# 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.

Title proof of convergence criterion for infinite product ProofOfConvergenceCriterionForInfiniteProduct 2013-03-22 15:35:36 2013-03-22 15:35:36 cvalente (11260) cvalente (11260) 10 cvalente (11260) Proof msc 26E99