It will be convenient to prove at this stage a number of elementary inequalities which will be useful to us later on.

(i) It is evident that if α>1 and r is a positive integer then rαr>αr1+αr2++1. Multiplying both sides of this inequality by α1, we obtain rαr(α1)>αr1; and adding r(αr1) to each side, and dividing by r(r+1), we obtain (1)αr+11r+1>αr1r(α>1). Similarly we can prove that (2)1βr+1r+1<1βrr(0<β<1).

It follows that if r and s are positive integers, and r>s, then (3)αr1r>as1s,1βrr<1βss. Here 0<β<1<α. In particular, when s=1, we have (4)αr1>r(α1),1βr<r(1β).

(ii) The inequalities (3) and (4) have been proved on the supposition that r and s are positive integers. But it is easy to see that they hold under the more general hypothesis that r and s are any positive rational numbers. Let us consider, for example, the first of the inequalities (3). Let r=a/b, s=c/d, where a, b, c, d are positive integers; so that ad>bc. If we put α=γbd, the inequality takes the form (γad1)/ad>(γbc1)/bc; and this we have proved already. The same argument applies to the remaining inequalities; and it can evidently be proved in a similar manner that (5)αs1<s(α1),1βs>s(1β), if s is a positive rational number less than 1.

(iii) In what follows it is to be understood that all the letters denote positive numbers, that r and s are rational, and that α and r are greater than 1, β and s less than 1. Writing 1/β for α, and 1/α for β, in (4), we obtain (6)αr1<rαr1(α1),1βr>rβr1(1β). Similarly, from (5), we deduce (7)αs1>sαs1(α1),1βs<sβs1(1β).

Combining (4) and (6), we see that (8)rαr1(α1)>αr1>r(α1). Writing x/y for α, we obtain (9)rxr1(xy)>xryr>ryr1(xy) if x>y>0. And the same argument, applied to (5) and (7), leads to (10)sxs1(xy)<xsys<sys1(xy).

Example XXVIII

1. Verify (9) for r=2, 3, and (10) for s=12, 13.

2. Show that (9) and (10) are also true if y>x>0.

3. Show that (9) also holds for r<0. [See Chrystal’s Algebra, vol. ii, pp. 43–45.]

4. If ϕ(n)l, where l>0, as n, then ϕklk, k being any rational number.

[We may suppose that k>0, in virtue of Theorem III of § 66; and that 12l<ϕ<2l, as is certainly the case from a certain value of n onwards. If k>1, kϕk1(ϕl)>ϕklk>klk1(ϕl) or klk1(lϕ)>lkϕk>kϕk1(lϕ), according as ϕ>l or ϕ<l. It follows that the ratio of |ϕklk| and |ϕl| lies between k(12l)k1 and k(2l)k1. The proof is similar when 0<k<1. The result is still true when l=0, if k>0.]

5. Extend the results of Exs. XXVII. 7, 8, 9 to the case in which r or k are any rational numbers.


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