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Given that k > 1 is a fixed constant, evaluate lim n→∞ 1− 1 k n . 1. 13. Example. Let a1 = 1, and for n ≥ 1 define an+1 = . Show that {an } is 7 + an convergent, and find its limit. Solution. 89 . . and it looks as though the sequence might be increasing. Let f (x) = 6(1 + x) , 7+x so f (an ) = an+1 . 2. BOUNDEDNESS AGAIN 25 By investigating f , we hope to deduce useful information about the sequence {an }. 6 − 6(1 + x) 36 = > 0. (7 + x)2 ((7 + x)2 Recall from elementary calculus that if f (x) > 0, then f is increasing; in other words, if b > a then f (b) > f (a).

Theorem (The monotone convergence principle). Let {an } be an increasing sequence which is bounded above; then {an } is a convergent sequence. Let {an } be a decreasing sequence which is bounded below; then {an } is a convergent sequence Proof. 2. Details will be given in third year, or you can look in (Spivak 1967) for an accurate deduction from the appropriate axioms for R . CHAPTER 3. MONOTONE CONVERGENCE 24 This is a very important result. It is the first time we have seen a way of deducing the convergence of a sequence without first knowing what the limit is.

Then g ◦ f is continuous at a Proof. Pick > 0. We must find δ > 0 such that if |x − a| < δ, then g(f (x)) − g(f (a))| < . We find δ using the given properties of f and g. Since g is continuous at f (a), there is some δ1 > 0 such that if |y − f (a)| < δ1 , then |g(y) − g(f (a))| < . Now use the fact that f is continuous at a, so there is some δ > 0 such that if |x − a| < δ, then |f (x) − f (a)| < δ1 . Combining these results gives the required inequality. 18. Example. 8 is continuous everywhere.