The mth Ratio Convergence Test and Other Unconventional Convergence Tests

Transcription

1 The mth Rtio Convergence Test nd Other Unconventionl Convergence Tests Kyle Blckburn My 14, 2012 Contents 1 Introduction 2 2 Definitions, Lemms, nd Theorems Defintions Lemms De Morgn s Hierrchy The Second Rtio Convergence Test Sttement nd Proof Corollries Strength Applictions The p-series The Hypergeometric Series Rbe s Test The mth Rtio Test 10 5 Bibliogrphy 11 1

2 1 Introduction The simplest rtio test for the convergence of infinite series is d Alembert s rtio test, which is tught in most undergrdute clculus courses. However, this test fils quite often, sometimes for complicted series such s the hypergeometric series nd sometimes for the simplest series such s 1. n p In these cses, mthemticins hve creted severl more rtio convergence tests to utilize when this simplest test is inconclusive. These further rtio convergence tests form the De Morgn Hierrchy of Rtio Tests which, in order from lowest to highest, re: d Alembert s rtio test, Rbe s test, Bertrnd s test, Guss s test, nd finlly, Kummer s test. These tests use the condition under which d Alembert s test fils ( n+1 = 1) to rewrite the series s (1 + b n ) for some sequence b n tht converges to 0. However, in this pper we will discuss different kind of rtio test which is more kin to d Alembert s test thn the other tests in De Morgn s hierrchy, the second rtio convergence test. This test cn be used to successfully check for convergence when d Alembert s test fils, without the possibly cumbersome ppliction of the higher rtio convergence tests. A specific troublesome sequence is the hypergeometric sequence, defined below. We will show tht this sequence converges for the rgument x = 1. We will lso use the second rtio convergence test to prove the convergence portion of Rbe s test. We will then conclude with the sttement of the mth rtio convergence test, the generliztion of the second rtio convergence test. 2 Definitions, Lemms, nd Theorems 2.1 Defintions Definition The hypergeometric series F (, b; c; x) is defined s F (, b; c; x) = 2.2 Lemms n=0 Lemm 2.1 For fixed m, ( + 1) ( + n 1)b(b + 1) (b + n 1) x n. c(c + 1) (c + n 1)n! [ ] n m tn + = e /t tn 2

3 Lemm 2.2 For fixed m, 2.3 De Morgn s Hierrchy [ ] n m tn + b = e (b c)/t tn + c Theorem 2.3 (d Alembert s Rtio Test) [3] Suppose { } is sequence of positive numbers.. If +1 < r for ll sufficiently lrge n, where r < 1, then the series 0 converges. On the other hnd, if +1 1 for ll sufficiently lrge n, then the series 0 diverges. b. Suppose tht l = n+1 exists. Then the series 0 converges if l < 1 nd diverges if l > 1. No conclusion cn be drwn if l = 1. Theorem 2.4 (Rbe s Test) [2] If > 0, ɛ n 0, nd +1 = 1 β n + ɛ n n, where β is independent of n, then n=1 converges if β > 1 nd diverges if β < 1. Theorem 2.5 (Bertrnd s Test) [4] Let { } be sequence of positive numbers. Suppose tht = n + ρ n n ln n. Then the series converges if inf ρ n > 1 nd diverges if sup ρ n < 1. Theorem 2.6 (Guss s Test) [5] Let { } be sequence of positive numbers nd B(n) bounded function of n s n, nd = 1 + h +1 n + B(n), r > 1. n r Then the series converges for h > 1 nd diverges for h 1. Theorem 2.7 (Kummer s Test) [2] If > 0, d n > 0, n=1 d n diverges, nd ( 1 ) n+1 1 = h, d n d n+1 then the series n=1 converges if h > 0 nd diverges if h < 0. 3

4 3 The Second Rtio Convergence Test The second rtio test is the min result of this pper. The generlized result, the mth rtio test, serves the sme purpose but is more cumbersome to pply in prctice. All results proven in this pper will follow from the second rtio test; the mth rtio test is presented to show tht the second rtio test cn be generlized to the third, fourth, nd so on. 3.1 Sttement nd Proof Theorem 3.1 (The Second Rtio Test) [1] Let { } be sequence of positive numbers. Let { } +1 L = mx sup, sup nd Then n=1 : 1. converges if L < 1 2 ; 2. diverges if l > 1 2 ; { l = mx inf, inf 3. my either converge or diverge if l 1 2 L. } +1 Proof 1. Suppose L < 1. Let r be such tht L < r < 1. Then there exists 2 2 n integer N such tht for ll n N. Now, r nd +1 r = ( N + N N 1 ) + ( 2N + 2N N 1 ) n=n + ( 4N + 4N N 1 ) + + ( 2 k N + 2 k N N 1) + k+1 = ( 2 k N + 2 k N N 1). k+1 k=0 4

