SELECTED CHAPTERS FROM ALGEBRA I. R. Shafarevich Preface

Selected chapters from algebra 79Thus, the left inequality (10)isproved for odd n with the constant c = 1 . So, the3left inequality (10) is valid for all n and c = 1.3We proceed to the proof of the right inequality in (10). We shall prove itbyinduction on n. Let, rst of all, n be even. We will write 2n instead of it. Takingthe inequality (11) for the coecient C2n n (i.e., substitute in Cn k n by 2n and k by n)together with the inequality (14), as a consequence we obtainand, passing to logarithms,n (2n);(n) 6 2 2n(21) (2n) ; (n) 6 2n2n (2n) 6 (n)+log n log n :In accordance with the inductive hypothesis, suppose that our inequality has beenproved: (n) 6 Cnlog nwith a constant C whose value we shall make more preciselater. Substituting in the formula (21), we obtain:(2n) 6 Cnlog n + 2n (C +2)n=log n log n :We would like toprove the inequality (2n) 6 C2nlog 2nand for that we have tochoosethe constant C in such away that the inequality(22)(C +2)nlog n6 2Cnlog 2nis valid for all n, starting from some limit.This is just a simple school exercise. Cancel in the inequality both sides by n,remark that log 2n = log 2 + log n = log n +1 and denote log n by x. Then theinequality (22)takes the formC +2x6 2Cx +1 :Multiplying both sides by x(x + 1) (as x>0) and transforming, we write it in theform (C ; 2)x > C +2. Obviously, C has to be chosen so that C ; 2 > 0. Setting,e.g., C =3,we obtain that it is valid for C =3andallx > 5. Since x denotes log n,this means that the necessary inequality would be valid if n > 2 5 = 32, 2n > 64.It remains to consider the case of odd values of the form 2n +1. Compare theinequality (11) (substituting in it n by 2n +1and k by n) with the inequality (15).We obtain the inequalityand, taking logarithms, the inequality2 2n+1 > (n +1) (2n+1);(n+1)2n +1> ((2n +1); (n + 1)) log(n +1):