How to prove it? (one of the Rogers-Ramanujan identities)

Question

Prove the following identity (one of the Rogers-Ramanujan identities) on formal power series by interpreting each side as a generating function for partitions:

$$1+\sum_{k\geq1}\frac{z^k}{(1-z)(1-z^2)\cdots(1-z^k)}=\prod_{k\geq1}\frac{1}{1-z^k}$$

Answer 1

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

Answer 2

Notations: Let $[n]=\{1,2,\ldots n\}$

$P(n)$ is the number of partitions of $n$.

$P(k,[n])$ is the number of partitions of $k$ taking values from $[n]$. for example-$P(n)=P(n,[n])$

Note that coefficients of $x^n$ in left hand side is $\displaystyle\sum_{k=0}^{n-1}P(k,[n-k])$.Easy enough the coefficients of $x^n$ in the right hand side is $P(n)$.

We shall establish a bijection between them.

For some $m$, we shall denote any partition $<\lambda_1,\lambda_2,\ldots \lambda_m>_n$ among the set of partitions of $n$. With $\lambda_i\le\lambda_j$ for all $i\ge j$

Take any such partition $<\lambda_1,\lambda_2,\ldots \lambda_m>_k$. Construct a new partition $<\lambda_1,\lambda_2,\ldots \lambda_m,\lambda_{m+1}=n-k>_n$.

I claim by constructing new partitions like this from all the partitions of of each $i$, where $i\in[n-1]$ we get unique partitions of $n$.

Remark- For the case $k=0$ we add $n$ to get the trivial partition $<n>_n$.

Hence, on the contrary assume some two of the constructed partitions are equal. But the maximum element is $n-k$ for some partition constructed from $k$. Hence the two equal constructed partitions must be from the same $k$.

Easy enough to prove the partitions are equal.

Now, it remains to show in the other way, i.e. to show by subtracting the maximum element we get all such partitions. Whhich we can do by a very similar arguement.