Advanced Calculus fi..

for when x = $, the numerator Ixn 1Chapter 6 Infinite Series 41 5has its largest value'arid the denominator 11-2 kl,has its smallest value in the interval. Accordingly,-- .- 1lim R, = lim - = 0;n+w n-rw 2"-1the maximum absolute error tends to 0 as n -+ oo, and the convergence is uniform.A similar reasoning applies to eachclosed interval -a 5 x 5 a (0 < a < 1 ) withinthe - interval of convergence. The worst absolute error occurs for x = a, and its valueis R, = an/(l - a); this tends to 0 as n + oo, so that the convergence is uniformin the interval.One might expect from this that the convergence would be uniform over theentire open interval - 1 < x < 1. This is not the case. Indeed, for each n the absoluteerror I Rn (x)( is unbounded for - 1 c x < 1, sinceThe least upper bound xn is always +oo! We can see the difficulty in detail byasking how many terms are needed to compute the sum with an absolute error lessthan E = 0.01, for example. For x = 0, one term is sufficient; for x = 0.5, one musthave(0.5)"- < 0.01 or 2"-' > 100;0.5hence n must be at least 8. For x = 0.9, one must havethis requires n > 65, as a computation with logarithms shows. As x increases toward1, more and more terms are needed; one can easily show that, as x += 1, thenumber of terms required approaches oo (see Problem 2 following Section 6.9).It appears from this discussion that the difficulty is due to the fact that the sumS(x) = 1 /(1 -x) becomes infinite as x approaches 1. However, the convergence is notuniform in the interval - 1 < x 5 0. For the absolute error I Rn(x)( lies between 0 andf in this interval; this is suggested graphically in Fig. 6.9 and can be proved strictlyby calculus. Since I Rn(x)l + as x -t -1, the least upper bound xn is always i;this is not the maximum absolute error, since x = - 1 is excluded from the intervalunder consideration. Accordingly, i?,, cannot converge to 0 as n + oo, and theconvergence is nonuniform for - 1 < x 5 0. Again we can verify that the number ofterms needed to compute S(x) with an error less than E = 0.01 approaches oo asx + -1.EXAMPLE 2 The sequence fn(x) = xn converges to 0 for -1 < x < 1 and to 1for x = 1; it diverges for all other values of x. The convergence for - 1 < x < 1 is