\(\alpha_{n+1}={\sqrt{2}\,}^{\alpha_{n}}\)
- Grigorios Kostakos
- Founder
- Posts: 461
- Joined: Mon Nov 09, 2015 1:36 am
- Location: Ioannina, Greece
\(\alpha_{n+1}={\sqrt{2}\,}^{\alpha_{n}}\)
Let \(\left({\alpha_{n}}\right)_{n\in\mathbb{N}}\) the sequence of real numbers with \[\alpha_{1}=\sqrt{2}\quad{\text{and}} \quad\alpha_{n+1}={\sqrt{2}\,}^{\alpha_{n}}\,,\quad n\in\mathbb{N}\,.\] Prove that the sequence converges and find the limit of the sequence.
Grigorios Kostakos
Re: \(\alpha_{n+1}={\sqrt{2}\,}^{\alpha_{n}}\)
Replied by ex-member aziiri:
\(1<a_n<2\) : we have \(1<a_1<2\). Suppose \(1<a_n<2\) then \(1<{\sqrt{2}\,}^{1}<a_{n+1}={\sqrt{2}\,}^{a_n}<{\sqrt{2}\,}^2=2\).
\((a_n)_{n\geq 0} \) is increases. On \([1,2)\) define \(f(x)={\sqrt{2}\,}^x-x\) then \(f'(x)= 2^{x/2-1} \ln 2 -1 \leq f'(2)=\ln 2 -1 <0\) then \(f\) is decreasing then \(f(x)\geq f(2)=0\) for any \(x\in [1,2)\) then \(a_{n+1}={\sqrt{2}\,}^{a_n}>a_n \) for any integer \(n\).
Then it converges to only root of \(f\) in \([1,2]\) which is clearly \(2\).
\(1<a_n<2\) : we have \(1<a_1<2\). Suppose \(1<a_n<2\) then \(1<{\sqrt{2}\,}^{1}<a_{n+1}={\sqrt{2}\,}^{a_n}<{\sqrt{2}\,}^2=2\).
\((a_n)_{n\geq 0} \) is increases. On \([1,2)\) define \(f(x)={\sqrt{2}\,}^x-x\) then \(f'(x)= 2^{x/2-1} \ln 2 -1 \leq f'(2)=\ln 2 -1 <0\) then \(f\) is decreasing then \(f(x)\geq f(2)=0\) for any \(x\in [1,2)\) then \(a_{n+1}={\sqrt{2}\,}^{a_n}>a_n \) for any integer \(n\).
Then it converges to only root of \(f\) in \([1,2]\) which is clearly \(2\).
admin
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 10 guests