The Omnibus Theorem of (Main) Convergence Tests Definition 1 Let F be either IR or C. Let an n∈IN be a sequence in F . (a) The sequence an n∈IN converges to A ∈ F if ∀ε > 0 ∃N ∈ IN s.t. an − A < ε whenever n ≥ N A sequence which does not converge to any A is said to diverge. (b) The sequence an n∈IN is said to be Cauchy if ∀ε > 0 ∃N ∈ IN s.t. am − an < ε whenever m, n ≥ N P∞ (c) The series n=1 an is said to converge to S if the sequence sn n∈IN of partial sums n P sn = ak converges to S. A series which does not converge to any S is said to diverge. k=1 (d) The series P∞ n=1 an is said to converge absolutely if the P∞ n=1 |an | converges. Theorem 2 Let an n∈IN , cn n∈IN and dn n∈IN be sequences of real or complex numbers n P and sn = ak . Then k=1 (a) (Cauchy Criterion) ∞ X k=1 m X ak < ε whenever m ≥ n ≥ N ak converges ⇐⇒ ∀ε > 0 ∃N ∈ IN s.t. k=n (b) (Divergence Test) ∞ P an converges =⇒ lim an = 0 n→∞ n=1 (c) If an ≥ 0 for all n ∈ IN, then ∞ X ak converges ⇐⇒ k=1 sn n∈IN is bounded (d) absolute convergence =⇒ convergence (e) (Comparison Test) Let N0 ∈ IN. ∞ ∞ P P (i) |an | ≤ cn ∀n ≥ N0 and cn converges =⇒ an converges absolutely n=1 (ii) an ≥ dn ≥ 0 ∀n ≥ N0 and ∞ P n=1 ∞ P dn diverges =⇒ c Joel Feldman. 2014. All rights reserved. an diverges n=1 n=1 October 8, 2014 (Main) Convergence Tests 1 (f ) (Integral Test) Let c ∈ IR. If f (x) is a real valued function which is defined and continuous for all x ≥ c and which obeys (i) f (x) ≥ 0 ∀x ≥ c (ii) f (x) is monotonically decreasing, i.e. y > x =⇒ f (y) ≤ f (x) (iii) f (n) = an ∀n ≥ c then y = f (x) y Z ∞ a1 ∞ X a2 an converges ⇐⇒ f (x) dx < ∞ a3 a4 c n=1 Furthermore, if N ≥ c, the truncation error Z ∞ X ≤ a − s n N n=1 1 2 3 4 x ∞ f (x) dx N (g) (Examples) ∞ P (i) (Geometric Series) (ii) (p Test) (iii) ∞ P n=2 ∞ P n=1 1 n(ln n)p r n converges to n=0 1 np 1 1−r if |r| < 1 and diverges if |r| ≥ 1. converges if p > 1 and diverges if p ≤ 1. converges if p > 1 and diverges if p ≤ 1. (h) (Root Test) Let α = lim sup n→∞ p n |an |. Then α < 1 =⇒ α > 1 =⇒ ∞ X n=1 ∞ X an converges absolutely an diverges n=1 α = 1 =⇒ nothing (i) (Ratio Test) If an ∞ X n=1 n∈IN is a sequence of nonzero numbers, then a an converges absolutely if lim sup n+1 an < 1 c Joel Feldman. n→∞ an+1 diverges if an ≥ 1 for all n bigger than some natural number N 2014. All rights reserved. October 8, 2014 (Main) Convergence Tests 2 (j) (Alternating Series Test) If (i) an ≥ 0 ∀n ∈ IN (ii) an+1 ≤ an ∀n ∈ IN (iii) limn→∞ an = 0 ∞ P then s = (−1)n−1 an converges and s − sn is between 0 and (the first dropped term) n=1 (−1)n an+1 . Proof: (a) ⇒ Let lim sn = s and let ε > 0. Then n→∞ ∃Ñ ∈ IN s.t. n ≥ Ñ =⇒ |sn − s| < ε 2 Set N = Ñ + 1. Then m ≥ n ≥ N =⇒ m > n − 1 ≥ Ñ =⇒ |sm − s| < 2ε , |sn−1 − s| < ε 2 =⇒ |sm − sn−1 | = |(sm − s) + (s − sn−1 )| < ε m X ak < ε =⇒ k=n (a) ⇐ By hypothesis ∀ε > 0 ∃Nε ∈ IN s.t. m ≥ n ≥ Nε =⇒ |sm − sn | < ε =⇒ sn n∈IN is bounded (Choosing ε = 1, n = N1 , we see that sm m ≥ N1 is bounded.) =⇒ sn n∈IN has a subsequential limit. Say lim sin = s. n→∞ =⇒ ∀ε > 0 n ≥ Nε/2 =⇒ |sn − s| = sn − sim + sim − s < 2ε + 2ε = ε if we choose m large enough that im ≥ Nε/2 and sim − s < 2ε . (b) Apply part (a) with m = n. (c) As an ≥ 0 for all n ∈ IN, the sequence sn converges if and only if it is bounded. (d) If ∞ P k=1 the series |ak | converges, then ∞ P ∞ P k=1 n∈IN is monotonically increasing. So it P m P m |ak |, ak ≤ |ak | obeys the Cauchy criterion. As k=n k=n ak then also obeys the Cauchy criterion and hence converges. k=1 c Joel Feldman. 