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3 Find the smallest natural number on such that n!> n³ for all natural numbers n=m, and then verify.

3 Find the smallest natural number on such that n!> n³ for all natural numbers n=m, and then verify.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Find the smallest natural number on such that n!>n²³ for all natural numbers nằm, and then verify. 7!=5040 343 1! = 1 2!= 28 3! = = 1 6 (27 4! = 24 / 64 إما 5!= 120 125 = 720 7214 3 Proof: For every nt R, 0716, let Pln);"n! >0³ " We prove by induction. We see for Plu) : " (0!>6³" is true because 61² = 720 > 216 = 6³. m² = 6 إما > (k+1) k3 > (6+1) K³ 7k³ Now let KE 2, k²6 and assume plie); "k! > K³. " 3 We will show P[K+1); " (K+1)! > K³. " We (K+1)! = (K+1) k! We see that > k³ Thus P(K) => P(k+1) is true. Therefore by PMI Pln) is true Vntz, n= 0. 2
Transcribed Image Text:Find the smallest natural number on such that n!>n²³ for all natural numbers nằm, and then verify. 7!=5040 343 1! = 1 2!= 28 3! = = 1 6 (27 4! = 24 / 64 إما 5!= 120 125 = 720 7214 3 Proof: For every nt R, 0716, let Pln);"n! >0³ " We prove by induction. We see for Plu) : " (0!>6³" is true because 61² = 720 > 216 = 6³. m² = 6 إما > (k+1) k3 > (6+1) K³ 7k³ Now let KE 2, k²6 and assume plie); "k! > K³. " 3 We will show P[K+1); " (K+1)! > K³. " We (K+1)! = (K+1) k! We see that > k³ Thus P(K) => P(k+1) is true. Therefore by PMI Pln) is true Vntz, n= 0. 2
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