Question

If (a_n)^infinity_n=1 is a sequence of real numbers satisfying a_n+1² =2a_n-1, then,

lim_n->infinity a_n = ?

Answer 1

Let $a_1=p,$ so

$a_2=\sqrt{2p-1},$

$a_3=\sqrt{2\sqrt{2p-1}-1},$

$a_4=\sqrt{2\sqrt{2\sqrt{2p-1}-1}-1}, ...,$ (can be seen, that $p\geqslant 1$)

$a_k\sim (2p)^{\frac{1}{k-1}}.$

So $\underset{n→\infin}{lim}a_n=\underset{n→\infin}{lim}(2p)^{\frac{1}{n-1}}=\underset{n→\infin}{lim}2^{\frac{1}{n-1}}p^{\frac{1}{n-1}}=1\cdot1=1.$