IMC 1997 · Day 1 · P4

a) We construct inductively the sequence $\{n_i\}$ and the ratios $$\theta_k = \frac{\alpha}{\prod_1^k (1 + \frac{1}{n_i})}$$ so that $$\theta_k > 1 \quad \text{for all } k.$$ Choose $n_k$ to be the least $n$ for which $$1 + \frac{1}{n} < \theta_{k-1}$$ ($\theta_0 = \alpha$) so that for each $k$, $$\tag{1} 1 + \frac{1}{n_k} < \theta_{k-1} \le 1 + \frac{1}{n_k - 1}.$$ Since $$\theta_{k-1} \le 1 + \frac{1}{n_k - 1}$$ we have $$1 + \frac{1}{n_{k+1}} < \theta_k = \frac{\theta_{k-1}}{1 + \frac{1}{n_k}} \le \frac{1 + \frac{1}{n_k - 1}}{1 + \frac{1}{n_k}} = 1 + \frac{1}{n_k^2 - 1}.$$ Hence, for each $k$, $n_{k+1} \ge n_k^2$.

Since $n_1 \ge 2$, $n_k \to \infty$ so that $\theta_k \to 1$. Hence $$\alpha = \prod_1^{\infty} \left( 1 + \frac{1}{n_k} \right).$$ The uniquness of the infinite product will follow from the fact that on every step $n_k$ has to be determine by (1).

Indeed, if for some $k$ we have $$1 + \frac{1}{n_k} \ge \theta_{k-1}$$ then $\theta_k \le 1$, $\theta_{k+1} < 1$ and hence $\{\theta_k\}$ does not converge to 1.

Now observe that for $M > 1$, $$\tag{2} \left( 1 + \frac{1}{M} \right) \left( 1 + \frac{1}{M^2} \right) \left( 1 + \frac{1}{M^4} \right) \cdots = 1 + \frac{1}{M} + \frac{1}{M^2} + \frac{1}{M^3} + \cdots = 1 + \frac{1}{M - 1}.$$ Assume that for some $k$ we have $$1 + \frac{1}{n_k - 1} < \theta_{k-1}.$$ Then we get $$\begin{align*} \frac{\alpha}{(1 + \frac{1}{n_1})(1 + \frac{1}{n_2}) \dots} &= \frac{\theta_{k-1}} {(1 + \frac{1}{n_k})(1 + \frac{1}{n_{k+1}}) \dots} \\ &\ge \frac{\theta_{k-1}} {(1 + \frac{1}{n_k})(1 + \frac{1}{n_k^2}) \dots} = \frac{\theta_{k-1}}{1 + \frac{1}{n_k - 1}} > 1 \end{align*}$$ – a contradiction.

b) From (2) $\alpha$ is rational if its product ends in the stated way.

Conversely, suppose $\alpha$ is the rational number $\dfrac{p}{q}$. Our aim is to show that for some $m$, $$\theta_{m-1} = \frac{n_m}{n_m - 1}.$$ Suppose this is not the case, so that for every $m$, $$\tag{3} \theta_{m-1} < \frac{n_m}{n_m - 1}.$$ For each $k$ we write $$\theta_k = \frac{p_k}{q_k}$$ as a fraction (not necessarily in lowest terms) where $$p_0 = p, \quad q_0 = q$$ and in general $$p_k = p_{k-1} n_k, \quad q_k = q_{k-1} (n_k + 1).$$ The numbers $p_k - q_k$ are positive integers: to obtain a contradiction it suffices to show that this sequence is strictly decreasing. Now, $$\begin{align*} p_k - q_k - (p_{k-1} - q_{k-1}) &= n_k p_{k-1} - (n_k + 1) q_{k-1} - p_{k-1} + q_{k-1} \\ &= (n_k - 1) p_{k-1} - n_k q_{k-1} \end{align*}$$ and this is negative because $\dfrac{p_{k-1}}{q_{k-1}} = \theta_{k-1} < \dfrac{n_k}{n_k - 1}$ by inequality (3).