Harmonic numbers problem

Prove that \sum_{i=1}^{n}\frac{(-1)^{i-1}}{i}{n\choose i}=H_n
 where H_n is the n-th harmonic number.

Proof. Denotе the sum by S. Consider f(t)=\sum_{i=1}^{n}{n\choose i}t^{i-1}=\frac{(1+t)^n-1}{t} Now integrate
\int_{0}^{x} f(t)\mathrm{d}t=\sum_{i=1}^{n}{n\choose i}\frac{x^{i}}{i}=\int_{0}^{x}\frac{(1+t)^n-1}{t}\mathrm{d}t Now plug x=-1 and we get \int_{0}^{-1}f(t)\mathrm{d}t=-S
On the other hand \int_{0}^{-1}f(t)\mathrm{d}t=\int_{0}^{-1}\frac{(1+t)^n-1}{t}\mathrm{d}t=-\int_{0}^{1}\frac{s^n-1}{s-1}\mathrm{d}s=\\=-\int_{0}^{1}1+s+s^2+\cdots+s^{n-1}\mathrm{d}s=-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)=-H_n

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