2014. All rights reserved. October 8, 2014 (Main) Convergence Tests 3 (e.i) The sequence n o N0 n ∞ P P P tn = |ak | is monotonic and bounded above by |ak | + ck k=1 and hence converges. So ∞ P k=1 k=1 ak converges absolutely and hence also converges by part (d). k=1 ∞ P (e.ii) If ∞ P ak were to converge, then k=1 dk would also converge by part (e.i). This is a k=1 contradiction. ∞ P (f) Let N be any fixed integer bigger than c + 1. Then ∞ P ak converges if and only if k=1 ak converges, which in turn is the case if and only if the sequence tn = k=N n P ak is k=N bounded, since ak ≥ 0 for all k ≥ N > c. As f (x) decreases as x increases, we have that f (x) ≥ ak when x ≤ k and f (x) ≤ ak when x ≥ k. Thus, for any k ≥ N Z k+1 f (x) dx ≤ ak ≤ Z k f (x) dx k−1 k Summing k from N to n, Z n+1 f (x) dx ≤ N ak n X ak ≤ Z y = f (x) n f (x) dx N−1 k=N So the tn ’s are bounded if and only if (g,i) Exercise. (g.ii) If p ≤ 0, the series ∞ P n=1 1 np = R∞ c ∞ P k−1 k+1 k f (x) dx < ∞. n|p| diverges by part (b). The remaining cases n=1 are easily proven using the integral test with c = 1 and f (x) = x1p , which is positive and monotonically decreasing for p > 0. We just have to observe that Z L ln L if p = 1 1 L1−p −1 xp dx = if p 6= 1 1−p 1 1 when p > 1. grows unboundedly as L → ∞ when p ≤ 1 and converges to p−1 Here is a second proof that does not use integrals. If p > 0, each successive term in the series is smaller than the term before. So each term with index n between 2m and 2m+1 obeys 1 1 ≤ n1p ≤ 2mp 2(m+1)p There are 2m+1 − 2m such terms. Hence 1 (m+1)(1−p) 2 2 = 2m+1 −2m 2(m+1)p ≤ X 1 np ≤ 2m+1 −2m 2mp = 1 2m(p−1) 2m ≤n<2m+1 c Joel Feldman. 2014. All rights reserved. October 8, 2014 (Main) Convergence Tests 4 If 0 ≤ p ≤ 1, the left hand bound 21 2(m+1)(1−p) ≥ 12 . Summing m from 0 to N − 1, we 2N −1 P 1 N have np ≥ 2 which forces divergence. If p > 1, then summing m from 0 to N − 1 n=1 in the right hand bound, we have N 2 P n=1 1 np ≤ N−1 P m=0 bounded and hence converges by part (c). m 1 2p−1 ≤ 1− −1 1 . 2p−1 So the series is |p| (g.iii) If p ≤ 0, then for all n ≥ 2, n(ln1n)p ≥ (ln n2) and the series diverges by the comparison test. Again, the remaining cases are easily proven using the integral test, this time with c = 2 and f (x) = x(ln1x)p . To compute the integral, we make the change of variables y = ln x, dy = dx x . Z L 2 1 x(ln x)p dx = Z ln L ln 2 1 yp ( ) ln(ln L) − ln(ln 2) if p = 1 dy = (ln L)1−p −(ln 2)1−p if p 6= 1 1−p ∞ if p ≤ 1 1−p as L → ∞ → (ln 2) if p > 1 p−1 Again, here is a second proof that does not use the integral test. If p > 0, each term with index n between 2m and 2m+1 obeys 1 2m+1 (m+1)p (ln 2)p ≤ 1 n(ln n)p ≤ 1 2m (m ln 2)p Hence 1 1 2(ln 2)p (m+1)p = 2m+1 −2m 2m+1 (m+1)p (ln 2)p ≤ X 1 n(ln n)p ≤ 2m+1 −2m 2m (m ln 2)p = 1 1 (ln 2)p mp 2m ≤n<2m+1 Summing m from 1 to N − 1, we have 1 2(ln 2)p N−1 X 1 (m+1)p m=1 ≤ N 2X −1 1 n(ln n)p ≤ n=2 1 (ln 2)p N−1 X 1 mp m=1 So if p > 1, the right hand inequality together with part (g.ii) shows that the series is bounded and hence converges, by part (c), and if p ≤ 1, the left hand inequality together with part (g.ii) shows that the series diverges. (h) If α < 1, pick any r obeying α < r < 1. There is a natural number N such that p n |an | < r, or equivalently, |an | < r n , for all n ≥ N . So convergence follows by comparison with the geometric series. On the other hand, if α > 1, there must be a subsequence with p nk |ank | ≥ 1, or equivalently, |ank | ≥ 1. So the series diverges by part (b). c Joel Feldman. 2014. All rights reserved. October 8, 2014 (Main) Convergence Tests 5 a ≥ 1 for all n ≥ N , then |an | ≥ |aN | > 0 for all n ≥ N and the series diverges (i) If an+1 n < 1, there is a r < 1 and an N ∈ IN such by part (b). On the other hand, if lim sup an+1 an n→∞ ≤ r for all n ≥ N . Consequently |aN+1 | ≤ r|aN |, |aN+2 | ≤ r|aN+1 | ≤ r 2 |aN | that aan+1 n and so on. Iterating gives |aN+p | ≤ r p |aN | which implies convergence by comparison with the geometric series. (j) Exercise c Joel Feldman. 2014. All rights reserved. October 8, 2014 (Main) Convergence Tests 6