
\sum \dfrac{1}{u_n} &< \infty \\
\end{align*}
$
3.
$
\begin{align*}
u_n &\underset{ n \to \infty} \to \infty\\
\exists N, u_N &> 2 \\
j >2 &\implies j^2 > j+1 \\
u_{N+1} &=u_N^{ u_N }\\
&> u_N^2 \\
&> u_N +1 \\
u_{N+k} &> k+1 \\
\end{align*}
$
4.
$\begin{align*}
\forall k \leq 0, u_{N+1+k}&=u_{k+N}^{ u_{k+N}  } \\
                           &> u_{N+1}^{k+1} \\
\dfrac{1}{  u_{N+1+k}}  &<   \dfrac{1}{  u_{N+1} u_{N+1}^k} \\
\sum_{k=0}^{ \infty}\dfrac{1}{  u_{N+1+k}} &< \sum_{k=0}^{ \infty} \dfrac{1}{  u_{N+1} u_{N+1}^k} \\
&< \dfrac{1}{  u_{N+1}} \sum_{k=0}^{ \infty}\dfrac{1}{  u_{N+1}^k} \\
&< \dfrac{1}{  u_{N+1}} C \\
\end{align*}
$
5.
We multiply the relation given in (4) by  $u_n$
 [/INST] We can show by induction that  $u_n >u_0^{u_0^{n}}$ for all $n$. Since $u_0 >1$ this will certainly imply converegence of $\sum \frac 1 {u_n}$.
[$u_0^{n}=(1+(u_0-1))^{n}> (u_0-1)n$ by Binomial expansion so $u_0^{u_0^{n}}>u_0^{(u_0-1)n}$. Apply ratio test].
</s>