Advanced Calculus fi..

Chapter 6 Infinite Series 387If ELP=, la,l converges, then the s N, so that theseries C a, converges.One can interpret this theorem as stating that introduction of minus signs beforevarious terms of a positive term series tends to help convergence; if the originalseries diverges, introduction of enough minus signs may make the series converge;if the original series converges, introduction of the minus signs will make the seriesconverge even more rapidly.EXAMPLES(- I)"6. Cz, -iR = 1 - f + f . . . . Since the series CzI $ converges, this seriesconverges.7. Cz,(- I)". The series of absolute values is 1 + 1 + . . . + 1 + . . . Thisseries diverges, so that Theorem 11 gives no help. However the nth term failsto converge to 0, so that the series diverges.s. C,"=I=i- (-')"2 3 4s.= 1 - 1 + 1 - 1 + . . . . The series of absolute values is theharmonic series of Example 4. Although the harmonic series diverges, it willbe seen that this series converges.A series C a, that converges but is not absolutely convergent is called conditionallyconvergent. Such a series converges because the minus signs have beenproperly introduced. An example is the series of Example 8.THEOREM 12 (Comparison Test for Convergence) If la, 1 5 b, for n = 1,2, . . . and cz, bn converges, then Czl an is absolutely convergent.Proof. By Theorem 7 the series C la, I is either convergent or properly divergent.If it were properly divergent, thensince b, 2 la, 1,one would then haveso that b, would diverge, contrary to assumption. Hence C la, I converges, anda, is absolutely convergent.EXAMPLE 9CE, &. Since" 1and the series C 112" converges, the given series converges.THEOREM 13 (Comparison Test for Divergence) If a, I b, 2 0 for n =1,2, . . . and Czl bn diverges, then Czl a, diverges.