Let P_{n}(x, z) = ∏_{1}^{n} (1 - z x^{i-1}) / (z - x^{i}). Prove that 1 + ∑_{1}^{∞} (1 + x^{n}) P_{n}(x, z) = 0 for |z| > 1 and |x| < 1.

**Solution**

If we calculate a few partial sums, we soon find a pattern emerging. Let S_{n} = 1 + ∑_{1}^{n} (1 + x^{i}) P_{i}(x, z). Then we find S_{1} = (1 - zx)/(z - x), S_{2} = S_{1} (1 - zx^{2})/(z - x^{2}), S_{3} = S_{2} (1 - zx^{3})/(z - x^{3}), ... . The general case follows by an easy induction.

But |x| < 1, so |1 - zx^{n}| → 1, and |z - x^{n}| → |z|. But |z| > 1, so we can find positive k < 1 such that for n sufficiently large |(1 - zx^{n})/(z - x^{n})| < k. Hence S_{n} → 0 as n tends to infinity.

© John Scholes

jscholes@kalva.demon.co.uk

1 Jan 2001