5 Let S k = 2 k N + 2 k N k+1 N 1 for k = 0, 1, 2, 3,.... Then, for k 1, S k = ( 2 k N + 2 k N+1) + ( 2 k N k N+3) + + ( 2 k+1 N k+1 N 1). Since r nd +1 r, we then obtin S k = ( 2 k N + 2 k N+1) + ( 2 k N k N+3) + + ( 2 k+1 N k+1 N 1) 2( 2 k 1 N)r + 2( 2 k 1 N+1)r + + 2( 2 k N 1)r = 2r( 2 k 1 N + 2 k 1 N k N 1) = 2rS k 1. So, by induction on k it follows tht for k 1. Thus, S k 2 k r k ( N + N N 1 ) = s k r k S 0 = n=n S k k=0 S 0 (2r) k < k=0 since r < 1 2. Therefore, n=1 converges if L < Suppose l > 1. Let r be such tht 1 < r < l. Then there is n integer 2 2 N such tht > r nd +1 > r for ll n N. Thus, > r nd +1 > r for ll n N. Let S k be s bove. We follow the sme induction process but use the previous two inequlities to obtin S k S 0 (2r) k for k 1. Therefore, since r > 1, we 2 hve = S k S 0 (2r) k =. n=n k=0 Thus, n=1 diverges if l > 1 2. k=0 3. The series n=1 where = 1 if p 1. But This completes the proof. n(ln n) p n(ln n) p = 2n[ln(2n)] = 1 p 2. 5 converges if p > 1 nd diverges

6 3.2 Corollries Some corollries re immeditely pprent. These serve to mke the second rtio test esier to pply. Corollry 3.2 Let n=1 be series with > 0 for n 1. Suppose +1 nd both exist. Let L 1 = 2n n, L 2 = 2n+1, L = mx{l 1, L 2 }, nd l = min{l 1, L 2 }. Then n=1 : 1. converges if L < 1 2 ; 2. diverges if l > 1 2 ; 3. my either converge or diverge if l 1 2 L. This first corollry is just the second rtio test when the its of the two rtios nd +1 both exist. The only difference is this presumption, which llows us to get rid of both instnces of sup found in the originl sttement. Corollry 3.3 If { } is positive decresing sequence of numbers, then n=1 n converges if 2n < 1 nd diverges if 2 +1 > 1. 2 The second corollry follows from the first one when given tht the sequence { } is decresing. 3.3 Strength To exmine the second rtio test nd wht mkes it so strong when compred to d Alembert s rtio test, we look t the reltions = = n The first thing to note is tht the first term in this product is +1. This mens tht if series converges by d Alembert s rtio test, then it converges by the second rtio test. Further, if series converges by d Alembert s test, then n+1 < 1, so then 2n = 2n+1 = 0. In the sme 6

7 wy, divergence by d Alembert s test implies divergence by the second rtio test; if n+1 > 1, then 2n = 2n+1 =. So the second rtio test is t lest s powerful s the ordinry rtio test. In the next section we will demonstrte this strength by proving the convergence of few series on which the ordinry rtio test fils. We begin with the simplest series nd then rmp up to proving the convergence portion of Rbe s test. 3.4 Applictions The p-series An immedite, esy ppliction of the second rtio test nd its corollries is the convergence of the series n=1 n p. Exmple (The p-series) Let = n p. Then = (2n) p n p = 1 2 p nd +1 = (2n + 1) p n p = 1 (2 + 1 n )p nd thus = +1 = 1 2 p. This it is less thn 1 if p > 1 nd greter thn 1 if p < 1. So by the 2 2 second rtio test, the series n=1 converges in the first cse nd diverges in the second. Thus this well-known result is proven without the need to use the integrl comprison test The Hypergeometric Series Guss s test stted in (2.6) ws originlly creted by him in order to test the convergence of the hypergeometric series (2.1) with the unit rgument x = 1, F (, b; c; 1). However, this test cn be done in greter simplicity with the second rtio test. In this exmple, we will use the results (2.2.1) nd (2.2.2). Exmple (The Hypergeometric Series) Let, b, nd c be positive numbers. Let = ( + 1)( + 2) ( + n 1)b(b + 1)(b + 2) (b + n 1). c(c + 1)(c + 2) (c + n 1)n! 7

