MATH 409 Advanced Calculus I Lecture 15: Tests for convergence. Convergence of infinite series ∞ [Cauchy Criterion] An infinite series n=1 an converges if and only if for every ε> 0 there existsPN N such that ∈ m n N implies an + an + + am <ε. ≥ ≥ | +1 ··· | [Trivial Test] If the terms of an infinite series do not converge to zero, then the series diverges. ∞ [Series with nonnegative terms] A series n=1 an with nonnegative terms converges if and only if theP sequence of partial sums sn = a + a + + an is bounded above. 1 2 ··· [Direct Comparison Test] Suppose that an, bn 0 for all ≥ n N and an bn for large n. Then convergence of the ∈ ∞ ≤ ∞ series n=1 bn implies convergence of n=1 an while ∞ ∞ n aPn = implies n bn = . P =1 ∞ =1 ∞ P P Harmonic series ∞ 1 1 1 1 =1+ + + + + ... diverges. • n 2 3 · · · n Xn=1 Indeed, 1 1 2 1 1 1 1 1 4 1 + > = , + + + > = ,... 3 4 4 2 5 6 7 8 8 2 2k+1 1 2k 1 In general, > = . n 2k+1 2 n=2Xk +1 2k 1 k It follows that > 1+ . n 2 Xn=1 Thus the sequence of partial sums is unbounded. Cauchy’s Condensation Test Theorem Suppose an is a nonincreasing sequence of { } ∞ positive numbers. Then the series an converges if n ∞=1 X k and only if the “condensed” series 2 a2k converges. k X =1 Proof: Since an is a nonincreasing sequence, for any i N i { } i ∈ we have 2 a2i+1 a2i +1 + a2i +2 + a2i +3 + + a2i+1 2 a2i . Summing this up≤ over i from 1 to k, we obtain··· ≤ k 2k+1 k i i 2 a2i+1 an 2 a2i . i n i X =1 ≤ X =3 ≤ X =1 Let sn be the partial sums of the original series and tk be the partial{ } sums of the condensed one. The latter inequaliti{ } es 1 are equivalent to (tk t ) s k+1 s tk . It follows 2 +1 − 1 ≤ 2 − 2 ≤ that the sequence tk is bounded if and only if the sequence { } s k is bounded. Since the sequence sn is nondecreasing, { 2 } { } it is bounded if and only if its subsequence s k is bounded. { 2 } Examples ∞ 1 1 1 1 =1+ + + + + ... , p > 0. n=1 np p p np • X 2 3 ··· p Since an =1/n , the condensed series is ∞ ∞ k ∞ k k 2 2 2 a k = = . 2 (2k )p 2p Xk=1 Xk=1 Xk=1 This is a geometric series. It converges if 2/2p < 1, i.e. p > 1, and diverges if 2/2p 1, i.e. p 1. ≥ ≤ ∞ 1 p , p > 0. n=2 n n • X log p −1 Since an =(n log n) , the condensed series is ∞ ∞ k ∞ ∞ k 2 1 1 1 2 a k = = = . 2 2k (log(2k ))p (k log 2)p logp 2 kp Xk=1 Xk=1 Xk=1 Xk=1 By the above it converges if p > 1 and diverges if p 1. ≤ d’Alembert’s Ratio Test Theorem Let an be a sequence of positive numbers. Suppose that the{ limit} a r = lim n+1 n→∞ an exists (finite or infinite). ∞ (i) If r < 1, then the series n=1 an converges. ∞ (ii) If r > 1, then the series P n=1 an diverges. P Remark. In the case r = 1, the Ratio Test is inconclusive. ∞ −p For example, consider a series n=1 n , where p > 0. Then P (n + 1)−p 1 p r = lim = lim 1 =1 n→∞ n−p n→∞ − n +1 for all p > 0. However the series converges for p > 1 and diverges for 0 < p 1. ≤ Theorem Let an be a sequence of positive numbers. { } Suppose that the limit r = lim an+1/an exists (finite or n→∞ infinite). ∞ (i) If r < 1, then the series n=1 an converges. ∞ (ii) If r > 1, then the series P n=1 an diverges. P Proof: If r > 1, then an+1/an > 1 for n large enough. It follows that the sequence an is eventually increasing. { } ∞ Then an 0 as n so that the series an 6→ →∞ n=1 diverges due to the Trivial Test. P In the case r < 1, choose some x (r, 1). Then ∈ an+1/an < x for n large enough. Consequently, n+1 n an+1/x < an/x for n large enough. That is, the sequence n an/x is eventually decreasing. It follows that this sequence { } n is bounded. Hence an Cx for some C > 0 and all n N. ≤ ∞ n ∈ Since 0 < r < x < 1, the geometric series n=1 x ∞ n converges. So does the series n=1 Cx .P By the Direct ∞ Comparison Test, the series Pn=1 an converges as well. P Cauchy’s Root Test Theorem Let an be a sequence of positive numbers and { } n r = lim sup √an. n→∞ ∞ (i) If r < 1, then the series n=1 an converges. ∞ (ii) If r > 1, then the series P n=1 an diverges. Pk N Proof: If r > 1, then supk≥n √ak r > 1 for all n . Therefore for any n N there exists≥k(n) n such that∈ 1/k(n) ∈ ≥ ak(n) > 1. In particular, ak(n) > 1. It follows that ak 0 ∞ 6→ as k so that the series ak diverges due to the →∞ k=1 Trivial Test. P In the case r < 1, choose some x (r, 1). Then k N ∈ k supk≥n √ak < x for some n . This implies that ak < x for all k n. Since 0 < r <∈x < 1, the geometric series ∞ k ≥ k x converges. By the Direct Comparison Test, the =1 ∞ Pseries k=1 ak converges as well. P Examples n 1 2 3 4 5 ∞ = + + + + + ... n=1 n • X 2 2 4 8 16 32 n If an = n/2 , then −1 an+1 n +1 n n +1 1 1 1 = n+1 n = = + an 2 2 2n 2 2n → 2 as n . By the Ratio Test, the series converges. →∞ x n x x 2 x 3 ∞ =1+ + + + ... , x > 0. n=0 n • X ! 1! 2! 3! n Let an = x /n!, n N. Then ∈ a x n+1 x n −1 x n+1 = = 0 as n . an (n + 1)! n! n +1 → →∞ By the Ratio Test, the series converges for all x > 0. Examples (n!)2 (1!)2 (2!)2 (3!)2 ∞ = + + + ... n=1 n • X (2 )! 2! 4! 6! 2 If an =(n!) /(2n)!, then 2 −1 an+1 (n + 1) n +1 1+ n 1 = = = −1 an (2n + 1)(2n + 2) 2(2n + 1) 4+2n → 4 as n . By the Ratio Test, the series converges. →∞ n n2 1 2 4 3 9 ∞ = + + + ... n=1 n • X +1 2 3 4 n2 If an =(n/(n + 1)) , then n n n +1 −n 1 −n 1 n √an = = = 1+ n +1 n n → e as n . By the Root Test, the series converges. →∞.