Let Pn(x, z) = ∏1n (1 - z xi-1) / (z - xi). Prove that 1 + ∑1∞ (1 + xn) Pn(x, z) = 0 for |z| > 1 and |x| < 1.
Solution
If we calculate a few partial sums, we soon find a pattern emerging. Let Sn = 1 + ∑1n (1 + xi) Pi(x, z). Then we find S1 = (1 - zx)/(z - x), S2 = S1 (1 - zx2)/(z - x2), S3 = S2 (1 - zx3)/(z - x3), ... . The general case follows by an easy induction.
But |x| < 1, so |1 - zxn| → 1, and |z - xn| → |z|. But |z| > 1, so we can find positive k < 1 such that for n sufficiently large |(1 - zxn)/(z - xn)| < k. Hence Sn → 0 as n tends to infinity.
© John Scholes
jscholes@kalva.demon.co.uk
1 Jan 2001