§2. Series expansions
385
lun(p)1 S; M(X)2 p +1 /(2p + I)! = v(p)
which suffices to prove normal convergence, since E v(p) < +00.
It thus remains to show that un(p) tends to x 2p + 1 /(2p + I)!. The factors
1 - l/n, ... , 1 - 2p/n present no problem: their product tends to 1. The
denominator (2p + I)! does not change and n. sin(x/n) tends to x as we saw
above; the product of the two last factors of (2) thus tends to x 2p +l /(2p+ I)!.
Since on the other hand cos(x/n) tends to 1, so does its -2p -1 th power for
all p.
We have almost reached our aim: it remains to prove that
(15.3)
limcosn(x/n) = 1
and this is the crucial point. It is clear that cos(x/n) tends to 1, but there
is no theorem to say that if Un tends to 1, so similarly does u~: this general
statement is grossly false and Euler was particularly well placed to know this,
since it very happily does not apply to the sequence Un = 1 + x/no He found,
miraculously, that cosn(x/n) really does tend to 1, but to prove this we have
to use a sharper result than cosO = 1, for example the inequality
(15.4)
1 - x 2 /2n 2 < cos(x/n) < 1,
valid for Ix/nl < 2, so for n large; this shows that the difference between 1
and cos x / n is of the order of magnitude of 1/ n 2 , in other words that cos( x / n)
tends to 1 much more rapidly than, for example, 1 +x/n. Still this argument
remains to be completed; the inequality
will not yield the desired result unless one can show that the left hand side
tends to 1; on putting x = yV2 this can be written (1 - y/n)n(1 + y/n)n
and so tends to exp( -y) exp(y) = 1; one might also use Theorem 21 with
zp =
x 2 /2p.
All this, one sees, is like a high-wire act (netless chez Euler) and even
breaks the vicious circle. The relation (14.9) which legitimates (4) reminds
one very strongly of the beginning of the power series of cosx, i.e. the result
that one hopes to establish! We could deduce it from the relation
cos x = (1 - sin 2 x) 1/2 2:: 1 - sin 2 (x)/2 2:: 1 - x 2 /2
so long as we know that I sin xl S; lxi, which can be taken as geometrically
obvious. Euler did not worry over these subtleties.
We said above that if a sequence Un tends to 1, the sequence u~ need not
do so too. The formula
385
lun(p)1 S; M(X)2 p +1 /(2p + I)! = v(p)
which suffices to prove normal convergence, since E v(p) < +00.
It thus remains to show that un(p) tends to x 2p + 1 /(2p + I)!. The factors
1 - l/n, ... , 1 - 2p/n present no problem: their product tends to 1. The
denominator (2p + I)! does not change and n. sin(x/n) tends to x as we saw
above; the product of the two last factors of (2) thus tends to x 2p +l /(2p+ I)!.
Since on the other hand cos(x/n) tends to 1, so does its -2p -1 th power for
all p.
We have almost reached our aim: it remains to prove that
(15.3)
limcosn(x/n) = 1
and this is the crucial point. It is clear that cos(x/n) tends to 1, but there
is no theorem to say that if Un tends to 1, so similarly does u~: this general
statement is grossly false and Euler was particularly well placed to know this,
since it very happily does not apply to the sequence Un = 1 + x/no He found,
miraculously, that cosn(x/n) really does tend to 1, but to prove this we have
to use a sharper result than cosO = 1, for example the inequality
(15.4)
1 - x 2 /2n 2 < cos(x/n) < 1,
valid for Ix/nl < 2, so for n large; this shows that the difference between 1
and cos x / n is of the order of magnitude of 1/ n 2 , in other words that cos( x / n)
tends to 1 much more rapidly than, for example, 1 +x/n. Still this argument
remains to be completed; the inequality
will not yield the desired result unless one can show that the left hand side
tends to 1; on putting x = yV2 this can be written (1 - y/n)n(1 + y/n)n
and so tends to exp( -y) exp(y) = 1; one might also use Theorem 21 with
zp =
x 2 /2p.
All this, one sees, is like a high-wire act (netless chez Euler) and even
breaks the vicious circle. The relation (14.9) which legitimates (4) reminds
one very strongly of the beginning of the power series of cosx, i.e. the result
that one hopes to establish! We could deduce it from the relation
cos x = (1 - sin 2 x) 1/2 2:: 1 - sin 2 (x)/2 2:: 1 - x 2 /2
so long as we know that I sin xl S; lxi, which can be taken as geometrically
obvious. Euler did not worry over these subtleties.
We said above that if a sequence Un tends to 1, the sequence u~ need not
do so too. The formula