Show that we can express any irrational number α ∈ (0, 1) uniquely in the form ∑_{1}^{∞} (-1)^{n+1} 1/(a_{1}a_{2} ... a_{n}), where a_{i} is a strictly monotonic increasing sequence of positive integers. Find a_{1}, a_{2}, a_{3} for α = 1/√2.

**Solution**

Answer: a_{1} = 1, a_{2} = 3, a_{3} = 8.

Let s_{n} be the sum of the first n terms. The terms alternate in sign and decrease in absolute value, so the odd terms of the sequence s_{n} decrease and the even terms increase. Every odd term exceeds every even term, so the odds and the evens must each converge. But s_{n} - s_{n+1} < 1/2^{n} which tends to zero, so they tend to a common limit.

Choose a_{n} as follows. Take a_{1} to be the smallest integer whose inverse exceeds α. Having chosen a_{2n-1}, take a_{2n} to be the largest integer such that s_{2n} < α. Having chosen a_{2n}, take a_{2n+1} to be the largest integer such that s_{2n+1} > α.

We have to show that these choices are always possible, or, in other words, that they yield a strictly increasing sequence a_{n}. This depends on the relation 1/k - 1/k(k+1) = 1/(k+1) (*). For suppose we have just chosen a_{2n-1}. Then we know that s_{2n-1} > α, but that if increased a_{2n-1} by 1, then s_{2n-1} would be < α. Hence, using (*), taking a_{2n} = a_{2n-1} + 1 certainly gives s_{2n} < α. On the other hand, if we take a_{2n} to be sufficiently large, then s_{2n} will be close to s_{2n-1} and hence exceed α (note that α is irrational so it cannot equal any s_{m}). So the choice of a_{2n} will exceed a_{2n-1}. A similar argument shows that a_{2n+1} exceeds a_{2n}.

So we have established that we can find a sequence a_{n} such that all the odd partial sums s_{n} exceed α and all the even partial sums are less than α. But we have also established that s_{n} tends to a limit, so that limit must be α. That establishes existence.

Suppose there is another expansion so that α = 1/a_{1} - 1/a_{1}a_{2} + ... = 1/b_{1} - 1/b_{1}b_{2} + ... . As above, we have that 1/(a_{1} + 1) < α < 1/a_{1} and also 1/(b_{1} + 1) < α < 1/b_{1}. But since a_{1} and b_{1} are both integers that implies that a_{1} = b_{1}. Suppose now that we have established that a_{i} = b_{i} for i ≤ n. Then we have that β = (-1)^{n} a_{1} ... a_{n}(α - 1/a_{1} + 1/a_{1}a_{2} - ... + (-1)^{n}/a_{1}...a_{n} ) = 1/a_{n+1} - 1/a_{n+1}a_{n+2} + ... . But we also have β = 1/b_{n+1} - 1/b_{n+1}b_{n+2} + ... . We now argue as before that β lies between 1/(a_{n+1} + 1) and 1/a_{n+1} and also between 1/(b_{n+1} + 1) and 1/b_{n+1}. Hence a_{n+1} = b_{n+1}. That establishes uniqueness.

Finally, consider α = 1/√2. We have 1/2 < 1/√2 < 1, so a_{1} = 1. We must pick a_{2} as the largest integer so that 1 - 1/a_{2} < 1/√2, or a_{2} < 2 + √2 = 3.4. So a_{2} = 3. We must pick a_{3} as the largest integer so that 1 - 1/3 + 1/3a_{3} > 1/√2 or 2 + 1/x > 3/√2 or x < 3√2 + 4 = 8.2. So a_{3} = 8.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002