proof of convergence criterion for infinite product

Consider the partial product $P_n={\prod }_{i=1}^n{p}_i$.

By definition we say that the infinite product ${\prod }_{n=1}^{\mathrm{\infty }}{p}_n$ is convergent iff $P_n$ is convergent.

Suppose every $p_n>0$

$\ln$ is a continuous bijection from ${ℝ}^{+}$ to $ℝ$, therefore ${lim}_{n\to \mathrm{\infty }}{a}_n=a\iff {lim}_{n\to \mathrm{\infty }}\ln (a_n)=\ln (a)$, provided $a_n>0$ and $a>0$.

so saying $P_n\to P>0$ is equivalent to saying that $\ln (P_n)$ converges.

Since $\ln (P_n)=\ln ({\prod }_{i=1}^n{p}_i)={\sum }_{i=1}^n\ln (p_i)$, the infinite product converges to a positive value iff the series ${\sum }_{n=1}^{\mathrm{\infty }}\ln (p_n)$ is convergent.

In particular, if the infinite product converges to a positive value, then $\ln (p_n)\to 0⟹p_n\to 1$.

$P_n\to 0$, is equivalent to saying ${\sum }_{n=1}^{\mathrm{\infty }}\ln (p_n)=-\mathrm{\infty }$

For the second part of the theorem:

${\prod }_{n=1}^{\mathrm{\infty }}(1+p_n)$ converges absolutely to a positive value iff ${\sum }_{n=1}^{\mathrm{\infty }}{p}_n$ converges absolutely.

as we have seen, $1+p_n\to 1⟹p_n\to 0$

consider: ${lim}_{x\to 0}\frac{\ln (1+x)}{x}=1$ (this is easy to prove since by Taylor’s expansion $\ln (1+x)=x+O(x^2)$)

Since $p_n\to 0$ we can say that ${lim}_{n\to \mathrm{\infty }}\frac{\ln (1+p_n)}{p_n}=1$ and by the limit comparison test, either both ${\sum }_{n=1}^{\mathrm{\infty }}\ln (1+p_n)$ and ${\sum }_{i=1}^n{p}_i$ converge or diverge.

Title	proof of convergence criterion for infinite product
Canonical name	ProofOfConvergenceCriterionForInfiniteProduct
Date of creation	2013-03-22 15:35:36
Last modified on	2013-03-22 15:35:36
Owner	cvalente (11260)
Last modified by	cvalente (11260)
Numerical id	10
Author	cvalente (11260)
Entry type	Proof
Classification	msc 26E99