Proof of convergence of an infinite product

Question

1. a) Show that $\Pi_{n=1}^\infty x_n$ converges if and only if for all $\varepsilon>0$ there exists an $N$ such that for all $m\ge n\ge N$, $\left|x_nx_{n+1}\cdots x_{m-1}x_m-1\right|<\varepsilon$. Also, (x-j)>0 for all j.

   b ) Show that there is an increasing sequence $x_n$ such that for all $n$, $0\le x_n<1$, and$$\Pi_{n=1}^\infty x_n=\frac12.$$(Hint: The existence of such a sequence $x_n$ can be established either constructively, with an explicit formula, or non-constructively, via an inductive construction.)

Answer 1

1. a) I know this can be done by taking logarithms and using the Cauchy criterion for sums but I'll try to do it using just definition of limit. Assume $x_j \neq 0$ for all $j$.

Assume $\prod_{j=1}^\infty x_j$ converges to $a \neq 0$. Let $\epsilon > 0$. Choose $N \in \mathbb{N}$ s.t. $|\prod_{j=1}^s x_j - a| < \max (\frac{|a|}{2}, \frac{\epsilon}{|a|})$ for all $s>N$. Then

$$ |\prod_{j=n}^m x_j - 1| = |\frac{\prod_{j=1}^m x_j}{\prod_{j=1}^{n-1} x_j} - 1 |$$

$$= |\frac{\prod_{j=1}^m x_j - \prod_{j=1}^{n-1} x_j - a +a}{\prod_{j=1}^{n-1} x_j-a+a} |$$

$$ \leq \frac{|\prod_{j=1}^m x_j - a|+ | \prod_{j=1}^{n-1} x_j - a|}{|a| - |\prod_{j=1}^{n-1} x_j-a|}$$,

$$ < \frac{\frac{\epsilon}{|a|} + \frac{\epsilon}{|a|}}{\frac{|a|}{2}} = \epsilon$$

when $n, m > N+1$.

I won't do the other direction from beginning, since proving cauchy $\Rightarrow $ convergent, involves proving that the sequence is bounded, getting a convergent subsequence by Bolzano-Weierstrass and then proving that the original sequence also converges to the same limit as the subsequence. So let's assume we know that a Cauchy sequence converges. We only need to show that the sequence is cauchy assuming that

for all $\epsilon > 0$ there is $N \in \mathbb{N}$ s.t. for all $m,n > N$ it holds that $$|\prod_{j=n}^m x_j - 1| < \epsilon.$$

Let $\epsilon > 0$. It follows from the assumption that the sequence $(\prod_{j=1}^m x_j)$ is bounded, let's say by $M$. Pick $N \in \mathbb{N}$ s.t. for all $m,n > N$ it holds that $|\prod_{j=n}^m x_j - 1| < \frac{\epsilon}{M}.$ Now

$$ |\prod_{j=1}^m x_j - \prod_{j=1}^n x_j| = |\prod_{j=1}^n x_j||\frac{\Pi_{j=1}^m x_j}{\prod_{j=1}^n x_j} - 1| < M \frac{\epsilon}{M} = \epsilon $$

whenever $m,n > N$.

1.b)

Let $q \in (0,\frac{1}{2})$. Consider the sequence $x_j = 1-q^j$. The product is then the Pochammer symbol $(q;q)_\infty$ (look: http://mathworld.wolfram.com/q-PochhammerSymbol.html ). When you vary $q$ from $0$ to $\frac{1}{2}$ the product changes from $1$ to something less than $\frac{1}{2}$ (since with $q=\frac{1}{2}$ the product has first term $\frac{1}{2}$ and others smaller than $1$). The product is also continous in $q$, so it must also obtain the value $\frac{1}{2}$ for some $q \in (0,\frac{1}{2})$.