proof of infinite product of sums $1-.1667em+-.1667em{a}_i$ result without exponentials

In this entry, we show how the proof presented in the parent entrymay be modified so as to avoid use of the exponential function.This modification makes it more elementary by not requiring thatone first develop the theory of the exponential function beforeproving this basic result about infinite products. Note that itis only necessary to redo the part of the result which states that,if the series converges, then the product also converges becausethe proof of the opposite implication did not involve the exponentialfunction.

We begin with a simple inequality. Suppose that $a$ and $b$ arereal numbers such that $0\le a$ and . Then wehave $2ab\le a$, hence

$(1+a)(1+2b)$$=1+2b+a+2ab$

$\le 1+2b+a+a$

$=1+2(a+b).$

Now suppose that the series $a_1+a_2+a_3+\mathrm{\cdots }$ convergesto a value $S$.Since the convergence of an infinite series or product is notaffected by removing a finite number of terms we may, withoutloss of generality, assume that . Then,since the terms $a_n$ are nonnegative for all $n$, for each partialsum $s_n$ we will have .

Clearly, $t_1\le 1+2s_1$. Suppose that, for some $n$, we have$t_n\le 1+2s_n$. Then, using the definitions of $t_n$ and$s_n$ along with the inequality demonstrated above, we conclude that

$t_{n+1}$$=t_n(1+a_{n+1})$

$\le (1+2s_n)(1+a_{n+1})$

$\le 1+2(s_n+a_{n+1})$

$=1+2s_{n+1}$

Hence, if $t_n\le 1+2s_n$, then$t_{n+1}\le 1+2s_{n+1}$ as well. By induction, we conclude that$t_n\le 1+2s_n$ for all $n$.

Thus, for all $n$, we have $t_n\le 1+2s_n\le 1+2S$. Substitutingthis inequality for the inequality $t_n\le e^{s_n}\le e^S$ in theparent entry, the rest of the proof proceeds in exactly the same manner.