Convergence of sequence

Number Theory Level 2

Given $a_{n+1}= \dfrac {a_n^2+1}2$ , for all natural number $n$ , and $a_1= \dfrac 12$ , is the sequence $\{a_n\}$ convergent?

No Depends on n Cant say Yes

1 solution

Simon Kaib
Sep 19, 2019

First, note that $a_{n+1}\geq a_{n}$ . Suppose that $\{a_n\}$ doesn‘t converge, then there exists some $p$ such that $a_p<1, \: a_{p+1}\geq 1$ . Now, $a_{p+1}=\frac{a_p^2+1}{2}<\frac{1^2+1}{2}=1$ , which is a contradiction. We conclude that the sequence converges.

And it has to converge to $1$ : let $L = \lim\limits_{n\to\infty} a_n;$ then $L = \frac{L^2+1}2,$ so $(L-1)^2 = 0,$ so $L=1.$

Patrick Corn - 1 year, 8 months ago

Convergence of sequence

1 solution

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