Problem Set Twelve: Raabe's Theorem Revisited Theorem: Let ( a n ) be a positive sequence for which ρ lim n [ a n /a n 1 1] exists. (1) If ρ 1 then there are constants c and p with 0 < c, 1 < p and a n c n p for all n. Moreover if ρ 1 then the sequence ( n a n ) is eventually decreasing. (2) If ρ 1 then ( n a n ) is eventually increasing. Lemma: If 1 < p then Γ(n) / (n p) n p . Proof. Use the log-convexity of the Gamma Function. n 1 α n β (n p) for β 1 / p and n Γ(n) Γ(n 1) Γ(n) α β Γ(n p) n Γ(n) 1 α β Γ(n p) Γ(n) / (n p) n p Proof of (1): Fix 1 p ρ . Since p lim n [ a n /a n 1 1] there is a m N so that n m implies p n[ an 1] a n 1 . In particular 1 n [ an 1] a n 1 makes (n 1) a n 1 n a n for all n m . Let c max { 1 p a 1 , 2 p a 2 , 3 p a 3 , ... , m p a m , a m Γ(m p) / Γ(m) } . Clearly a n c n p for 1 n m . For n > m p n[ an 1] a n 1 a a n n a n 1 Γ(m) Γ(m p) np 1 n a n 1 a n2 an p n am2 ... a m 1 an a n 1 a m 1 a m a n 1 an n np , (n 1)(n 2) ... (m 1)(m) a m a m , and (n p 1)(n p 2) ... (m p 1)(m p) (n 1)(n 2) ... (m 1)(m) Γ(m) (n p 1)(n p 2) ... (m p 1)(m p) Γ(m p) (n 1)(n 2) ... (m 1) Γ(m 1) (n p 1)(n p 2) ... (m p 1) Γ(m p 1) ... (n 1) Γ(n 1) (n p 1) Γ(n p 1) am Γ(n) Γ(n p) am am a m a mn p so that a n cn p for each n > m. Proof of (2): Since lim n [ a n /a n 1 1] 1 there is a m N so that n m implies n [ that makes n a n (n 1) a n 1 for all n m . an a n 1 1] 1 and Corollary (Raabe’s Test): Let ( a n ) be a positive sequence for which ρ lim n [ a n /a n 1 1] exists. (1) If ρ 1 then (2) If ρ 1 then k 1 converges by comparison with a convergent p-series ak k 1 diverges by comparison with the harmonic series. ak PROBLEMS Problem 12-1: Use a change of variable to check each of these identities. 1 x 1 0 ln(1/t) d t (a, Alternate Definition) For x > 0, Γ(x) 0 (b, Laplace Transform of t x ) For x 1 and s > 0, (c, Connection with Gaussian) Γ(1/2) Problem 12-2: Γ(1/2) G 2 x 0 0 0 e 2 G d x and Γ(1/2) 2 G π 2 0 t 1/2 e . To see why let G 0 (x y ) 2 e π e x 2 t dt 2 0 x t e e x s t d t Γ(x 1)/s 2 (x ) 0 e x 1 dx 2 d x and work through this calculation. y2 e dydx 0 π/2 r 2 dydx e r d r d θ π/4, 0 0 . 1 Problem 12-3: Show that Γ(n ) 2 (1)(3)(5). ..(2n 1) 2 n π (2n)! 2 2n π . (n! ) Problem 12-4: For positive a and b, show (b)(a b)(2 a b)(3 a b)...( (n 1) a b) a n Γ(n (b/a)) Γ(b/a) . Problem 12-5: We proved these consequences of the log-convexity of the Gamma function: (i) Γ(n p) Γ(n) n p for 0 < p < 1 (ii) (n 1) p Use those inequalities to estimate the size of a n Γ(n p) Γ(n) for 0 < p. (1)(6)(11) (16)...(5n 4) (2)(7)(12) (17)...(5n 3) . Does the series converge or diverge? Problem 12-6: Prove c 0 x 0 , Γ(x) c/x . Any such equality implies lim Γ(x) . x 0 n 1 a n