Interesting (double...) series !

The key idea to solve this problem is that for any positive integer $n$ and any strictly positive odd integer $k, \ d(2^n k)=k$ .

I will denote the set of all the odd positive integer as $\mathbb{I}$ (i for "impair", odd in french, not a conventional notation).

You probably know that the function bellow is a bijection (it is a consequence of the fundamental theorem of arithmetic):

$\begin{array}{lrcl} \varphi : & \mathbb{N}\times\mathbb{I} & \longrightarrow & \mathbb{N^*} \\ & (n,k) & \longmapsto & 2^n k \end{array}$

Therefore,

$\begin{aligned} \displaystyle \sum_{n=1}^{+\infty} \dfrac{1}{nd(n)} & = \displaystyle \sum_{(n,k)\in \mathbb{N}\times\mathbb{I}} \dfrac{1}{2^n k \times k} \\ & = \displaystyle \sum_{k\in\mathbb{I}}\sum_{n=0}^{+\infty} \dfrac{1}{2^n k^2} \\ & = \displaystyle \sum_{k\in\mathbb{I}} \dfrac{2}{k^2} \\ & = 2\displaystyle \sum_{n=1}^{+\infty} \dfrac{1}{n^2} - 2\displaystyle \sum_{n=1}^{+\infty} \dfrac{1}{(2n)^2}\\ & = 2\displaystyle \sum_{n=1}^{+\infty} \dfrac{1}{n^2} - \dfrac{1}{2}\displaystyle \sum_{n=1}^{+\infty} \dfrac{1}{n^2}\\ & = \dfrac{3}{2} \displaystyle \sum_{n=1}^{+\infty} \dfrac{1}{n^2}\\ & = \dfrac{3}{2} \times \dfrac{\pi^2}{6}\\ & = \boxed{\dfrac{\pi^2}{4}} \end{aligned}$