In mathematics, for a sequence of complex numbers a₁, a₂, a₃, ... the infinite product

${\displaystyle \prod _{n=1}^{\infty }a_{n}=a_{1}a_{2}a_{3}\cdots }$

is defined to be the limit of the partial products a₁a₂...a_n as n increases without bound. The product is said to converge when the limit exists and is not zero. Otherwise the product is said to diverge. A limit of zero is treated specially in order to obtain results analogous to those for infinite sums. Some sources allow convergence to 0 if there are only a finite number of zero factors and the product of the non-zero factors is non-zero, but for simplicity we will not allow that here. If the product converges, then the limit of the sequence a_n as n increases without bound must be 1, while the converse is in general not true.

The best known examples of infinite products are probably some of the formulae for π, such as the following two products, respectively by Viète (Viète's formula, the first published infinite product in mathematics) and John Wallis (Wallis product):

${\displaystyle {\frac {2}{\pi }}={\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdot \;\cdots }$

${\displaystyle {\frac {\pi }{2}}=\left({\frac {2}{1}}\cdot {\frac {2}{3}}\right)\cdot \left({\frac {4}{3}}\cdot {\frac {4}{5}}\right)\cdot \left({\frac {6}{5}}\cdot {\frac {6}{7}}\right)\cdot \left({\frac {8}{7}}\cdot {\frac {8}{9}}\right)\cdot \;\cdots =\prod _{n=1}^{\infty }\left({\frac {4n^{2}}{4n^{2}-1}}\right).}$

Convergence criteria

The product of positive real numbers

${\displaystyle \prod _{n=1}^{\infty }a_{n}}$

converges to a nonzero real number if and only if the sum

${\displaystyle \sum _{n=1}^{\infty }\log(a_{n})}$

converges. This allows the translation of convergence criteria for infinite sums into convergence criteria for infinite products. The same criterion applies to products of arbitrary complex numbers (including negative reals) if the logarithm is understood as a fixed branch of logarithm which satisfies ln(1) = 0, with the proviso that the infinite product diverges when infinitely many a_n fall outside the domain of ln, whereas finitely many such a_n can be ignored in the sum.

For products of reals in which each ${\displaystyle a_{n}\geq 1}$, written as, for instance, ${\displaystyle a_{n}=1+p_{n}}$, where^{[clarification needed]} ${\displaystyle p_{n}\geq 0}$, the bounds

${\displaystyle 1+\sum _{n=1}^{N}p_{n}\leq \prod _{n=1}^{N}\left(1+p_{n}\right)\leq \exp \left(\sum _{n=1}^{N}p_{n}\right)}$

show that the infinite product converges if the infinite sum of the p_n converges. This relies on the Monotone convergence theorem. We can show the converse by observing that, if ${\displaystyle p_{n}\to 0}$, then

${\displaystyle \lim _{n\to \infty }{\frac {\log(1+p_{n})}{p_{n}}}=\lim _{x\to 0}{\frac {\log(1+x)}{x}}=1,}$

and by the limit comparison test it follows that the two series

${\displaystyle \sum _{n=1}^{\infty }\log(1+p_{n})\quad {\text{and}}\quad \sum _{n=1}^{\infty }p_{n},}$

are equivalent meaning that either they both converge or they both diverge.

If the series ${\textstyle \sum _{n=1}^{\infty }\log(a_{n})}$ diverges to ${\displaystyle -\infty }$, then the sequence of partial products of the a_n converges to zero. The infinite product is said to diverge to zero.^[1]

For the case where the ${\displaystyle p_{n}}$ have arbitrary signs, the convergence of the sum ${\textstyle \sum _{n=1}^{\infty }p_{n}}$ does not guarantee the convergence of the product ${\textstyle \prod _{n=1}^{\infty }(1+p_{n})}$. For example, if ${\displaystyle p_{n}={\frac {(-1)^{n}}{\sqrt {n}}}}$, then ${\textstyle \sum _{n=1}^{\infty }p_{n}}$ converges, but ${\textstyle \prod _{n=1}^{\infty }(1+p_{n})}$ diverges to zero. However, if $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}|p_n$Zdroj:https://en.wikipedia.org?pojem=Infinite_product
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