8 Then = = ( + n)( + n + 1) ( + 2n 1)(b + n)(b + n + 1) (b + 2n 1) (n + 1)(n + 2) (2n)(c + n)(c + n + 1) (c + 2n 1) [ ( + n)(b + n) ( + n + 1)(b + n + 1) ( + n + 2)(b + n + 2) 2n(c + n) (n + 1)(c + n + 1) (n + 2)(c + n + 2) ] ( + 2n 1)(b + 2n 1). (2n 1)(c + 2n 1) We note tht ech of the expressions inside of the brckets hs the form (+x)(b+x). So we let f(x) = (+x)(b+x). Then x(c+x) x(c+x) f(x) = ( + x)(b + x) x(c + x) = 1 + ( + b c)x + b. x(c + x) So if + b < c, then there is some N > 0 such tht f(x) is incresing for ll x > N. Therefore, if + b < c, ( + n)(b + n) 2n(c + n) [ ] n 2 ( + 2n 1)(b + 2n 1) (2n 1)(c + 2n 1) for ll n > N. Similrly, with this sme process s ll of the bove, we cn write, for +1 : +1 ( + n)(b + n) (2n + 1)(c + n) [ ( + 2n)(b + 2n) 2n(c + 2n) for ll n > N. Then using the lemms (2.2.1) nd (2.2.2), ] n 1 [ ] n 2 ( + 2n 1)(b + 2n 1) = e ( 1+1)/2 e (b 1 c+1)/2 = e +b c 2 nd (2n 1)(c + 2n 1) So then [ ( + 2n)(b + 2n) 2n(c + 2n) ] n 1 = e /2 e (b c)/2 = e +b c 2. sup = sup e +b c 2 < 1 2 when + b < c. Thus, by the second rtio test, the hypergeometric series converges under the condition + b < c. 8

9 3.4.3 Rbe s Test Rbe s test (2.4) is the second test in the De Morgn Hierrchy nd so its deliccy is outdone by the tests further up the hierrchy. We will now present proof of the convergence portion Rbe s test bsed on the second rtio test. The divergence portion is excluded becuse of irrelevncies. Proof (Rbe s Test) Suppose +1 = 1 β n + ɛn n, where > 0 nd ɛ n 0. Assume β > 1, nd choose r such tht β > r > 1. Then there is some N such tht +1 < 1 r n for n N. Then = ( 2n < r ) ( 1 r ) +1 1 n 2n 1 for n N. Since 1 x e x for 0 < x < 1, ( r ) ( 1 r ) e ( r n + n 2n 1 nd since r n + Thus we hve r n n 1) r, r + + r > r ln ( ) 2n n+1 2n 1 n = r ln 2, we hve ) ( 2n < r ) n sup ( 1 r 2n r < 1 2. < e r ln 2 = 1 2 r. Similrly, so ( 2n+1 < r ) ( 1 r ) 2n+1 r ln < e n, n 2n sup +1 2n+1 r ln sup e n < e r ln 2 = 1 2 < 1 r 2. Thus by the second rtio test, the series n=1 converges. 9

10 4 The mth Rtio Test Theorem 4.1 (The mth Rtio Test) Let { } be positive sequence nd let m > 1 be fixed positive integer. Let L k nd l k be such tht for 1 k m, L k = sup l k = inf mn+k 1, mn+k 1. Let L = mx{l k } nd l = min{l k }. Then the series n=1 1. converges if L < 1 m ; 2. diverges if l > 1 m ; 3. my either converge or diverge if l 1 m L. This test is interesting for theoreticl purposes but serves very little prcticl purpose, especilly when m is lrge. It is for this reson tht we hve focused primrily on the second rtio test insted of this more generl mth rtio test. 10

11 5 Bibliogrphy References [1] The mth Rtio Test: New Convergence Test for Series, Syel A. Ali, The Americn Mthemticl Monthly, Vol. 115, No. 6 (Jun-Jul., 2008), pp , Mthemticl Assocition of Americ. [2] E. Rinville, Infinite Series, Mcmilln, New York, [3] Gerld B. Follnd, Advnced Clculus, Prentice Hll, New Jersery, [4] Bromwich, T. J. I A. nd McRobert, T. M., An Introduction to the Theory of Infinite Series, 3rd ed., New York: Chelse, p. 40, [5] Cournt, R. nd John, F. Introduction to Clculus nd Anlysis, Vol. 1. New York: Springer-Verlg,