Convergence of recursive sequence $a_{n+1} =\frac{ 1}{k} \left(a_{n} + \frac{k}{a_{n}}\right)$

Let $$ a_{n+1} = \frac{1}{k} \left(a_{n} + \frac{k}{a_n}\right) ; k>1, a_1>0 $$ The problem is to show that it converges.

Attempt: The sequence is not monotone but it has a lower bound. It seems that odd terms subsequence and even term subsequence are monotonic sequences (I wrote some basic code to make this observation) though I am not able to prove it analytically. I also know that if odd subsequence and even subsequence converge to same limit then the sequence also converges. So, that tells me that I am on the right track.

Please provide any hints.

If it converges, one can solve immediately for $a_\infty = \frac{\sqrt k}{\sqrt{k-1}}$.

So define $b_n = a_n \frac{\sqrt{k-1}}{\sqrt k}$. This gives the recursion

$b_{n+1} = \frac{1}{k} \left(b_{n} + \frac{{k-1}}{b_{n}}\right)$

which, if convergent, gives $b_\infty = 1$.

We will show that a lower bound and an upper bound both converge to $b_\infty = 1$ from an arbitrary starting point. Hence $b_\infty = 1$ is a unique fixed point and convergence is global.

Let's begin with a lower bound.

Using $1/x = 1/(1 + x-1) \ge 1 - (x-1) = 2 -x$ we have

$b_{n+1} = \frac{1}{k} \left(b_{n} + \frac{{k-1}}{b_{n}}\right) \ge \frac{1}{k} \left(b_{n} + {{(k-1)}}{(2-b_{n})}\right) = \frac{1}{k} \left(b_{n} (2-k) + {(2k-2)}\right)$

and the RHS establishes a lower bound $b^-_n \le b_n$ given by the recursion $b^-_{n+1} = \frac{1}{k} \left(b^-_{n} (2-k) + {(2k-2)}\right) $. We can inspect this recursion to see if and where the lower bound converges to.

According to Banach's contraction theorem for iterating fixed points in the form $x_{n+1} = f(x_n)$ convergence is guaranteed if the absolute value of the derivative is less than 1, i.e. $|f'(x)| < 1$, for some interval of $x$ containing the fixed point. Here, for $b^-_n$ in particular, $f(x_n)$ is a line equation and convergence is global if the absolute slope is less than 1.

Since the absolute slope $|2-k|/k$ is always $< 1$ for $k > 1$, iterating the lower bound $b^-_n$ converges, and solving $b^- = \frac{1}{k} \left(b^- (2-k) + {(2k-2)}\right)$ results in the lower bound limit $b^-_\infty= 1$. So the lower bound converges to the actual value that the original series should attain (if convergent), which proves convergence from below.

Now for the upper bound.

Suppose $b_1 <1$. Then $b_2 >1$. So let us suppose w.l.o.g. that we start with some $b_1 >1$.

Then we have that

$b_{n+1} = \frac{1}{k} \left(b_{n} + \frac{{k-1}}{b_{n}}\right) < \frac{1}{k} \left(b_{n} +{{k-1}}\right)$

provided that $b_{n} > 1$. This is certainly true for $n=1$, and we would like to keep it that way.

So we have an upper bound $b^+_n$ given by the recursion

$b^+_{n+1} = \frac{1}{k} \left(b^+_{n} + {{k-1}}\right)$.

Clearly, if $b^+_{n} > 1$, so is $b^+_{n+1} > 1$, so our condition $b^+_{n} > 1$ holds for all $n$, starting with $b^+_{1} > 1$, and we can work with that upper bound.

Now apply again the contraction theorem. Convergence is globally established since the slope is $\frac{1}{k} <1$, and the fixed point can immediately by computed as $b^+_\infty= 1$.

Hence upper and lower bound converge globally to the same value $b_\infty= 1$, which establishes that the original series also converges, from any starting point, to $b_\infty= 1$. $\quad \Box$