If $n$ is a positive integer, then the above identity holds. I tried to solve this question using generating function.
$$A(x)=\sum_n\left(\sum_{k=1}^n\dfrac{1}{k}\right)x^n=-\dfrac{\log(1-x)}{1-x}$$ $$B(x)=\sum_n\left(\sum_{k=1}^{n}\dfrac{(-1)^{k+1}}{k}{n\choose k}\right)x^n=\log(x+1)(1+x)^n$$
But each side doesn't equal to each other. Did I make any mistake?