# the limit of infinite product $(1+y)(1+y^2)(1+y^3)(1+y^4)\cdots$

I wonder if the function $(1+y)(1+y^2)(1+y^3)(1+y^4)\cdots, 0< y<1$, converges to some well-known function.

If we let $(1+y)(1+y^2)(1+y^3)(1+y^4)\cdots = \prod_{i=1}^\infty (1+y^i) = \sum_{i=0}^\infty a_i y^i$ then $a_i$ satisfies the following relation: $$a_0=1, i > 0$$ $a_i$ is the the number of partitions $(b_1, \cdots , b_s)$ of natural number $i$ such that $\sum_{t=1}^s b_t =i$ and $b_t < b_{t+1}$. For instance $a_{10} = 9$

$$(1,2,3,4) \; (1,2,7) \; (1,3,6) \; (1,4,5) \; (1,9) \; (2,8) \; (3,7) \; (4,6) \; (10)$$

$a_{10}$ is like Ramanujan's function $p(n)$. Is there anything in number theory related with $a_i$? At any rate the infinite product converges the function we know well? That is to say, it is a

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I found the material :

Let $p(n)$ is a Ramanujan function. For instance p(4) =4 :

1+1+1+1 = 1+ 3  =2+2 = 2+ 1+1


$\sum^\infty_{n=0} p(n) x^n = \Pi^\infty_{n=1} \frac{1}{1-x^n}$

Here the convenience is $p(0)=0$

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en.wikipedia.org/wiki/Q-Pochhammer_symbol – user17762 Oct 16 '12 at 6:53

## 2 Answers

There is not a simple closed function for $$\prod_{i=1}^\infty (1+y^i)$$ but there are the alternatives $$1\left/\prod_{m =0}^\infty \left(1-y^{2m+1}\right)\right.$$ and $$\sum_{k= 0}^\infty \prod_{j=1}^k \frac{y^j}{1-y^j}.$$

OEIS A000009 has more information.

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+1 I had some difficulty reading your sentence, you are saying the three expressions are all equal, but none of them qualifies as closed form. – Marc van Leeuwen Oct 16 '12 at 7:54
It is nice to notice that the equality of the first two expressions here is precisely the theorem of Euler. – Mariano Suárez-Alvarez Oct 17 '12 at 3:21

One extraordinarily beautiful result, due to Euler, is that the number of partitions of a number with odd parts is always equals the number of partitions of that number with distinct parts. Your $a_i$ is the latter.

This is the beginning of a beautiful part of combinatorics.

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I do not understand the number of partitions of a number with odd parts. What is odd parts ? – HK Lee Oct 16 '12 at 13:32
@user37116: You can decompose 7=1+1+5=1+3+3 and other ways, but not 7=1+2+4 because 2 and 4 are even. You might start with Wikipedia on partitions – Ross Millikan Oct 16 '12 at 13:47
It is a partition with odd parts, not a number with odd parts. Maybe clearer to say partitions of a number into odd parts. – GEdgar Oct 16 '12 at 14:44