Limit of a product I

Question

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 $k \geq 1$.

1. Prove $\lim_{k \to \infty} a_{k} = 1$ for every $a_{1}$
2. Prove \begin{align} \lim_{n \to \infty} \, \prod_{k=1}^{n} \{ a_{k} \} \end{align} exists and is different from zero for every $a_{1}$
3. Express the limit in (2) as a function of $a_{1}$.

It is to be noted that the problem is trivial for $a_{1} = 1$.

Answer 1

PART 1:

We are given a sequence described by the recursive relationship

$$a_{k+1}^2=\frac{2a_k}{1+a_k}\tag 1$$

We will first prove that the sequence defined by $(1)$ converges.

To that end, we form the difference

$$a_{k+1}^2-a_k^2=\frac{a_k(1-a_k)(2+a_k)}{1+a_k} \tag 2$$

and examine convergence under $3$ cases.

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Case 1: $a_k=1$

If $a_1=1$, then $a_k=1$ for all $k$ and the series converges.

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Case 2: $a_k<1$

Note from $(2)$ that for $a_k<1$, the sequence is increasing. Furthermore from $(1)$ we see that

$$a_{k+1}^2=2\frac{a_k}{1+a_k}\le2\implies a_{k}\le \sqrt{2}$$

Therefore, by The Monotone Convergence Theorem the sequence converges.

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Case 3: $a_k>1$

Note from $(2)$ that for $a_k>1$, the sequence is decreasing. Furthermore from $(1)$ we see that

$$a_{k+1}^2=2\frac{a_k}{1+a_k}>0\implies a_{k}0$$

Therefore, by The Monotone Convergence Theorem the sequence converges.

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Given that the sequence converges, suppose it converges to $L$. Then from $(1)$

$$L^2=\frac{2L}{L+1}\implies L=1\,\,\text{or}\,\,L=0$$

Since $a_k$ is a positive sequence we see

$$\bbox[5px,border:2px solid #C0A000]{\lim_{k\to \infty}a_k=1}$$

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PART 2:

Following the comment from @Did, let $a_k=b_k+1$, where $b_k\to 0$ as $k\to \infty$. Then, we have

$$a_{k+1}^2=1+2b_{k+1}+O(b_{k}^2) \tag 3$$

and

$$\frac{2a_k}{1+a_k}=\frac{1+b_k}{1+\frac12b_k}=1+\frac12 b_k+O(b_k^2) \tag 3$$

whereupon equating $(3)$ and $(4)$ we see that asymptotically

$$b_{k+1}=\frac14 b_k\implies a_k\sim 1+A4^{-k}$$

for a constant $A$ that depends on $a_1$. Now, we know that if the product $\prod_{k=1}^{\infty} a_k$ converges, then the sum $\sum_{k=1}^{\infty}\log a_k$ converges also.

Inasmuch as asymptotically $a_k\sim 1+A4^{-k}$, and $\sum_{k=1}^{\infty}\log(1+A4^{-k})\le A\sum_{k=1}^{\infty}4^{-k}=\frac43 A$ converges, then $\prod_{k=1}^{\infty} a_k$ converges.

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PART 3:

Also, following the comment from @Did, if we "guess" that the solution to the recurrence relationship in $(1)$ has the form $a_k=\sec(\phi_k)$, for $a_1>1$ (and thus, $a_k>1$), then we have

$$\begin{align} \sec^2(\phi_{k+1})&=\frac{2\sec(\phi_k)}{1+\sec(\phi_k)}\\\\ &\implies\cos^2(\phi_{k+1})=\frac{1+\cos(\phi_k)}{2}\\\\ &\implies \phi_{k+1}=\frac12\phi_k\\\\ &\implies \phi_k=\phi_1/2^{k-1}\\\\ &\implies a_k=\sec\left(\phi_1/2^{k-1}\right) \end{align}$$

with $a_1=\sec(\phi_1)$, $a_1>1$ and $0<\phi_1<\pi/2$.

Finally, we have using the so-called Morrie's Law in reverse

$$\prod_{k=1}^{\infty}\sec\left(\phi_1/2^{k-1}\right)=\frac{2\phi_1}{\sin(2\phi_1)}$$

We can prove this form of Morrie's Law simply by noting that

$$\sin 2x=2\sin x\cos x\implies \sec x=\frac{2\sin x}{\sin 2x}$$

and then iterating with $x\to x/2$ in each iteration. Thus,

$$\begin{align} \prod_{k=1}^{N}\sec(\phi_12^{-(k-1)})&=\frac{2\sin \phi_1}{\sin(2\phi_1)}\frac{2\sin(\phi_1 2^{-1})}{\sin \phi_1}\frac{2\sin(\phi_12^{-2})}{\sin(\phi_12^{-1})}\cdots\frac{2\sin(\phi_1 2^{-(N-1)})}{\sin(\phi_1 2^{-N})}\\\\ &=\frac{2^N\sin(\phi_1 2^{-(N-1)})}{\sin (2\phi_1)}\to \frac{2\phi_1}{\sin (2\phi_1)} \end{align}$$

as was to be shown.

For large $a_1>>1$, it is better to write the final result using $\sin (2\phi_1)=2\sin (\phi_1)\cos (\phi_1)=2\frac{\sqrt{a_1^2-1}}{a_1^2}$ and $\phi_1=\text{arcsec}(a_1)$. Then, we have for $a_1>1$

$$\bbox[5px,border:2px solid #C0A000]{\prod_{k=1}^{\infty}a_k=\frac{a_1^2\text{arcsec}(a_1)}{\sqrt{a_1^2-1}}}$$

In the limit as $a_1\to \infty$ the product approaches $\frac{\pi \,a_1}{2}$.

Answer 2

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 $a_1$.

Answer 3

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}