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IV - Powers, Exponentials, Logarithms, Trigonometric Functions
§3. Infinite products
17 - Absolutely convergent infinite products
In analysis we meet infinite products
to which we have to give a meaning. The only reasonable solution is to suppose
that the partial products
Pn = al·· .an
tend to a limit P, which by convention will be the value of the product. Taken
literally, the definition is of no interest if some of the an are zero; so we agree
to omit them from the product.
Experience shows that, with exceptions, the problem is again uninteresting (or that we do not know how to treat it ... ) if the limit P is zero. Suppose
therefore that P f=. o. Then an = Pn/Pn-l tends to pip = 1, which allows us
to put an = 1 + Un where Un tends to o. Then
log IPnl
log 11 + ull + ... + log 11 + unl ::;
< 10g(1 + lUll) + ... + 10g(1 + Iunl)
since the function log is increasing. Now we have shown in Chap. II, nO 10,
that log x ::; x-I for all x > 0, i.e.
10g(1 + x) ::; x for all x > -1.
Thus
log IPnl ::; lUll + ... + Iunl·
Suppose now that the series L Un is absolutely convergent. The preceding
inequality shows that the sequence (log IPnl) is bounded above by a number
M > o. The relation log IPnl ::; M implies IPnl ::; eM = M', so the sequence
(Pn) is bounded.
The inequality
IPn - Pn-ll = IPn-1Unl ::; M'lunl
now shows that the series with general term Pn - Pn-l is absolutely convergent, so convergent. Whence the existence of P = limpn.
It remains to check that P f=. O. For this, one replaces the product of the
an by that of their reciprocals a;:;-l = 1 + Vn , which replaces Pn by qn = I/Pn.
The relation (1 + un )(1 + vn ) = 1 shows that
Ivnl = Iun l/ll + unl ::; 21un l for n large
since 11 + unl, which tends to 1, is ~ 1/2 for n large. The series Vn therefore
converges absolutely. Consequently qn tends to a limit q, and since Pnqn = 1
for all n, it follows that pq = 1, whence P f=. o. In conclusion:
IV - Powers, Exponentials, Logarithms, Trigonometric Functions
§3. Infinite products
17 - Absolutely convergent infinite products
In analysis we meet infinite products
to which we have to give a meaning. The only reasonable solution is to suppose
that the partial products
Pn = al·· .an
tend to a limit P, which by convention will be the value of the product. Taken
literally, the definition is of no interest if some of the an are zero; so we agree
to omit them from the product.
Experience shows that, with exceptions, the problem is again uninteresting (or that we do not know how to treat it ... ) if the limit P is zero. Suppose
therefore that P f=. o. Then an = Pn/Pn-l tends to pip = 1, which allows us
to put an = 1 + Un where Un tends to o. Then
log IPnl
log 11 + ull + ... + log 11 + unl ::;
< 10g(1 + lUll) + ... + 10g(1 + Iunl)
since the function log is increasing. Now we have shown in Chap. II, nO 10,
that log x ::; x-I for all x > 0, i.e.
10g(1 + x) ::; x for all x > -1.
Thus
log IPnl ::; lUll + ... + Iunl·
Suppose now that the series L Un is absolutely convergent. The preceding
inequality shows that the sequence (log IPnl) is bounded above by a number
M > o. The relation log IPnl ::; M implies IPnl ::; eM = M', so the sequence
(Pn) is bounded.
The inequality
IPn - Pn-ll = IPn-1Unl ::; M'lunl
now shows that the series with general term Pn - Pn-l is absolutely convergent, so convergent. Whence the existence of P = limpn.
It remains to check that P f=. O. For this, one replaces the product of the
an by that of their reciprocals a;:;-l = 1 + Vn , which replaces Pn by qn = I/Pn.
The relation (1 + un )(1 + vn ) = 1 shows that
Ivnl = Iun l/ll + unl ::; 21un l for n large
since 11 + unl, which tends to 1, is ~ 1/2 for n large. The series Vn therefore
converges absolutely. Consequently qn tends to a limit q, and since Pnqn = 1
for all n, it follows that pq = 1, whence P f=. o. In conclusion: