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Let $a_{1} >0$ and let $a_{n+1} = \frac{1}{2} ( a_{n} + \frac{1}{a_{n}})$ for all $n \geq 1$. Show that $a_{n}$ converges and find it's limit.

The Attempt: I am going to use the Monotone Convergence Theorem to be able to show the sequence converges. First I am going to show the sequence is monotonically decreasing by induction. However, I am having a hard time figuring out if the sequence is monotonically decreasing.

Please give me hints to solve the problem. Please do not solve the problem completely.

Thank you very much.

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  • $\begingroup$ No matter what $a_0$ is, we definitely have $a_1\geq1$. From that point on, the sequence is decreasing. $\endgroup$ – Arthur Mar 27 '16 at 20:53
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Note that for all $n$ $$\frac{a_{n+1}}{a_n}=\frac{1}{2}+\frac{1}{2}\frac{1}{a_n^2}\leq \frac{1}{2}+\frac{1}{2}=1$$ since $a_n\geq 1$ (for the last, use the classic inequality $\displaystyle x+\frac{1}{x}\geq 2$ for $x>0$ and apply it for $a_n$, which is positive for all $n$).

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$a_{n+1} - a_n = \dfrac{\dfrac{1}{a_n}-a_n}{2}\leq 0$,and you can prove this by using AM-GM inequality to prove $a_n \geq 1$, or completing square would be sufficient: $a_{n+1} - 1 = \dfrac{(a_n -1)^2}{2a_n} \geq 0, \forall n \geq 1\Rightarrow a_{n+1} \geq 1, \forall n \geq 1$. Thus $a_{n+1}\leq a_n$, and the sequence $\{a_n\}$ is monotonically decreasing, and is bounded below by $1$, hence is convergent to $L$ which satisfies the equation: $L = \dfrac{L+\dfrac{1}{L}}{2} \to L = 1$.

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First, if $x>0$, then $x+\frac1x-2=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\geq 0$, hence $\frac{1}{2}\left(x+\frac1x\right)\geq1$. And the equality holds only when $x=1$. This means that if $a_0>0$ and $a_0\neq 1$, then $a_1>1$, and then $a_n>1$ for all $n\geq1$.

Therefore, for $n>0$, you can define $u_n=\mathrm{argcoth}\; a_n$, i.e. $a_n=\coth u_n$, with $u_n>0$, and

$$a_{n+1}=\frac12(\coth u_n+\tanh u_n)=\coth (2u_n)$$

That is, $a_n=\coth (2^{n-1}u_1)=\coth (2^{n-1}\mathrm{argcoth}\; a_1)$.

Then, since $2^{n-1}u_1 \to \infty$ as $n\to\infty$, you get $a_n\to1$.

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