infinite product of sums $1-.1667em+-.1667em{a}_i$

Lemma.  Let the numbers $a_i$ be nonnegative reals.  The infinite product

${\displaystyle \prod _{i=1}^{\mathrm{\infty }}}(1+a_i)=(1+a_1)(1+a_2)(1+a_3)\mathrm{\cdots }$ (1)

converges iff the series  $a_1+a_2+a_3+\mathrm{\dots }$  is convergent.

Proof.  Denote

$$\sum _{i=1}^n{a}_n:=s_n,\prod _{i=1}^n(1+a_i):=t_n\mathit{\hspace{1em}\hspace{1em}}(n=1,\mathrm{\hspace{0.17em}2},\mathrm{\dots }).$$

Now  $1+a_i\leqq \mathrm{\hspace{0.17em}1}+{\displaystyle \frac{a_i}{1!}}+{\displaystyle \frac{a_i^2}{2!}}+\mathrm{\dots }=e^{a_i}$,  whence

$t_n\leqq {\displaystyle \prod _{i=1}^n}{e}^{a_i}=e^{s_n}.$ (2)

We can estimate also downwards:

$t_n=(1+a_1)(1+a_2)\mathrm{\cdots }(1+a_n)=\mathrm{\hspace{0.33em}1}+{\displaystyle \sum _{i=1}^n}{a}_i+\mathrm{\dots }+a_1{a}_2\mathrm{\cdots }{a}_n>{\displaystyle \sum _{i=1}^n}{a}_i=s_n$ (3)

If the series is convergent with sum $S$, then by (2),

$$t_n\leqq e^{s_n}\leqq e^S,$$

and since the monotonically nondecreasing sequence  $⟨t_1,t_2,t_3,\mathrm{\dots }⟩$  thus is bounded from above, it converges (cf. limit of nondecreasing sequence).  So (1) converges.

If, on the other hand, the series is divergent, then  $\underset{n\to \mathrm{\infty }}{lim}{s}_n=\mathrm{\infty }$  and by (3), also  $\underset{n\to \mathrm{\infty }}{lim}{t}_n=\mathrm{\infty }$,  i.e. the (1) diverges.

Title	infinite product of sums $1-.1667em+-.1667em{a}_i$
Canonical name	InfiniteProductOfSums1ai
Date of creation	2013-03-22 18:40:01
Last modified on	2013-03-22 18:40:01
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	6
Author	pahio (2872)
Entry type	Theorem
Classification	msc 40A20
Classification	msc 26E99
Related topic	LimitOfRealNumberSequence
Related topic	DeterminingSeriesConvergence
Related topic	InfiniteProductOfDifferences1A_i
Related topic	AbsoluteConvergenceOfInfiniteProductAndSeries