Let xn=2nan, and an+1=1−1−an22, a0=1, how to prove xn converges?

Get the solution to "Let xn=2nan, and an+1=1−1−an22, a0=1, how..."

Miguel Davenport

Answered question

2022-01-27

Let

x_{n} = 2^{n} a_{n}

, and

a_{n + 1} = \sqrt{\frac{1 - \sqrt{1 - a_{n}^{2}}}{2}}, a_{0} = 1

, how to prove

x_{n}

converges?

Answer & Explanation

Fallbasisz8

Beginner2022-01-28Added 9 answers

Since
$\frac{a_{n + 1}^{2}}{a_{n}^{2}} = \frac{1 - \sqrt{1 - a_{n}^{2}}}{2 a_{n}^{2}} = \frac{1}{2 (1 + \sqrt{1 - a_{n}^{2}})} \leq \frac{1}{2}$
we have
$a_{n}^{2} = a_{0}^{2} \prod_{k = 1}^{n} \frac{a_{k}^{2}}{a_{k - 1}^{2}} \leq 2^{- n} \Rightarrow \sum_{k = 0}^{\infty} a_{k}^{2} \leq \sum_{k = 0}^{\infty} 2^{- k} = 2$
We find
$1 \leq \frac{x_{n + 1}^{2}}{x_{n}^{2}} \leq 1 + a_{n}^{2} \leq e^{a_{n}^{2}}$
LHS tells us the sequence $(x_{n})$ is increasing while RHS tells us it is bounded from above.
$x_{n}^{2} = x_{0}^{2} \prod_{k = 1}^{n} \frac{x_{k}^{2}}{x_{k - 1}^{2}} \leq \prod_{k = 1}^{n} e^{a_{k - 1}^{2}} = e^{\sum_{k = 0}^{n - 1} a_{k}^{2}} < e^{\sum_{k = 0}^{\infty} a_{k}^{2}} = e^{2}$
As a result, sequence $(x_{n})$ converges to some number $\leq e$

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