Evaluate $\sum_{n=1}^{\infty}{\frac{1}{n 2^{n-1}}}$

I need to find $$S = \sum_{n=1}^{\infty}{\frac{1}{n 2^{n-1}}}$$

Attempt:

$$f'(x) = \sum_{n=1}^{\infty}\frac{x^n}{2^n} = \frac{x}{2-x}$$

Which is just evaluating geometric series

$$f(x) = \sum_{n=1}^{\infty}\frac{x^{n+1}}{(n+1)2^{n}}$$

Now, by finding antiderivative of $$\frac{x}{2-x}$$

$$f(x) = -x-2\ln(x-2)$$

Finding sum should be just $$S = 1 + f(1)$$

But $$f(x)$$ is undefined as a real-valued function for $$x \leq 2$$