106
II - Convergence: Discrete variables
x = [1 + un+d(n + l)r+ 1 < [1 + un/(n + l)r+ 1 .
Since, for u, v > 0 and m ~ 1, the relation urn < v rn implies u < v, one
deduces that Un+l < Un. The sequence (8) is therefore decreasing, and, since
all its terms are> 0, it tends to a limit f(x) as stated.
For 0 < x < 1, one puts x = l/y, whence
The factor yl/n (Example 7, nO 5) tends to 1 and n(1 - yl/n) converges to
- f(y) by what we have just established, whence convergence again.
To establish the first relation (7) one observes that
f(xy) - f(x) - f(y)
limn [(Xy)l/n _ xl/n _ yl/n + 1] =
limn(x 1 / n _1)(yl/n -1)
since (xy)l/n = xl/nyl/n. Now n(xl/n - 1) tends to f(x) and yl/n - 1 to
O. The right hand side thus tends to 0, whence (7). Note that (7) implies
f(l) = 0 and f(l/x) = - f(x).
To establish the differentiability of f and the formula f' (a) = 1/ a one
observes first that
f(a + h) - f(a) = f(1 + h/a)
by the first formula (7), whence
f(a + h) - f(a)
h
1 f(l+h/a)
a
h/a
.!. f(x) - f(l)
a
x-I
on putting x = 1 + h/a. Since h/a = k tends to 0 with h it thus suffices to
establish that f(x)/(x -1) tends to 1 when x tends to 1, Le. that f possesses
a derivative equal to 1 at the point 1, in order to show that f' (a) = 1/ a,
Consider first the ratio (x - l)/un . On putting xl/n = y one finds
(x - l)/un = (yn - 1) /n(y - 1) = (1 + y + ... + yn-l) In.
Whether y is < 1 or > 1, the n numbers yk whose arithmetic mean is being
calculated all belong to the closed interval with end points 1 and yn = x, so
their mean (x - l)/un does too. The quotient un/(x - 1) thus belongs, for
all n, to the closed interval with end points 1 and l/x. It is thus the same
for its limit f(x)/(x - 1). Since l/x tends to 1 when x tends to 1, the ratio
f (x) / (x - 1) tends to 1. Hence the existence of the derivative and the second
relation (7). Moreover, f is Coo (Le. has derivatives of all orders) because f'
is.
This little calculation in fact shows that
(10.10)
1 - l/x = (x - l)/x :::; f(x) :::; x-I for all x > 0