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Show that for $n\geq 1, \sum_{i=1}^n \dfrac{(-1)^i}i {n\choose i} = -\sum_{k=1}^n \frac{1}k$ by using the fact that $\int_0^1 x^i dx = \frac{1}{i+1}.$

I tried first using the binomial theorem to get the equality $\displaystyle\sum_{i=1}^n (-1)^i x^{i-1} {n\choose i} = \dfrac{(1-x)^n-1}{x},$ but I'm not sure how I can integrate $\dfrac{(1-x)^n - 1}x$ from $0$ to $1$. Integrating the LHS from $0$ to $1$ yields $\sum_{i=1}^n \dfrac{(-1)^i}i {n\choose i}$, so if I could integrate that, it'd be very useful.

Fred Jefferson
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You can take into account that, for $y=1-x$ $$ \int_0^1\frac{(1-x)^n-1}{x}dx=-\int_0^1\frac{1-y^n}{1-y}dy $$ and one of the relations that Harmonic numbers satisfy $$ H_n=\sum_{k=1}^n\frac{1}{k}=\int_0^1\frac{1-y^n}{1-y}dy $$ The last one can be easily proved $$ \int_0^1\frac{1-y^n}{1-y}dy= \int_0^1\sum_{k=0}^{n-1}y^kdy= \sum_{k=0}^{n-1}\left.\frac{y^{k+1}}{k+1}\right|_0^1= \sum_{k=0}^{n-1}\frac{1}{k+1}= \sum_{k=1}^{n}\frac{1}{k} $$

Vincenzo Tibullo
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