§1. Convergent sequences and series
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for any z E C. By taking absolute values we can restrict to the case where
z > O. Writing Un for the general term, we first remark that, for n > p, we
have
zpzn-p
Z
Z
Un = p! (p + 1) ... n = Up . P + 1 ... ;;:.
Let us choose for p the integer part of lOz, so that p ~ 10z < p+ 1, and keep
p fixed. We have z/q < 1/10 for q :::: p + 1, and the relation above shows that
Un ~ up/lO n - p = 10 P u p/lO n for all n > p,
whence Un < r as soon as n is so large that IOn > lOPup/r, qed.
Example 7. On writing xl/n = yX, we have
(5.5)
lim x 1 / n = 1 for all x > O.
Suppose first that x > 1, whence x 1 / n = 1 + Xn with Xn > o. All the terms
in the binomial formula
x = (1 + xn)n = 1 + n.Xn + ... ,
are > 0, whence 0 < Xn < (x - l)/n and so limxn = 0, which proves (5).
The case where x = 1 is trivial. If 0 < x < 1, one puts x = l/y with y > 1,
whence x 1 / n = l/yl/n. It remains to show that, generally,
lim Un = U =I- 0 implies lim l/un = l/u,
which we shall do in a little while.
This example arose in the construction of the first tables of logarithms
by Napier and then Briggs, Kepler, etc. The fundamental relation log(xy) =
log x + log y shows that 10g(x P ) = p.log x, whence
log(xl/n) = log(x)/n.
Suppos(' that x > 1 and, as above, put
(5.6)
xl/n = 1 + x n , whence 0 < Xn < (x - 1)/n < x/no
Then log x = n 10g(1 + x n ), whence
log(x)/nxn = 10g(1 + xn)/xn .
When n increases indefinitely, the right hand side is of the form log ( 1 + h) / h
where h tends to o. If one assumes that the function log, which clearly satisfies
log 1 = 0, is very "regular", it is simplest, so as not to make the calculations
too intricate, to assume that
loge 1 + h) rv hash --+ 0