Limit of a product I While reviewing old problems in American Mathematical Monthly the following problem was encountered. What are some methods to solving the problem ?

Proposed by L. S. Johnston, 1929. 
Consider the infinite sequence $\{ a_{n} \}$ of real positive numbers with the recurrent relation
  \begin{align}
a_{k+1}^{2} = \frac{2 \, a_{k}}{a_{k} + 1}
\end{align}
  for $k \geq 1$. 
  
  
*
  
*Prove $\lim_{k \to \infty} a_{k} = 1$ for every $a_{1}$ 
  
*Prove 
  \begin{align}
\lim_{n \to \infty} \, \prod_{k=1}^{n} \{ a_{k} \}
\end{align}
  exists and is different from zero for every $a_{1}$
  
*Express the limit in (2) as a function of $a_{1}$.
  
  
  It is to be noted that the problem is trivial for $a_{1} = 1$.

 A: L. S. Johnston's solution:
obtained from: American Mathematical Monthly vol 36, issue 4, 1929, p. 235, problem 3313.  
If $a_{1} = 1$ it is evident that $a_{k} = 1$ for all $k$'s, and that 
$\lim_{n\to\infty} \, \prod_{k=1}^{n} \{a_{k}\} =1$.
Consider next the case for which $a_{1} > 1$ and set $a_{1} = \sec(\omega)$, $0 < \omega < \frac{\pi}{2}$. Then from the identity
\begin{align}
\sec^{2}\left(\frac{\omega}{2}\right) = \frac{2 \, \sec(\omega)}{1 + \sec(\omega)},
\end{align}
we may set $a_{2} = \sec\left(\frac{\omega}{2}\right)$. Repeating this reasoning we obtain $a_{n} = \sec(2^{1-n} \, \omega)$. Hence it follows that $a_{n}$ approaches the limit unity, and it decreases to this limit except in the trivial case $a_{1} =1$. We may now set
\begin{align}
\frac{1}{\prod_{k=1}^{n} \{a_{k}\}} &= \cos(\omega) \, \cos(2^{-1} \omega) \cdots \cos(2^{1-n} \, \omega) \\
&= \frac{\sin(2 \omega)}{2^{n} \, \sin(2^{1-n} \omega)} = \frac{\sin(2 \omega)}{2 \, \omega} \, \frac{2^{1-n} \, \omega}{\sin(2^{1-n} \, \omega)}, 
\end{align}
where the second form results by thetransformation of each factor by means of the formula 
\begin{align}
\cos(A) = \frac{\sin(2A)}{2 \, \sin(A)}.
\end{align}
Hence we have
\begin{align}
\lim_{n \to \infty} \prod_{k=1}^{n} \{ a_{k} \} = \frac{2 \, \omega}{\sin(2 \omega)} = \frac{a_{1}^{2} \, \sec^{-1}(a_{1})}{\sqrt{a_{1}^{2} - 1}}.
\end{align}
For the case in which $a_{1} < 1$ we may set $a_{1} = sech(\omega)$, $\omega > 0$. We have merely to replace the trigonometric formulae by the corresponding hyperbolic formulae, and the resoning follows in a similar manner. We thus find that $a_{n}$ approaches unity as a limit and increases toward this limit, while
\begin{align}
\lim_{n \to \infty} \prod_{k=1}^{n} \{ a_{k} \} = \frac{a_{1}^{2} \, sech^{-1}(a_{1})}{\sqrt{1 - a_{1}^{2}}}.
\end{align}
A: I'll handle (1) here
You might consider this a counter example $a_1=0$, otherwise we'll consider $0$ outside the range of positive real numbers.
Assuming the series converges it's easy to prove which values it will converge to. Say the value is $k$. Substitute that value into the recurrence.
$$k=\sqrt{{{2 \cdot k} \over {k+1}}}$$
We can do this since the relation converges.
Solving for k, we get
$k=0$ or $1$
Thus excluding $a_1=0$, any choice of $a_1$ converges to 1.
This part gives a start for handling (2)
The recurrence relation is
$$a_{n+1}=f(a_n)$$
Defining $f(x)$ according to
$$f(x)=\sqrt{{{2 \cdot x} \over {x+1}}}$$
Using the fixed point theorem from the theory of dynamical systems, we can see the global behavior of the recurrence relation. Taking the derivative of $f$ at $k$ we get a value of $1/4$. This means the fixed point $k=1$ is an attractive fixed point. It also means that the convergence is monotone. In other words values of $a_1 \lt 1$ have $a_n$ that are strictly increasing. While $a_1 \gt 1$ have $a_n$ that are strictly decreasing. Putting this all together, we automatically know the product of $a_n$ with $a_1 \gt 1$ will have a value that is real and not equal to $0$. A similar proof works for the other $a_1$.
