How to show $1-\sum_{n=1}^{\infty}\frac{24ne^{-2\pi n/5}i^{8n/5}}{1-e^{-2\pi n/5}i^{8n/5}}=\frac{15}{\pi}.$

Let$$\tau = \frac{2+i}{5}, \qquad q = e^{2\pi i\tau}$$Then the desired sum is$$E_2(\tau) = 1 - 24 \sum_{n=1}^{\infty} \frac{nq^n}{1-q^n} $$where $E_2$ is the Eisenstein series. Now, this remarkable identity is effectively enough to get the desired result: $$E_2\left(\frac{a\tau+b}{c\tau+d}\right) = (c\tau+d)^2 E_2(\tau) - \frac{6ic}{\pi}(c\tau+d)$$ For any $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \text{SL}_2(\mathbb{Z})$. Specifically, choosing the matrix $\gamma=\begin{pmatrix} 2 & -1 \\ 5 & -2 \end{pmatrix}$ because $\gamma\tau = \tau$ we immediately have$$E_2(\tau) = (5\tau - 2)^2 E_2(\tau) - \frac{6i \cdot 5}{\pi} (5\tau - 2)= -E_2(\tau) + \frac{30}{\pi}$$ $$\implies E_2(\tau)=\frac{15}{\pi}$$ The Eisenstein series identity follows from the Dedekind functional equation proven here $(1.2)$ and then simply using the relation between the functions:$$\frac{d}{d\tau} \log \eta(\tau) = \frac{\pi i}{12} E_2(\tau)$$