MATH 2300-004 QUIZ 6 Name: 1. What is a sequence? What does it mean for a sequence to converge? Give an example of a convergent sequence and of a divergent sequence. A sequence (an)n is an ordered list of real numbers. A sequence converges, lim an = L; n!1 if there is a real number L to which the an are approaching as n increases, i.e. an becomes arbitrarily close to L for n sufficiently large. [More precisely, the sequence an converges to L if for any error > 0 (no matter how small!), there is an index N > 0 (possibly very big!) such that jan − Lj < when n ≥ N. In other words, when you go out further than N terms in the sequence, the terms of the sequence are within the specificed small error around L.] Here are some convergent sequences and their limits: 1 1 1 1 1 1 1 (an)n=1 = = 1; ; ; ; ;::: ; n n=1 2 3 4 5 1 lim an = lim = 0 n!1 n!1 n 0 p q p p 1 1 p 1 π 1 2 + 2 2 + 2 + 2 (bn)n=0 = cos = @−1; 0; p ; ; ;:::A ; 2n n=0 2 2 2 π π lim bn = lim cos = cos lim = cos(0) = 1; n!1 n!1 2n n!1 2n n 1 1 (−1) 3 2 5 4 (cn)n=1 = 1 + = 0; ; ; ; ;::: ; n n=1 2 3 4 5 (−1)n lim cn = lim 1 + = 1: n!1 n!1 n Here are some divergent sequences: 1 n 1 (an)n=0 = (2 )n=0 = (1; 2; 4; 8; 16;:::) exponential growth; p p p p ! 1 1 nπ 3 3 3 3 (bn)n=0 = sin = 0; ; ; 0; − ; − ;::: 3 n=0 2 2 2 2 periodic sequence; 1 n 2 1 (cn)n=0 = (−1) (n + 1) n=0 = (1; −2; 5; −10; 17;:::) grows in absolute value, alternating in sign: 2. What is a series? What does it mean for a series to converge? Give an example of a convergent series and of a divergent series. 1 X A series an is (a formal expression for) an attempt to add infinitely many numbers n=0 an, the terms of the series. A series is convergent if the limit of the partial sums N X lim an = lim (a0 + a1 + ::: + aN−1 + aN ) N!1 N!1 n=1 exists. In other words, we add up finitely many terms (in order) from a given sequence an and see if the partial sums approach a limiting value. Here are some convergent series: 1 X 1 1 1 1 = 1 + + + ::: = = 2; 2n 2 4 1 − 1=2 n=0 a convergent geometric series, r = 1=2, jrj < 1. 1 X 1 1 1 1 + + + ::: = 1; n(n + 1) 2 6 12 n=1 X 1 a telescoping series, or convergent by comparison to . n2 n 1 X 1 1 1 π2 = 1 + + + ::: = ; n2 4 9 6 n=1 a convergent p-series, p = 2 > 1. [The fact that the sum is π2=6 is a non-trivial fact we won't show in class.] Here are some divergent series: 1 X 2n; a divergent geometric series; r = 2; jrj ≥ 1; n=0 1 X 1 p ; a divergent p-series; p = 1=2 ≤ 1; n n=1 1 X 2n2 − 1 X 1 ; divergent by comparison to the harmonic series : 3n3 + 1 n n=0 n 3. Use the integral test to determine the convergence or divergence of the series 1 X 1 . n(ln n)2 n=2 1 The function f(x) = x(ln x)2 is positive and decreasing on the interval [2; 1) (positive because x and (ln x)2 are positive there, decreasing because the numerator is fixed and the denominator is increasing). The series therefore behaves like the improper integral Z 1 dx Z T dx Z ln T du 2 = lim 2 = lim 2 2 x(ln x) T !1 2 x(ln x) T !1 ln 2 u 1 ln T 1 1 1 = lim − = lim − + = < 1: T !1 u ln 2 T !1 ln T ln 2 ln 2 The improper integral converges, therefore the series converges as well. 4. Use the comparison test to determine the convergence or divergence of the series 1 X 3n − n3 . 5n + n5 n=0 We can use either the direct comparison test or the limit comparison test, in both cases 1 X comparing to the convergent geometric series (3=5)n. n=0 For the direct comparison test, we have 3n − n3 3n 0 ≤ ≤ ; 5n + n5 5 so the series in the problem statement is dominated by the convergent geometric series P n n(3=5) . For the limit comparison test, we have 3n−n3 n3 n 5 1 − lim 5 +n = lim 3n = 1 n!1 (3=5)n n!1 n5 1 + 5n since n5 n3 lim = lim = 0 n!1 5n n!1 3n e.g. using L'H^opital'srule a few times. Therefore the series in the problem statement P n has the same convergence behavior as the convergent geometric series n(3=5) ..