# infinite product of differences $1\tmspace-{.1667em}-\tmspace-{.1667em}a_{i}$

We consider the infinite products of the form

 $\displaystyle\prod_{i=1}^{\infty}(1\!-\!a_{i})\,=\,(1\!-\!a_{1})(1\!-\!a_{2})(% 1\!-\!a_{3})\cdots$ (1)

and the series  $a_{1}\!+\!a_{2}\!+\!a_{3}\!+\ldots$  where the numbers $a_{i}$ are nonnegative reals.

• If the series converges, then also the product converges and has a value which does not depend on the order of the factors and vanishes only when some of the factors is 0.

• If  $\displaystyle\lim_{i\to\infty}a_{i}=0$  but the series diverges, then the value of the infinite product is always zero though no of the factors were zero.

Example.$(1\!-\!\frac{1}{2})(1\!-\!\frac{1}{3})(1\!-\!\frac{1}{4})\cdots\;=\;0$;  see the harmonic series  .

Proof.$1^{\circ}$.  Now we have  $\displaystyle\lim_{i\to\infty}a_{i}=0$  (see the necessary condition of convergence of series), and so  $a_{i}<\frac{1}{2}$  when  $i\geqq i_{0}$.  We write

 $\displaystyle\prod_{i=1}^{\infty}(1\!-\!a_{i})\,=\,\prod_{i=1}^{i_{0}-1}(1\!-% \!a_{i})\prod_{i=i_{0}}^{\infty}(1\!-\!a_{i})$ (2)

and set in the last product

 $1\!-\!a_{i}\;=\;\frac{1}{\frac{1}{1\!-\!a_{i}}}\,=\,\frac{1}{1+\frac{a_{i}}{1% \!-\!a_{i}}},$

whence

 $\displaystyle\prod_{i=i_{0}}^{n}(1\!-\!a_{i})\,=\,\frac{1}{\prod_{i=i_{0}}^{n}% \left(1+\frac{a_{i}}{1\!-\!a_{i}}\right)}.$ (3)

As  $a_{i}<\frac{1}{2}$, we have  $\displaystyle\frac{1}{1\!-\!a_{i}}<2$  and thus  $\displaystyle 0<\frac{a_{i}}{1\!-\!a_{i}}<2\cdot a_{i}$,  and therefore the series $\displaystyle\sum_{i=i_{0}}^{\infty}\frac{a_{i}}{1\!-\!a_{i}}$ with nonnegative is absolutely convergent.  The theorem of the http://planetmath.org/node/6204parent entry then says that the product in the denominator of the right hand side of (3) tends, as  $n\to\infty$,  to a finite non-zero limit, which don’t depend on the order of the factors.  Consequently, the same concerns the product of the left hand side of (3).  By (2), we now infer that the given product (1) converges, its value is on the order and it vanishes only along with some of its factors.
$2^{\circ}$.  There is an $i_{0}$ such that  $a_{i}<1$  when  $i\geqq i_{0}$,  whence  $\frac{a_{i}}{1\!-\!a_{i}}>a_{i}$  and the series  $\displaystyle\sum_{i=i_{0}}^{\infty}\frac{a_{i}}{1\!-\!a_{i}}$ diverges.  The denominator of the right hand side of (3) tends, as  $n\to\infty$,  to the infinity and thus the product of the left hand side to 0.  Hence the value of (1) is necessarily 0, also when all factors were distinct from 0.

## References

• 1 E. Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset III.2.  Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1940).
Title infinite product of differences $1\tmspace-{.1667em}-\tmspace-{.1667em}a_{i}$ InfiniteProductOfDifferences1ai 2013-03-22 18:39:45 2013-03-22 18:39:45 pahio (2872) pahio (2872) 12 pahio (2872) Theorem msc 40A20 msc 26E99 HarmonicSeriesOfPrimes InfiniteProductOfSums1A_I InfiniteProductOfSums1a_i