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How to transform this infinite sum


to an infinite product


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You are forgetting a $x^i$ in the numerator of the summand in the first expression. – anon Dec 10 '12 at 14:26
@anon. Thank you i will correct it – Adi Dani Dec 10 '12 at 14:30
Are you sure they're equal? Putting $x = 0$ suggests not. – arkeet Dec 10 '12 at 14:39
My bad, it should be $x^{i-1}$. I am using a different description of the sum. – anon Dec 10 '12 at 14:40
The expression in the problem is not quite correct; the sum needs to start at $i=0$, or else the two expressions differ by $1$. – mjqxxxx Dec 10 '12 at 15:32
up vote 7 down vote accepted

Let $$a_{n}=\frac{x^{n}}{(1-x)(1-x^2)\cdots(1-x^{n})}$$ and $S_{n}=\sum_{i=0}^{n}a_i$. We want to show that $$ S_{n}=\frac{1}{(1-x)(1-x^2)\cdots(1-x^{n})}.$$ Clearly this is true for $n=1$, since $S_0=a_0=1$; and assuming it's true for $n=k$, we have $$ \begin{eqnarray} S_{k+1}&=&S_{k}+a_{k+1} \\ &=&\frac{1}{(1-x)\cdots(1-x^{k})}+\frac{x^{k+1}}{(1-x)\cdots(1-x^{k})(1-x^{k+1})} \\ &=&\frac{1-x^{k+1}}{(1-x)\cdots(1-x^{k})(1-x^{k+1})}+\frac{x^{k+1}}{(1-x)\cdots(1-x^{k})(1-x^{k+1})} \\ &=& \frac{1}{(1-x)(1-x^2)\cdots(1-x^{k+1})}, \end{eqnarray} $$ i.e., it's true for $n=k+1$. So it's true for all $n$ by induction. In particular, the partial products are equal to the partial sums, and $$ \sum_{i=0}^{\infty}\frac{x^{i}}{(1-x)(1-x^2)\cdots(1-x^{i})}=\prod_{i=1}^{\infty}\frac{1}{1-x^{i}}. $$

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Great and simple answer – Adi Dani Dec 10 '12 at 18:32

Combinatorical arguments are a nice way to manipulate generating functions (and vice-versa); we can make convergence a non-issue by working over formal polynomial series rings if need be.

The $q$-Pochhammer symbol is defined by $(a,q)_n:=(1-a)(1-aq)\cdots(1-aq^{n-1})$. The inverse of $(q,q)_n$ is a generating function for partitions of a certain type:



The coefficient of $q^\ell$ above will be the number of ways it can be expressed as

$$q^\ell=q^{1\cdot j_1}q^{2\cdot j_2}\cdots q^{n\cdot j_n}$$

for nonnegative integers $j_i$, $1\le i\le n$, which are in bijective correspondence with partitions

$$(\underbrace{n,\cdots,n}_{j_n},\cdots,\underbrace{2,\cdots,2}_{j_2},\underbrace{1,\cdots,1}_{j_1})\vdash n.$$

These are precisely the integer partitions of $\ell$ with parts having size at most $n$.

The function $(a,q)_\infty^{-1}$ will also be a generating function:


Evidently the coefficient of $a^nq^\ell$ above will be the number of ways to express it as

$$a^nq^\ell=a^{(j_0+j_1+j_2+\cdots)}(q^{0\cdot j_0}q^{1\cdot j_1}q^{2\cdot j_2}\cdots).$$

(Obviously almost all $j_i$s will be $0$ in these expressions.) These correspond to the partitions

$$(\cdots,\underbrace{3,\cdots,3}_{j_3},\underbrace{2,\cdots,2}_{j_2},\underbrace{1,\cdots,1}_{j_1})\vdash \ell.$$

These are precisely the integer partitions of $\ell$ having at most $n=\cdots+j_1+j_0$ parts. The partitions of $\ell$ into at most $n$ parts are in bijective correspondence with the partitions of $\ell$ into parts of size at most $n$ - the bijection is given by conjugating, or flipping the axes of, the Young diagrams that are associated to the partitions.) Thus the $q$-coefficient of $a^n$ in $(a,q)_\infty^{-1}$ is the generating function (in $q$) for partitions of integers $\ell$ into parts of size at most $n$, which we have already established is $(q,q)_n^{-1}$.



