2015_III_2_Pure

(i)

True. Note that n^2 \ge 1000n iff n(n - 1000) \ge 0. This is the case for n \ge 1000, so m = 1000 works here.

(ii)

False. Let s_n = (-1)^n and t_n = (-1)^{n + 1}. Then s_{2n} > t_{2n} but s_{2n+1} < t_{2n+1} for all n \in \mathbb N so no such m can exist.

(iii)

True. Since (s_n) \le (t_n) there exists some m_1 such that s_n \le t_n for all n \ge m_1. Similarly since (t_n) \le (u_n) there exists some m_2 such that t_n \le u_n for all n \ge m_2. Then for n \ge \max\{m_1, m_2\} both of these inequalities hold, ie. we have s_n \le t_n \le u_n, so s_n \le u_n. So (s_n) \le (u_n).

(iv)

True. Note that 4^2 = 2^4. Suppose that 2^n \ge n^2. Then:

2^{n + 1} = 2 \times 2^n = 2n^2 = n^2 + n^2

Note that:

n^2 \ge 2 n + 1

iff

(n - 1)^2 - 2 \ge 0

It suffices that:

n \ge 1 + \sqrt 2

so certainly for n \ge 4 we have n^2 + n^2 \ge n^2 + 2n + 1 = (n + 1)^2.

So 2^{n + 1} \ge (n + 1)^2.

So by induction 2^n \ge n^2 for all n \ge 4. So m = 4 works here, and (n^2) \le (2^n).

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