About the closed form of $\sum _{k=1}^{n}\binom{n}{k}\frac{1}{k}$

Question

Suppose we are going to evaluate:

$$\sum _{k=1}^{n}\binom{n}{k}\frac{1}{k}$$

Using Pascal's rule we have:

$$\sum _{k=1}^{n}\binom{n}{k}\frac{1}{k}=\sum _{k=1}^{n}\binom{n-1}{k-1}\frac{1}{k}+\underbrace{\sum _{k=1}^{n}\binom{n-1}{k}\frac{1}{k}}_{(1)}$$$$=\sum _{k=1}^{n}\binom{n-1}{k-1}\frac{1}{k}+\underbrace{\sum _{k=1}^{n}\binom{n-2}{k-1}\frac{1}{k}+\sum _{k=1}^{n}\binom{n-2}{k}\frac{1}{k}}_{(1)}$$$$=\sum _{k=1}^{n}\binom{n-1}{k-1}\frac{1}{k}+\sum _{k=1}^{n}\binom{n-2}{k-1}\frac{1}{k}+\sum _{k=1}^{n}\binom{n-3}{k-1}\frac{1}{k}+\sum _{k=1}^{n}\binom{n-3}{k}\frac{1}{k}$$ On the other hand: $$\sum _{k=1}^{n}\binom{n-r}{k-1}\frac{1}{k}=\frac{1}{n-r+1}\sum _{k=1}^{n}\binom{n-r+1}{k}$$$$=\frac{1}{n-r+1}\left[\color{red}{\sum _{k=0}^{n-r+1}\binom{n-r+1}{k}}+\sum _{k=n-r+2}^{n}\binom{n-r+1}{k}-1\right]$$$$=\frac{\color{red}{2^{n-r+1}}-1}{n-r+1}\tag{I}$$

Continuing this way:$$\sum _{k=1}^{n}\binom{n}{k}\frac{1}{k}=\sum _{k=1}^{n}\binom{n-1}{k-1}\frac{1}{k}+\sum _{k=1}^{n}\binom{n-2}{k-1}\frac{1}{k}+...+\sum _{k=1}^{n}\binom{n-(n-1)}{k-1}\frac{1}{k}+\color{blue}{\sum _{k=1}^{n}\binom{n-(n-1)}{k}\frac{1}{k}}$$

Using (I) implies:

$$=\sum_{k=0}^{n-2}\frac{2^{\left(n-k\right)}-1}{n-k}+\color{blue}{1}$$

Hence: $$\bbox[5px,border:2px solid #00A000]{\sum _{k=1}^{n}\binom{n}{k}\frac{1}{k}=\sum_{k=0}^{n-2}\frac{2^{\left(n-k\right)}-1}{n-k}+\color{blue}{1}}$$

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My questions:

1) Is there any elementary way to find a closed form for this expression?

2) Can my last relation be considered as a closed form?

3) Is the last relation new or it has been done before?

Answer 1

$$\sum_{k=1}^{n}\binom{n}{k}\frac{1}{k} = \int_{0}^{1}\sum_{k=1}^{n}\binom{n}{k}x^{k-1}\,dx = \int_{0}^{1}\frac{(1+x)^n-1}{x}\,dx $$ Most of the mass of the integral is concentrated near the right endpoint of the integration range, so it makes sense to enforce the substitution $x\mapsto 1-x$, then expand around the origin: $$ \sum_{k=1}^{n}\binom{n}{k}\frac{1}{k}=\int_{0}^{1}\frac{(2-x)^n-1}{1-x}\,dx = \int_{0}^{1}\sum_{h=0}^{n-1}(2-x)^h\,dx=\sum_{h=0}^{n-1}\frac{2^{h+1}-1}{h+1}=\sum_{m=1}^{n}\frac{2^m}{m}-H_n. $$ A closed form necessarily depends on a closed form for $$ \sum_{m=1}^{n}\frac{2^m}{m} = \frac{2^n}{n} \sum_{m=0}^{n-1}\frac{1}{\left(1-\frac{m}{n}\right)2^m} $$ which I do not believe to exist.