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For an infinite product $(1+a_1)(1+a_2)(1+a_3)\cdots$,

  1. whats the definition of convergence and absolute convergence?

  2. Why the absolute convergence corresponding to the absolute convergence of sum of an infinite series $a_1+a_2+a_3+\cdots$?

  3. If the infinite product absolutely convergent, does it mean it is convergent?

  4. If the sum of $a_1+a_2+a_3+\cdots$ convergent, does it mean the corresponding product convergent?

I know the meaning of the convergent and absolute convergent of an infinite sum $\sum_{n=1}^\infty x_n$, it can be understand very intuitively.

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  • $\begingroup$ We could say that $\displaystyle\prod_{n = 1}^{\infty}(1+a_n)$ converges absolutely if and only if $\displaystyle\sum_{n = 1}^{\infty}\ln(1+a_n)$ converges absolutely. $\endgroup$
    – JimmyK4542
    Commented Jul 31, 2015 at 6:22
  • $\begingroup$ @This is the definition? what about other questions? $\endgroup$
    – an offer can't refuse
    Commented Jul 31, 2015 at 6:27
  • $\begingroup$ For 3, the answer is no, take $a_k=1$ for all $k$. $\endgroup$
    – wythagoras
    Commented Jul 31, 2015 at 6:48
  • $\begingroup$ @wythagoras if $a_k=1$, the product not absolutely convergent. $\endgroup$
    – an offer can't refuse
    Commented Jul 31, 2015 at 9:01
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    $\begingroup$ @wythagoras if all $a_k=1$, $a_1a_2a_3\cdots$ is convergent, it also absolute convergent because \sum_{n} \log a_n absolute convergent. See the answer. $\endgroup$
    – an offer can't refuse
    Commented Jul 31, 2015 at 10:53

1 Answer 1

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You may know that convergence of products is not defined just in terms of "limit of partial products exists", for example $\prod \frac 1n$ is by definition not convergent, even though the partial products converge to $0$. Instead the definition is such that there may be finitely many factors $=0$ and for the remaining factors the sum of the logs is convergent. If we get rid of the logarithm, this gives us the direct definition:

$\prod x_n$ is said to be convergent if there exists $N$ such that the sequence of partial products $\prod_{n=N}^Mx_n$ converges to a nonzero(!) limit as $M\to\infty$.

This is also why absolut convergence of $\prod (1+a_n)$ infinite products is defined as absolute convergence of $\sum\ln(a+a_n)$ (ignoring the possibly finitely many factors $=0$ here). (After all, the signs of the factors themselves cannot play a role: they must converge to $+1$, hence there can only be finitely many negative factors.) As a consequence, we have the same properties as with sums, i.e., absolute convergence implies that the order of factors does not matter.

Now assume $\sum \ln(1+a_n)$ is convergent. Then specifically $a_n\to 0$. As $\ln(1+x)=x+O(x^2)$ we see that the behaviour of $\sum \ln(1+a_n)$ is essentially the same as that of $\sum a_n$. Therefore the convergence/absolute convergence of $\prod(1+a_n)$ can be related to convergence/absolute convergence of $\sum a_n$.

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