Advanced Calculus fi..

400 Advanced Calculus, Fifth EditionIt will now be shown how such a monotone decreasing sequence Tn can be foundwhen the series converges by one of the following tests: comparison test, integraltest, ratio test, root test, alternating series test.THEOREM 22 If la, I ( bn for n 2 nl and C:, bn converges, thenfor n 2 nl; the sequence Tn is monotone decreasing and converges to 0.,ProoJ: By definitionSince, for n 2 n 1,"+PR. =an+l +..-+an+, + --. = lim x a,.P+oom=n+llan+l + . -+ an+,, I 5 Ian+~ 1 + .. + lan+plt i"52 e, , .m=n+l, IJone concludes that1tninqx"+P 00IR~I= Iim aml 5 bm = T".P+w m=n+l m=n+~Since Tn is the remainder, after n terms, of a convergent series of positive terms, itis necessarily a monotone sequence converging to 0.The theorem can be stated as follows: If a series converges by the comparisontest, then the remainder is in absolute value at most equal to that of the comparisonseries.rrnEXAMPLE 1 The series zz, & can be compared with the geometric seriesC(1/2"). Hence the remainder after five terms is at mostIn general, Tn = 2-"-I + 2"-2inequality 2" > l/r, so that+ . . . = 2-". l'hgkondition T, < r leads to then > -(log E/ log 2);we can thus choose N(E) as the smallest integer n satisfying this inequality. "THEOREM 23 If the series xzl an converges by the integral test of Theorem 14,with the function f (x) decreasing for x > c, then' .%*>t+YIR~I f(x)dX = Tnnfor n 2 c; the sequence Tn is monotone decreasing and converges to 0.< SW"E tf