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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.