Partitions into distinct parts and into odd parts

Idea. Compare two infinite products as formal power series.

Distinct parts. Choosing, for every $k\ge1$, whether $k$ appears (0 or 1 times) shows that
$$D(x)=\sum_{n\ge0}d_nx^n=\prod_{k\ge1}(1+x^k),$$
where $d_n$ is the number of partitions of $n$ into distinct parts.

Odd parts. A part $2k-1$ can be used any number of times: it contributes the factor $\dfrac1{1-x^{2k-1}}$. So
$$O(x)=\sum_{n\ge0}o_nx^n=\prod_{k\ge1}\frac1{1-x^{2k-1}}.$$

Equality. Since $1+x^k=\dfrac{1-x^{2k}}{1-x^k}$,
$$D(x)=\prod_{k\ge1}\frac{1-x^{2k}}{1-x^k}=\frac{\prod_{k\ge1}(1-x^{2k})}{\prod_{k\ge1}(1-x^{k})}.$$
The numerator contains exactly the factors of the denominator with even index, which cancel; what remains in the denominator is $\prod_{k\ge1}(1-x^{2k-1})$. Therefore $D(x)=O(x)$ and $d_n=o_n$.

Check. The values $d_n=o_n$ for $n=1,\dots,12$ are $1,1,2,2,3,4,5,6,8,10,12,15$.

Lesson. Encode a counting problem as a product of simple series and compare the products algebraically.