Let f be a positive integer be define recursively by $f(1)=1$ and $f(n+1)=\sqrt{2+f(n)}$ for all integers n. Prove that $f(n) = (2^n)-1$.

Question

Let f: be a positive integer defined recursively by $f(1)=1$ and $f(n+1)=\sqrt{2+f(n)}$ for all integers n. Prove that $f(n)<2$.

I am supposed to prove this by using proof of induction. I've tested the base case of 1, and I've let $P(n)$ be the statement that $f(n)<2$. I'm not sure how to go about the rest. I can't seem to prove that it is true for the case of $n+1$.

See this closely related answer for a standard approach to these problems. — Andrés E. Caicedo, Sep 29 '13 at 18:17

score 1 · Answer 1 · answered Sep 29 '13 at 18:14

1

Hint: $\sqrt{2+2}=2$. ${}{}{}{}{}{}$

answered Sep 29 '13 at 18:14

njguliyev

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Let f be a positive integer be define recursively by $f(1)=1$ and $f(n+1)=\sqrt{2+f(n)}$ for all integers n. Prove that $f(n) = (2^n)-1$.

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