Plugging in $a=q$ gives your identity as the particular case


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don't know the truth, but the sum wants to be a collapsing sum as $\frac{x^i}{(1-x)(1-x^2)\cdots(1-x^i)} = - \frac{1}{(1-x)(1-x^2)\cdots(1-x^{i-1})} + \frac{1}{(1-x)(1-x^2)\cdots(1-x^i)} $ – mike Dec 10 '12 at 15:22
@mike: Yes, the series is telescoping in the way you describe, which actually makes for a slick answer requiring no combinatorial thinking. If you spruced up your answer a bit, it would be nice and we'd benefit from you undeleting it. – anon Dec 10 '12 at 15:26
Nevermind, it looks like mjqxxxx has posted it. – anon Dec 10 '12 at 15:27

Start by taking the difference $$ \begin{align} \prod_{i=1}^k\frac1{1-x^i}-\prod_{i=1}^{k-1}\frac1{1-x^i} &=\left(\frac1{1-x^k}-1\right)\prod_{i=1}^{k-1}\frac1{1-x^i}\\ &=\frac{x^k}{1-x^k}\prod_{i=1}^{k-1}\frac1{1-x^i}\\ &=\frac{x^k}{\prod_{i=1}^k(1-x^i)}\tag{1} \end{align} $$ Summing both sides of $(1)$ yields $$ \begin{align} \prod_{i=1}^n\frac1{1-x^i}-1 &=\sum_{k=1}^n\left(\prod_{i=1}^k\frac1{1-x^i}-\prod_{i=1}^{k-1}\frac1{1-x^i}\right)\\ &=\sum_{k=1}^n\frac{x^k}{\prod_{i=1}^k(1-x^i)}\tag{2} \end{align} $$ This proves that $$ \prod_{i=1}^n\frac1{1-x^i}-1=\sum_{k=1}^n\frac{x^k}{\prod_{i=1}^k(1-x^i)}\tag{3} $$ Equation $(3)$ says that the product and sum in the question differ by $1$, which agrees with computation when $x=0$. Noting that the term for $k=0$ on the right side of $(3)$ is $1$, we get $$ \prod_{i=1}^n\frac1{1-x^i}=\sum_{k=0}^n\frac{x^k}{\prod_{i=1}^k(1-x^i)}\tag{4} $$

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The usual technique is to study the function $$f(a, q) = \prod_{n = 1}^{\infty}\frac{1}{1 - aq^{n - 1}}\tag{1}$$ and then express it as a series $$f(a, q) = \sum_{n = 0}^{\infty}x_{n}a^{n}\tag{2}$$ Now we can see that $$f(aq, q) = (1 - a)f(a, q)\tag{3}$$ so that $$\sum_{n = 0}^{\infty}x_{n}a^{n}q^{n} = (1 - a)\sum_{n = 0}^{\infty}x_{n}a^{n}\tag{4}$$ and thus we have $$x_{n}q^{n} = x_{n} - x_{n - 1}$$ or $$x_{n} = \frac{x_{n - 1}}{1 - q^{n}} = \frac{x_{0}}{(1 - q)(1 - q^{2})\cdots(1 - q^{n})}$$ Putting $a = 0$ we can see that $x_{0} = 1$ and hence we have $$\prod_{n = 1}^{\infty}\frac{1}{1 - aq^{n - 1}} = \sum_{n = 0}^{\infty}\frac{a^{n}}{(1 - q)(1 - q^{2})\cdots(1 - q^{n})}\tag{5}$$ Putting $a = q$ we get $$\prod_{n = 1}^{\infty}\frac{1}{1 - q^{n}} = \sum_{n = 0}^{\infty}\frac{q^{n}}{(1 - q)(1 - q^{2})\cdots(1 - q^{n})}$$ which is what we need to prove (question uses $x$ in place of $q$).

The identity $(5)$ is a special case of the $q$-binomial theorem which says that $$\sum_{n = 0}^{\infty}\frac{(a;q)_{n}}{(q;q)_{n}}z^{n} = \frac{(az; q)_{\infty}}{(z; q)_{\infty}}\tag{6}$$ where $$(a;q)_{n} = \prod_{i = 1}^{n}(1 - aq^{i - 1}),\, (a;q)_{\infty} = \prod_{n = 1}^{\infty}(1 - aq^{n - 1})$$ Identity $(5)$ is derived when we put $a = 0$.

The above technique is a formal one (i.e. it does not deal with issues of convergence) which was perhaps championed by Euler and it can be used to prove many identities related to $q$-series.

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