Question 549775
Is your sequence kn or k^n? You typed kn, which is interpreted as "k times n" but in question iii) but you implied that the limit as n goes to infinity of kn/k^n (where 0 < k < 1) is zero.


Either way, you can use limits or any other technique. If you mean k^n, part i) is simple, because 1^n is always 1 and 0^n is always 0.


For part ii), we can actually prove it by assuming that on the other hand k^n converges to some number X. If this is so, then k(k^n) must also converge (to kX). However, k(k^n) is equivalent to k^n (since we are evaluating where n approaches infinity) so kX = X. This implies X = 0 (since k is not 1), contradiction. Hence k^n diverges.


For part iii), assume on the other hand it diverges. We have |k(k^n)| < |k^n| (since |k| < 1). This is also a contradiction because we assumed it diverges, so it converges. By the same argument in ii), it converges to 0.