The sequence an is defined by a1 = 2, an+1 = an2 - an + 1. Show that any pair of values in the sequence are relatively prime and that ∑ 1/an = 1.
Solution
We show by induction on k that an+k = 1 (mod an). Obviously true for k = 1. Suppose it is true for k. Then for some m, an+k = m an + 1. Hence an+k+1 = an+k(m an + 1 - 1) + 1 = an+km an + 1 = 1 (mod an). So the result is true for all k. Hence any pair of distinct an are relatively prime.
We show by induction that ∑1n 1/ar = 1 - 1/(an+1 - 1). For n = 2, this reduces to 1/2 = 1 - 1/(3 - 1), which is true. Suppose it is true for n. Then ∑1n+1 1/ar = 1 - k, where k = 1/(an+1 - 1) - 1/an+1 = 1/(an+12 - an+1) = 1/(an+2 - 1), so it is true for n + 1. But an → ∞, so ∑ 1/an = 1.
© John Scholes
jscholes@kalva.demon.co.uk
4 Dec 1999