Question 549775: Let k > 0 be a constant and consider the important sequence {kn}. It’s behaviour as n tends to infinity will depend on the value of k.
(i) State the behaviour of the sequence as n tends to infinity when k = 1 and when k = 0.
(ii) Prove that if k > 1 then kn tends to infinity as n teds to infinity
(hint: let k = 1 + t where t > 0 and use the fact that (1 + t)n > 1 + nt.
(iii) Prove that if 0 < k < 1 then kn tends to 0 as n tends to infinity .
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! 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.
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