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 $\\prod_{n=1}^{\\infty}p_{n}$ is convergent iff $P_{n}$ is convergent.

Suppose every $p_{n}>0$

$\\ln$ is a continuous bijection from $\\mathbb{R}^{+}$ to $\\mathbb{R}$ , therefore $\\lim_{n\\to\\infty}a_{n}=a\\iff\\lim_{n\\to\\infty}\\ln(a_{n})=\\ln(a)$ , 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 $\\sum_{n=1}^{\\infty}\\ln(p_{n})$ is convergent.

In particular, if the infinite product converges to a positive value, then $\\ln(p_{n})\\to 0\\implies p_{n}\\to 1$ .

$P_{n}\\to 0$ , is equivalent to saying $\\sum_{n=1}^{\\infty}\\ln(p_{n})=-\\infty$

For the second part of the theorem:

$\\prod_{n=1}^{\\infty}(1+p_{n})$ converges absolutely to a positive value iff $\\sum_{n=1}^{\\infty}p_{n}$ converges absolutely.

as we have seen, $1+p_{n}\\to 1\\implies p_{n}\\to 0$

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 $\\lim_{n\\to\\infty}\\frac{\\ln(1+p_{n})}{p_{n}}=1$ and by the limit comparison test, either both $\\sum_{n=1}^{\\infty}\\ln(1+p_{n})$ and $\\sum_{i=1}^{n}p_{i}$ converge or diverge.

单词	ProofOfConvergenceCriterionForInfiniteProduct
释义	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 $\\prod_{n=1}^{\\infty}p_{n}$ is convergent iff $P_{n}$ is convergent. Suppose every $p_{n}>0$ $\\ln$ is a continuous bijection from $\\mathbb{R}^{+}$ to $\\mathbb{R}$ , therefore $\\lim_{n\\to\\infty}a_{n}=a\\iff\\lim_{n\\to\\infty}\\ln(a_{n})=\\ln(a)$ , 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 $\\sum_{n=1}^{\\infty}\\ln(p_{n})$ is convergent. In particular, if the infinite product converges to a positive value, then $\\ln(p_{n})\\to 0\\implies p_{n}\\to 1$ . $P_{n}\\to 0$ , is equivalent to saying $\\sum_{n=1}^{\\infty}\\ln(p_{n})=-\\infty$ For the second part of the theorem: $\\prod_{n=1}^{\\infty}(1+p_{n})$ converges absolutely to a positive value iff $\\sum_{n=1}^{\\infty}p_{n}$ converges absolutely. as we have seen, $1+p_{n}\\to 1\\implies p_{n}\\to 0$ 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 $\\lim_{n\\to\\infty}\\frac{\\ln(1+p_{n})}{p_{n}}=1$ and by the limit comparison test, either both $\\sum_{n=1}^{\\infty}\\ln(1+p_{n})$ and $\\sum_{i=1}^{n}p_{i}$ converge or diverge.
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