Lab 2 Convergence/Divergence Testing - Some Guidelines Sample Solutions 1. If a series is expressed using sigma notation, write out the first few terms to get a feel for what the series looks like. Try: ∞ ∑ (-1)n (n/[n+1]) = -1/2 + 2/3 – 3/4 + 4/5 – 5/6 + … n=1 2. Is the series a geometric series? If so, it’s easy to see whether it converges or diverges just by determining the magnitude of the constant ratio. If it converges, you can calculate the sum by a formula, page 408. Try: ∞ a. ∑ (e)-n n=1 This is a geometric series with first term 1/e and constant ratio 1/e (whose magnitude is less than 1), so the series 1 1 converges to e = . 1 e 1 1− − e ∞ b. ∑ (1/2)-n n=1 This is a geometric series with first term 2 and constant ratio 2 (whose magnitude is greater than 1), so the series diverges. 3. Is the limit of the nth term of the series equal to zero? If not, the series diverges (page 413, Thm 9.2 #3). Try: ∞ ∑ [n+2]/ [5n+17] n=1 2 1 n + 2 + 1 The limit of the nth term, [n+2]/ [5n+17], is lim = lim n = . n→∞ 5n 17 n→∞ 17 5 + 5 + n Since 1/5 is not equal to 0, the series diverges. 4. Is each term of the series positive? If so, a comparison test (page 417) may be used. Try: ∞ a. ∑ 1 / [2n + 1] n=1 The nth term, 1 / [2n + 1], is less than 1/2n , the nth term of a ∞ convergent geometric series, so the series ∑ 1/[2n+1] n=1 converges. ∞ ∑ [(e)-n - (3)-n] n=1 ∞ ∞ Both the series ∑ [(e)-n and ∑ (3)-n converge, so the series n=1 n=1 ∞ -n -n 1 1 ∑ [(e) - (3) ] converges to their difference − . n=1 e − 1 3 − 1 5. Is the series an alternating one in which the absolute values of the terms are decreasing? If so, the alternating series test (page 420) may be used. Try: ∞ a. ∑ (-1)n (1/[n+1]) n=1 The factor (-1)n insures that the series is alternating. Since the absolute value of the (n+1)st term, 1/(n+1), is less than the absolute value of the (n)th term, 1/n, the absolute values of 1 the terms are strictly decreasing. Finally, lim = 0 , so by the n→∞ n + 1 Alternating Series Test, the series converges. ∞ b. ∑ (-1)n+1 ([n+2]/ [n+1]) n=1 2 1 n + 2 + lim = lim n = 1 , which is not zero, so the series must diverge. n→∞ n 1 n→∞ 1 + 1+ n ∞ c. ∑ (-1)n (1/3) n n=1 As in part a, the factor (-1)n insures that the series is n+1 n 1 1 n n+1 alternating. Since ( ) < ( ) ((3) < (3) ), i.e., the absolute value 3 3 of the (n+1)st term is less than the absolute value of the (n)th term, the absolute values of the terms are strictly n 1 decreasing. Finally, lim , so by the Alternating Series Test, n ( ) →∞ 3 the series converges. 6. If none of the above works, try the Ratio Test (page 419). Try: ∞ a. ∑ (-1)n (n/5 n) n=1 The absolute value of the ratio of the (n+1)st term to the nth n 1 + 1 n+1 n 1 n + 1 + 1 term is 5 = ⋅ 5 = n ⋅ , which approaches 1/5 as n n n n+1 1 5 n 5 5 approaches infinity. By the ratio test, the series converges absolutely. ∞ b. ∑ (n !)/(2n)! n=1 The absolute value of the ratio of the (n+1)st term to the nth (n + 1)! 1 1+ [2(n + 1)]! (n + 1)! (2n)! (n + 1) 1 1 term is = ⋅ = ⋅ = n ⋅ , n! n! (2n 2)! 1 (2n 2)(2n 1) 1 2 + + + (2 + )(2n + 1) (2n)! n which approaches 0 as n approaches infinity. By the ratio test, this series also converges absolutely. ∞ c. ∑ [(n)n /n !] (Hint: The limit, as n ∞, of [1 + (1/n)] n is a n=1 ‘famous’ limit; if you don’t recognize it, try looking at the expression for large values of n.) The absolute value of the ratio of the (n+1)st term to the nth (n+1) (n+1) (n+1) n n (n + 1)! (n+1) n! (n + 1) n+1 1 1 term is n = n ⋅ = ⋅( ) ⋅ = (1+ ) , which (n) (n) (n + 1)! 1 n n + 1 n n! approaches the number e as n approaches infinity. Since e>1, the ratio test implies that this series diverges..