I'm looking for a closed-form solution to this infinite series:
$$S(\alpha):=\sum_{n=1}^{\infty}\frac{\log{(1+n)}}{(1+n)^{\alpha}-1},~~~\Re(\alpha)>1.$$
My attempt
All I've really been able to do is confirm this series converges for the specified values of $\alpha$ via the integral test:
$$\begin{align} S(\alpha)\leq I(\alpha)&=\int_{0}^{\infty}\frac{\log{(1+x)}}{(1+x)^{\alpha}-1}\mathrm{d}x\\ &=\int_{0}^{\infty}\frac{u\,e^u}{e^{\alpha u}-1}\mathrm{d}u\\ &=\int_{0}^{\infty}ue^u\sum_{n=1}^{\infty}e^{-n\alpha u}\mathrm{d}u\\ &=\sum_{n=1}^{\infty}\int_{0}^{\infty}ue^ue^{-n\alpha u}\mathrm{d}u\\ &=\sum_{n=1}^{\infty}\frac{1}{(n\alpha-1)^2},~~~\text{for }\Re(\alpha)>1\\ &=\frac{1}{\alpha^2}\psi^{(1)}{\left(1-\frac{1}{\alpha}\right)} \end{align}$$
where $\psi^{(1)}(x)$ is the first derivative of the digamma function.
I thought about expanding the logarithmic part in terms of its power series to write $S(\alpha)$ as a double-sum, and then try switching the order of summation, e.g.:
$$\begin{align} S(\alpha)&=\sum_{n=1}^{\infty}\frac{\log{(1+n)}}{(1+n)^{\alpha}-1}\\ &=-\sum_{n=1}^{\infty}\frac{1}{(1+n)^{\alpha}-1}\sum_{k=1}^{\infty}\frac{(-1)^kn^k}{k}\\ &=-\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\frac{n^k}{(1+n)^{\alpha}-1}\\ &=-\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\sum_{n=1}^{\infty}\frac{n^k}{(1+n)^{\alpha}-1}. \end{align}$$
But I'm not sure how to proceed from there. Suggestions?