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I have to evaluate an infinite product in the form

$$\prod_{i=0}^{\infty} \left( 1 + f_i \right) $$

where the $f_i$ are real numbers. I came out with the following trick: by noting that, calculating all the products, I will get all the possible 0-order terms (the only term of that kind being $1$), all possible 1st-order terms ($f_1 + \cdots + f_n$), all possible 2nd-order terms ($f_1 \cdot f_2 + f_1 \cdot f_3 \cdots $) and so on... and by noting that each term shows up exactly one time, and that permutations won't show up in the product, I wrote: $$ \prod_{i=0}^{\infty} \left( 1 + f_i \right) = \left(1 + \sum_i^\infty f_i + \frac{1}{2!} \sum_i^\infty f_i \sum_j^\infty f_j + \cdots \right) = \exp \left( \sum_i^\infty f_i\right) $$

the combinatorial factors $\frac{1}{n!}$ arise from the no-permutations condition... However Wikipedia says that this is not the correct result, but only an upper bound... I can't see why it is so. (first question)

(Second question) I would like to demonstrate that, given the right assumptions, what I found is the right result. The extra assumption I can make on $f_i$ is that they are positive real numbers; also I know for sure that the result will converge. Is it enough?

Thanks in advance!

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In your $\frac{1}{2!} \sum_i^\infty f_i \sum_j^\infty f_j$ term, what happens when $i=j$? – Henry Jan 9 '12 at 1:26
They are fully independent, maybe my notation is confusing, think of it as $\frac{1}{2!} \left( \sum_i^\infty f_i \right) \left( \sum_j^\infty f_j \right) = \frac{1}{2!} \left( \sum_i^\infty f_i \right)^2$ – zakk Jan 9 '12 at 1:29
Then your first equals sign is not correct. – Henry Jan 9 '12 at 1:30
Care to explain why, please? I wrote down a few terms, like ($f_1 \cdot f_2 + f_1 \cdot f_3 + f_2 \cdot f_3 \cdots $) and I'm quite convinced that this is equal to what I wrote... Edit: ok I can see it now! – zakk Jan 9 '12 at 1:34
You still have an upper bound (under certain conditions). But note that if the $f_i$ are rational then the left hand side also is, while the right hand side is probably not. – Henry Jan 9 '12 at 1:44
up vote 3 down vote accepted

Let's take a simple case where $f_0=0$, $f_1=1$, $f_2=2$ and $f_i=0$ for $i \ge 3$.

Then $\prod_{i=0}^{\infty} \left( 1 + f_i \right) = 2 \times 3 =6$

while $\left(1 + \sum_i^\infty f_i + \frac{1}{2!} \sum_i^\infty f_i \sum_j^\infty f_j + \cdots \right) = 1 + 3 + \frac{3^2 }{2!} + \cdots$, which is clearly more.

To make the first equality work, you need summation terms like $$\frac{1}{2!} \sum_i f_i \sum_{j \not = i} f_j$$ but if you do then the second equality will fail.

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So what I found is only an upper bound... Is there any way to find the conditions on which this upper bound will be the exact result of the product? (I'm working on a physics calculation, so most probably leaving out a term out an infinite sum would be ok and yield the same result, so I would like to know if there is a way of determining quantitatively how wrong I was) – zakk Jan 9 '12 at 1:46
$\exp(\sum f_i) = \prod \exp(f_i)$. For all real $f_i$ we have $1+f_i\leq \exp(f_i)$ with equality only when $f_i=0$. So in general you would never expect equality unless all the $f_i$'s are zero. – Ben Blum-Smith Jan 9 '12 at 4:10

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