# Cost Analysis in Primary Health Care: A Training Manual for by Andrew Creese, David Parker By Andrew Creese, David Parker

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Example text

Thus, we want 3/(k + 3) < Ik/(k + 3) — 11 = — 3/(k + or k> (3 — 3E)/E for k sufficiently large. This suggests that N be chosen as the first integer larger than (3 — 3E)/E. Then for k N, we have I k = k+3 as 3 < 3 k+3N+3

Theorem 22. If Xk then { Xk } converges to sup { Xk: k E }. Proof If { Xk } is bounded, then Theorem 11 gives the result. Assume that { Xk } is not bounded. Then since { Xk } is bounded below, for such that XN M. If k N, then Xk M; XN each M> 0, there is N E The following proposition gives some of the algebra for infinite limits. Sequences 33 Proposition 23. Let Xk (i) If { Yk } is bounded below, then lim (xk + Yk) cia. (ii) If t > 0, then tXk (iii) l/xk 0. Proof. For (i), suppose Yk b for all k.

By the definition } of supremum, there is N such that x — to i/i. Choose x1 > 0 We'll inductively construct a sequence that converges arbitrarily and set 0. Xk Xk+1 + a/xk 2 for kEN. We first show that (Xk) is bounded below. Since Xk is a real root of the quadratic equation — 4a must — 2Xk+1Xk + a = 0, the discriminant be nonnegative, that is, a. Using the definition of Xk+ we obtain Xk Xk — Xk+1 = Xk — + a/xk 2 = — a)/2xk 0, so that { Xk } is decreasing for k 2.