A First Course in Mathematical Analysis by J. C. Burkill

By J. C. Burkill

This easy path in response to the belief of a restrict is meant for college students who've received a operating wisdom of the calculus and are prepared for a extra systematic therapy which additionally brings in different restricting tactics, corresponding to the summation of endless sequence and the growth of trigonometric services as strength sequence. specific consciousness is given to readability of exposition and the logical improvement of the subject material. lots of examples is incorporated, with tricks for the answer of lots of them.

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Converges to sum t, then the series (u + vi) + (u2+ v2) + . . converges to sum s + t. Exercise. Prove, more generally, that the series whose nth term is au„+ bv„, where a and b are constants, converges to sum as+ bt. (3) If ul+ u2+ ... converges, then lira un = 0. Proof. un = sn — sn-1. Both sn and s,1tend to the same limit s. 53 (i), lim u„ exists and lim u„ = s—s = O. I Note carefully that the converse of (3) is false. The example un, = 1In shows that it is possible to have lim un = 0 and Zun divergent.

2 2 sn_a —s„ If s„ > a, the right-hand side is negative and so Also 4+1 < sn• 4+1 — a2 = (sn+a)— (a +a) = sn — a. This shows that, if s,, > a, then sn+i > a. Suppose that b = s1 > a. Then, by induction, s„ forms a decreasing sequence with s„ > a for all n. 6, 4, tends to a limit s where s a, and, by the argument of remark (ii), s = a. Similarly if si< a, then s„ forms an increasing sequence with limit a. If si= a, every s„ = a. Graphical representation. For recurrence relations of the special form 4+1 = f(s„) the movements of snmay be traced by drawing the graphs of the curve C and the line L whose respective equations are y = f(x) and y = x and making the following construction.

Converges to sum s + t. Exercise. Prove, more generally, that the series whose nth term is au„+ bv„, where a and b are constants, converges to sum as+ bt. (3) If ul+ u2+ ... converges, then lira un = 0. Proof. un = sn — sn-1. Both sn and s,1tend to the same limit s. 53 (i), lim u„ exists and lim u„ = s—s = O. I Note carefully that the converse of (3) is false. The example un, = 1In shows that it is possible to have lim un = 0 and Zun divergent. In other words (see exercise 2 (g), 1) The condition lim un = 0 is necessary but not sufficient for the convergence of Zun.

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