How to solve this summation (Lerch Transcendent)?

Question

How is it possible to deduce the closed form of the following?

$$\sum_{i = 0}^{n - 1} \frac{2^i}{n - i} = ?$$

Answer 1

$\frac{2^i}{n-i} = \frac{1}{2^{-i}(n-i)} = \frac{2^n}{2^{n-i}(n-i)}$

So, $\sum_{i=0}^{n-1}\frac{2^i}{n-i} = \sum_{i=0}^{n-1}\frac{2^n}{2^{n-i}(n-i)} = 2^n \sum_{i=0}^{n-1}\frac{1}{2^{n-i}(n-i)} = 2^n \sum_{i=1}^{n}\frac{1}{i2^{i}}\tag1$

Now, $\frac{d}{dx}\sum_{i=1}^n\frac{x^i}{i} = \sum_{i=1}^{n-1}x^i = \frac{1-x^n}{1-x}\tag2$

To achieve a closed form of (1), integrate (2) and substitute $\frac{1}{2}$ for $x$.