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Homeproof of infinite product of sums $1\!+\!a_i$ result without exponentials

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# proof of infinite product of sums $1\!+\!a_{i}$ result without exponentials

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

We begin with a simple inequality. Suppose that $a$ and $b$ are real numbers such that $0\leq a$ and $0\leq b<1/2$. Then we have $2ab\leq a$, hence

$\displaystyle(1+a)(1+2b)$ | $\displaystyle=1+2b+a+2ab$ | ||

$\displaystyle\leq 1+2b+a+a$ | |||

$\displaystyle=1+2(a+b).$ |

Now suppose that the series $a_{1}+a_{2}+a_{3}+\cdots$ converges to a value $S$. Since the convergence of an infinite series or product is not affected by removing a finite number of terms we may, without loss of generality, assume that $S<1/2$. Then, since the terms $a_{n}$ are nonnegative for all $n$, for each partial sum $s_{n}$ we will have $0\leq s_{n}<1/2$.

Clearly, $t_{1}\leq 1+2s_{1}$. Suppose that, for some $n$, we have $t_{n}\leq 1+2s_{n}$. Then, using the definitions of $t_{n}$ and $s_{n}$ along with the inequality demonstrated above, we conclude that

$\displaystyle t_{{n+1}}$ | $\displaystyle=t_{n}(1+a_{{n+1}})$ | ||

$\displaystyle\leq(1+2s_{n})(1+a_{{n+1}})$ | |||

$\displaystyle\leq 1+2(s_{n}+a_{{n+1}})$ | |||

$\displaystyle=1+2s_{{n+1}}$ |

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

Thus, for all $n$, we have $t_{n}\leq 1+2s_{n}\leq 1+2S$. Substituting this inequality for the inequality $t_{n}\leq e^{{s_{n}}}\leq e^{S}$ in the parent entry, the rest of the proof proceeds in exactly the same manner.

## Mathematics Subject Classification

40A20*no label found*26E99

*no label found*

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