Eisenstein-type series

Question

Is the series, $$1 - 24\sum_{n = 1}^\infty \frac{q^{2n}}{(1 - q^{2n})^2}, \quad q = e^{\pi i \tau}, \quad \textbf{I}[\tau] > 0,$$ somehow related to $$E_2(q) = 1 - 24\sum_{n = 1}^\infty \frac{nq^{2n}}{1 - q^{2n}}, \quad q = e^{\pi i \tau}, \quad \textbf{I}[\tau] > 0,$$ the Eisenstein series of weight 2?

Answer 1

It appears that they are one and the same. In fact, it is very easy to establish.

Let $\sum$ denote the summation that runs over all the natural numbers. Then \begin{align*} \sum_n \frac{q^{2n}}{(1 - q^{2n})^2} &= \sum_n \sum_m q^{2n} m q^{2n(m - 1)} \\ &= \sum_m \sum_n m q^{2nm} \\ &= \sum_m \frac{m q^{2m}}{1 - q^{2m}}. \end{align*}