$u_{n+1}=u_n^{u_n}$

Question

• $u_0>1$
• $u_{n+1}=u_n^{u_n}$

We want to show that :

1. the sequence $u_n$ diverges
2. $ \sum \dfrac{1}{u_n}$ converges
3. $ \exists N \in \mathbb{N} , \forall k : u_{N+k} \geq k+2$
4. $\exists C >0, \forall n \geq N : \sum_{k=n+1}^{ \infty} \dfrac{1}{u_k} \leq \dfrac{C}{u_{n+1}} $
5. $\forall n \geq N : u_n \sum_{k=n+1}^{ \infty} \dfrac{1}{u_k} \leq \dfrac{C} {u_n^{u_n-1} }$

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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 $u_n$

Answer 1

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].