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.
 A: 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 $b_\infty= 1$. $\quad \Box$
