Verified answer

a) Since |exp{ix}| = 1 for all real x, the sequence {exp(2πik/K): k ∈ N} is bounded by 1. Let a_k = exp(2πik/K), for all k ∈ N. The subsequence a_{kK} is the constant sequence 1, so it converges to 1. On the other hand, the subsequence a_{kK + 1} is the constant sequence exp(2πi/K), so it converges to exp(2πi/K) ≠ 1. Since the two subsequences of a_k do not have the same limit, a_k is divergent.

b) Fix a positive integer n. Let w = exp(2πi/K). Then w^k = a_k and ∑(k = 0 to n) a_k = ∑(k = 0 to n) w^k = (w^n - 1)/(w - 1). Since |w^n - 1| ≤ |w|^n + 1 = 2 and |w - 1| = √[(cos(2π/K) - 1]^2 + (sin(2π/K))^2] ≥ |cos(2π/K) - 1|,

|(w^n - w)/(w - 1)| ≤ 2/|cos(2π/K) - 1|,

It follows that the sequence of partial sums of a_k is bounded by 2/|(cos(2π/K) - 1|.

Let s_n denote the nth partial sum of a_k. To prove that s_n is divergent, we follow an argument similar to the one given in part a). We have s_{Kn} = (w^{Kn} - 1)/(w - 1) = 0 since w^K = 1, and similarly s_{Kn + 1} = (w^{Kn + 1} - 1)/(w - 1) = (w - 1)/(w - 1) = 1. Hence s_{Kn} converges to 0 whereas s_{Kn + 1} converges to 1. Since the two subsequences of s_n do not have the same limit, s_n diverges.