Denote the $q$-expressions $[q]\_n=(1-q)\cdots(1-q^n)$ for $n\geq1$ and $[q]\_0:=1$. Also, $[q]\_{\infty}=(1-q)(1-q^2)(1-q^3)\cdots$.



> 
> **QUESTION.** Is this identity true? It seems to be.
>  $$\sum\_{n=0}^{\infty}\left(1-\frac{[q]\_{\infty}}{[q]\_n}\right)
> =\sum\_{k=1}^{\infty}\frac{q^k}{1-q^k}.$$
> 
> 
> 



A $q$-series identity: proof request
As darij grinberg says, the result is equivalent to the main result in the linked paper. On the right-hand side we have


$$\sum\_{k=1}^\infty \frac{q^k}{1-q^k} = \sum\_{k=1}^\infty (q^k + q^{2k} + \cdots ) = \sum\_{n=1}^\infty \bigl( \sum\_{d \mid n} 1 \bigr) q^n = \sum\_{n=1}^\infty \tau(n) q^n $$


where $\tau(n)$ is the number of divisors of $n$. 


For the left-hand side, observe that $-[q]\_\infty/[q]\_{m-1} = -(1-q^m)(1-q^{m+1}) \cdots $ is the generating function for partitions with distinct parts not less than $m$, counting partitions with an even number of parts with sign $-1$. Therefore in $\sum\_{m=1}^\infty (1- [q]\_\infty/[q]\_{m-1})$, a partition with smallest part $b(\lambda)$ is counted precisely $b(\lambda)$ times, once in each of the first $b(\lambda)$ summands, and always with the same sign. Therefore for $n \in \mathbb{N}$, the coefficient of $q^n$ in the left-hand side is


$$\sum\_{\lambda \vdash n} (-1)^{\ell(\lambda)-1} b(\lambda). $$


By the [main result](https://www.cs.vu.nl/~wanf/pubs/partition.pdf) of Bing, Fokkink and Fokkink, this is $\tau(n)$. Since the constant terms on either side are $0$, this proves the identity.


*Remark.* A well-known recurrence for the partition function is $np(n) = \sum\_{m < n} \sigma(m)p(n-m)$, where $\sigma$ is the sum of divisors function. MacMahon published [two](https://londmathsoc.onlinelibrary.wiley.com/doi/10.1112/plms/s2-19.1.305) [papers](https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/plms/s2-19.1.75) in the 1920s with many related identities, replacing $\sigma(n)$ with $\sigma\_s(n) = \sum\_{d \mid m} d^s$. But I cannot find this result in his work.


