An introduction to the study of the elements of the differential and integral calculus. From the German of the late Axel Harnack, With the permission of the author.

~ 162. 163. Condition that an infinite product may be convergent.

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investigations apply also in case there are complex factors. If, in an
infinite product, we give the form:
(] +.1^) (1 + t2). (1 + - n) (1 + 2un+)...
to the factors that can be unrestrictedly continued in accordance with
some law, the successive values obtained by multiplying, first n factors,
then n + 1,...n + 7,...:
Pn (1 + t) (1 + 2). (1 + tI),
PnJ+1 (1 + t) (1 + t2). (1 -+ un) (1 + Ui+1l),
Pn+k = 1)(1 L + na2) * * ) (1 -+ un) (1 + Ua+l).*. (1 n+ n+k)
must form a sequence of numbers with a determinate finite limiting
value. For this it is requisite: first that none of the products P, and
therefore also none of the terms u shall become infinite; second that for
any number 8 however small, there shall be a place n such that:
abs [Pn+k - P,] < (d
for every value of k. From this inequality follows, provided Pn does
not fall below any finite assignable limit, that:
abs [ P L < -, or: abs [P ] < 1 + abs p;
in other words: There must be a place n from which onwards the
ratio of the values P differs inappreciably from unity; this must also
be the case in particular for:
Pn +l' P. -=-1 + Un+1t,
i. e. the terms u must necessarily converge to zero.
The case, that the quantities P sink below any finite amount, or
that separate factors vanish and the limiting value of the product
is therefore zero, must be excluded both here and in the following
investigations.
If a product still converge even when we give all the terms u
their absolute values, it is called absolutely convergent.
This definition requires a preliminary proof, that the convergence
of the product TT(1 + un) is always a necessary consequence of the
convergence of the product formed of the absolute values v of the
terms u; the case that even a single quantity u is = - 1 is excluded.
Denoting by vr the absolute amount of Ur, and by Qn the product:
we have:
(Q - ) = (1 + Vu+l) (1 + v+2).. (1 + n+k) - 1,
( P, n) K(1 + n+l, (1 + Un+2). (1+ 1k)- 1.
HArKNACK, Calculus. 19