Prove that the sequence $a_{n+1} =\frac{1}{2}\left(a_{n}+\frac{c}{a_{n}}\right)$ is convergent and find its limit

Let $c>0$, $a_{1} = 1$, and $$a_{n+1} =\frac{1}{2}\left(a_{n}+\frac{c}{a_{n}}\right)$$

I need to:

1. Show that $a_{n}$ is defined for every $n\geq 1$
2. Show that this sequence is convergent.
3. Find its limit.

I proved the first part by showing by induction that this sequence is positive for every $n$. To show that this sequence is convergent I'm thinking of showing that this sequence is a Cauchy series, yet can't figure out how.

For the third part I'm clueless at the moment.

Solution 1:

For (2). You have to show that the sequence is bounded and decreasing, therefore convergent.

For (3). what you need to do is set $\lim{a_n} = L$ and solve. We have
$L = (1/2)L+\frac{c}{L}$, since $\lim{a_{n+1}} = \lim{a_n}$. Solving yields $L = \sqrt{c}$.

Solution 2:

It might be interesting that the recursion

$$a(n+1) = \frac{1}{2} (a(n) + \frac{c}{a(n)})\tag{1}$$

can be solved explicitly.

The solution is

$$a(n) = \sqrt{c}\; \coth {(2^{n-1}\;\text{arccoth}( a(1)/\sqrt{c}))}\tag{2} $$

For large $n$ the argument of the $\text{coth}$ gets large as well so that the $\text{coth} \to 1$ and $a(n) \to \sqrt{c}$, as expected.

Derivation

First we normalize the problem letting

$$d(n) = a(n) \sqrt{c}$$

which leads to the recursion

$$d(n+1) = \frac{1}{2} (d(n) + \frac{1}{d(n)})\tag{3}$$

The sum of $d(n)$ and its inverse suggests the ansatz

$$d(n) = \coth(f(n)) = \frac { e^{f(n)} + e^{- f(n)} } { {e^{ f(n)} - e^{- f(n)} }}\tag{4}$$

Inserting this into the r.h.s of (3) gives

$$\frac{ e^{2 f(n)} + e^{-2 f(n)} } { {e^{2 f(n)} - e^{-2 f(n)} }}$$

which has the same form as $d(n+1)$ if we let $f(n)$ obey the recursion

$$f(n+1) = 2 f(n)\tag{5}$$

But this can be solved immediately

$$f(n) = 2^{n-1} f(1) $$

so that the solution to (3) is given by

$$d(n) = \coth {(2^{n-1}f(1))}$$

The constant $f(1)$ has to be adapted to the initial condition leading to

$$f(1) = \text{arccoth}( d(1)) $$

Hence the solution to (3) becomes

$$d(n) = \coth {(2^{n-1}\text{arccoth}( d(1)))} $$

And, reversing the nomalization, we find the solution (2).

Solution 3:

Hints:

1. One can prove that if $a_1>0$ and $n \geq 2$ then $a_n \geq \sqrt{c}$.
2. One can prove that if $a_1>0$ and $n \geq 2$ then $a_{n+1} \leq a_n$. Knowing that and the bound in 1, convergence follows.
3. Once you know that a recursive sequence is convergent, its limit can only be a fixed point of the recursion mapping, i.e. in your case a solution to $\frac{x}{2}+\frac{c}{2x}=x$.