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

Lemma.  Let the numbers $a_{i}$ be nonnegative reals.  The infinite product

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

converges iff the series  $a_{1}\!+\!a_{2}\!+\!a_{3}\!+\ldots$  is convergent   .

Proof.  Denote

 $\sum_{i=1}^{n}a_{n}\;:=\;s_{n},\quad\prod_{i=1}^{n}(1\!+\!a_{i})\;:=\;t_{n}% \qquad(n=1,\,2,\,\ldots).$

Now  $\displaystyle 1\!+\!a_{i}\,\leqq\,1\!+\!\frac{a_{i}}{1!}\!+\!\frac{a_{i}^{2}}{% 2!}\!+\ldots\;=\;e^{a_{i}}$,  whence

 $\displaystyle t_{n}\;\leqq\;\prod_{i=1}^{n}e^{a_{i}}\;=\;e^{s_{n}}.$ (2)

We can estimate also downwards:

 $\displaystyle t_{n}\;=\;(1\!+\!a_{1})(1\!+\!a_{2})\cdots(1\!+\!a_{n})\;=\;1\!+% \!\sum_{i=1}^{n}a_{i}\!+\ldots+\!a_{1}a_{2}\cdots a_{n}\;>\;\sum_{i=1}^{n}a_{i% }\;=\;s_{n}$ (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  $\langle t_{1},\,t_{2},\,t_{3},\,\ldots\rangle$  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  $\displaystyle\lim_{n\to\infty}s_{n}=\infty$  and by (3), also  $\displaystyle\lim_{n\to\infty}t_{n}=\infty$,  i.e. the (1) diverges.

Title infinite product of sums $1\tmspace-{.1667em}+\tmspace-{.1667em}a_{i}$ InfiniteProductOfSums1ai 2013-03-22 18:40:01 2013-03-22 18:40:01 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 40A20 msc 26E99 LimitOfRealNumberSequence DeterminingSeriesConvergence InfiniteProductOfDifferences1A_i AbsoluteConvergenceOfInfiniteProductAndSeries