PureMath.pdf

# Increases thus log x 0 requires x 1 log 2 x 0

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increases: thus log x > 0 requires x > 1, log 2 x > 0 requires x > e , log 3 x > 0 requires x > e e , and so on; and it is easy to see that e e > 10, e e e > e 10 > 20 , 000, e e e e > e 20 , 000 > 10 8000 . The reader should observe the extreme rapidity with which the higher ex- ponential functions, such as e e x and e e e x , increase with x . The same remark of course applies to such functions as a a x and a a a x , where a has any value greater than unity. It has been computed that 9 9 9 has 369 , 693 , 100 figures, while 10 10 10 has of course 10 , 000 , 000 , 000. Conversely, the rate of increase of the higher log- arithmic functions is extremely slow. Thus to make log log log log x > 1 we have to suppose x a number with over 8000 figures. * 5. Prove that the integral Z a 0 1 x log 1 x s dx , where 0 < a < 1, is convergent if s < - 1, divergent if s = - 1. [Consider the behaviour of Z a 1 x log 1 x s dx as +0. This result also may be refined upon by the introduction of higher logarithmic factors.] * See the footnote to p. 450 .

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[IX : 211] THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS 470 6. Prove that Z 1 0 1 x log 1 x s dx has no meaning for any value of s . [The last example shows that s < - 1 is a necessary condition for convergence at the lower limit: but { log(1 /x ) } s tends to like (1 - x ) s , as x 1 - 0, if s is negative, and so the integral diverges at the upper limit when s < - 1.] 7. The necessary and sufficient conditions for the convergence of Z 1 0 x a - 1 log 1 x s dx are a > 0, s > - 1. Examples LXXXIX. 1. Euler’s limit. Show that φ ( n ) = 1 + 1 2 + 1 3 + · · · + 1 n - 1 - log n tends to a limit γ as n → ∞ , and that 0 < γ 5 1. [This follows at once from § 174 . The value of γ is in fact . 577 . . . , and γ is usually called Euler’s constant .] 2. If a and b are positive then 1 a + 1 a + b + 1 a + 2 b + · · · + 1 a + ( n - 1) b - 1 b log( a + nb ) tends to a limit as n → ∞ . 3. If 0 < s < 1 then φ ( n ) = 1 + 2 - s + 3 - s + · · · + ( n - 1) - s - n 1 - s 1 - s tends to a limit as n → ∞ . 4. Show that the series 1 1 + 1 2(1 + 1 2 ) + 1 3(1 + 1 2 + 1 3 ) + . . . is divergent. [Compare the general term of the series with 1 / ( n log n ).] Show also that the series derived from n - s , in the same way that the above series is derived from (1 /n ), is convergent if s > 1 and otherwise divergent. 5. Prove generally that if u n is a series of positive terms, and s n = u 1 + u 2 + · · · + u n ,
[IX : 212] THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS 471 then ( u n /s n - 1 ) is convergent or divergent according as u n is convergent or divergent. [If u n is convergent then s n - 1 tends to a positive limit l , and so ( u n /s n - 1 ) is convergent. If u n is divergent then s n - 1 → ∞ , and u n /s n - 1 > log { 1 + ( u n /s n - 1 ) } = log( s n /s n - 1 ) ( Ex. lxxxii . 1); and it is evident that log( s 2 /s 1 ) + log( s 3 /s 2 ) + · · · + log( s n /s n - 1 ) = log( s n /s 1 ) tends to as n → ∞ .] 6. Prove that the same result holds for the series ( u n /s n ). [The proof is the same in the case of convergence. If u n is divergent, and u n < s n - 1 from a certain value of n onwards, then s n < 2 s n - 1 , and the divergence of ( u n /s n ) follows from that of ( u n /s n - 1 ). If on the other hand u n = s n - 1 for an infinity of values of n , as might happen with a rapidly divergent series, then u n /s n = 1

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