limit of $a_{n+1}= \sqrt{ \frac{1+a_n}{2}}$ with $a_1 \ge -1$

Question

I'm given this recursive succession:

$a_{n+1}= \sqrt{ \frac{1+a_n}{2}}$ with $a_1 \ge -1$

I have to find the limit.

The condition $a_1 \ge -1$ makes $a_2$ definited, and as a conseguence all the terms of the sequence are definited.

Besides this, the sequence is definitely positive.

If I suppose there is a limit: $L=\sqrt{ \frac{1+L}{2}}$ from which I have the two values $L_1=1$ and $L_2=- \frac{1}{2}$

Being always positive $L_1=1$ is impossible.

I have tried numericaly to find the limit and it seems that if I take $a_0 \ge - \frac{1}{2}$ the limit is 1 from right, while if I take $a_0 \le - \frac{1}{2}$ the limit is 1 from left, but I don't know how to prove it.

Answer 1

In terms of finding the limit, you can always ignore the first term $a_1$, because $\forall a_1\geq-1$ we have $$a_1+1\geq0\Rightarrow a_2=\sqrt{\frac{1+a_1}{2}}\geq0$$ so the $2$nd term is always positive. Thus, we can reduce the problem to $a_{n+1}=\sqrt{\frac{1+a_n}{2}}$ and $a_1\geq0$.

One extra observation is that

• if $0\leq a_1 \leq 1$ then $1\leq\frac{a_1+1}{2}\leq1$ and by induction $0\leq a_n\leq1, \forall n$.
• if $a_1 > 1$, then $a_1>\frac{a_1+1}{2}>1\Rightarrow a_1>\sqrt{a_1}>a_2>1$ and by induction $a_1>a_n>1, \forall n$.

That being written ... use Banach fixed-point theorem, for $a_{n+1}=f(a_n)$ where $f(x)=\sqrt{\frac{1+x}{2}}$ and $f'(x)=\frac{1}{2\sqrt{2(1 + x)}}$. Using MVT $$|f(x)-f(y)|=|f'(\varepsilon)|\cdot |x-y|= \frac{1}{2\sqrt{2(1 + \varepsilon)}}\cdot|x-y|< \frac{1}{2\sqrt{2}}\cdot|x-y|$$ for

• $x,y\in[0,1]$ in the $1$st scenario above.
• $x,y\in[1,a_1]$ in the $2$nd scenario above.

Similar to another question you asked. So, the limit exists because $\frac{1}{2\sqrt{2}}<1$ and you can legitimately use $L=\sqrt{\frac{L+1}{2}}$ to find it. Because the sequence is always positive (except $a_1$, but we decided to ignore this case), the limit must be non-negative, i.e. $L=1$.

Answer 2

Note that $a_n\gt1\implies a_{n+1}\gt1$ and $a_n\lt1\implies a_{n+1}\lt1$ for all $n\in\mathbb{N}$. So if $a_1\gt1$ then $a_n\gt1$ for all $n\in\mathbb{N}$ and if $a_1\lt1$ then $a_n\lt1$ for all $n\in\mathbb{N}$. Also we have that $a_n\gt1\implies a_{n+1}\lt a_n$ and $a_n\lt1\implies a_{n+1}\gt a_n$ for all $n\in\mathbb{N}$. These two conditions imply that $a_n$ converges as $n\to\infty$ by the monotone convergence theorem. (There is also the trivial case $a_1=1$ in which $a_n=1$ for all $n\in\mathbb{N}$). One can then say that $a_n\to1$ as $n\to\infty$ because of the fixed point that you found by solving $a_{n+1}=a_n$.

Answer 3

Note that if $\lim_{n\to\infty}a_n=a$ exists, then because of the continuity of the square root function, $a$ must satisfy $$ a = \sqrt{\frac{1+a}2} \implies a = 1. $$ Now, if $a_1=1$ then $a_{n+1} = \sqrt{(1+1)/2} = 1$ for all $n$, so clearly $\lim_{n\to\infty} a_n=1$. Now consider the function $f:[-1,\infty)\to\mathbb R$ given by $f(x) = \sqrt{\frac{1+x}2}$. Since $\frac{f(x)}x>1$ for $-1<x<1$ and $\frac{f(x)}x<1$ for $1<x<\infty$, we see that the sequence $a_n$ is increasing when $-1\leqslant a_1<1$ and decreasing when $1<a_1<\infty$.

If $-1\leqslant a_1<1$, then clearly $a_1\leqslant 1$. Assume that $a_n\leqslant 1$ for some positive integer $n$, then because the square root function is increasing, $$ a_{n+1} = \sqrt{\frac{1+a_n}{2}}\leqslant\sqrt{\frac{1+1}2}=1. $$ Therefore the sequence $a_n$ is bounded above by $1$. A monotone increasing sequence that is bounded above converges to its least upper bound, so we have $\lim_{n\to\infty} a_n = \sup_n a_n = 1$.

If $a_n>1$, we may similarly show that $a_n$ is decreasing and bounded below by $1$, and therefore also converges to $1$.

Answer 4

There is an easy exact soluiton to this problem.

$$ a_n=\sqrt{\frac{a_{n-1}+1}{2}} $$

Let $a=\cos b_n$, so that

$$ \cos b_n=\sqrt{\frac{\cos b_{n-1}+1}{2}}=\cos \frac{b_{n-1}}{2} $$

Thus

$$ b_n=\frac{b_{n-1}}{2}\\ b_n=b_02^{-n} $$

and finally,

$$ a_n=\cos \bigg(\cos^{-1}(a_0)2^{-n} \bigg) $$

Clearly, $L=\lim_{n\to \infty}a_n=1$.