Value of $\ \sum_{r=1}^{n} (-1)^{r-1}(1+ \dfrac{1}{2} + \dfrac{1}{3} + \ ...\ +\dfrac{1}{r}) {n \choose r}$

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My approach:

$(1){n \choose 1} - (1+ \dfrac{1}{2}){n \choose 2} + (1+\dfrac{1}{2} + \dfrac{1}{3}){n \choose 3} - (1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}){n \choose 4} \ ...$

Grouping the terms:

$1[{n \choose 1}-{n \choose 2}+{n \choose 3}...] - \dfrac{1}{2}[{n \choose 2}-{n \choose 3}+{n \choose 4}...] + \dfrac{1}{3}[{n \choose 3}-{n \choose 4}+{n \choose 5}...] \ ...$

However, I can't seem to proceed any further. My school material offers a hint regarding this problem, saying to 'use integration' without any other context, but I don't see how integration can be used in the first place since the terms associated with each binomial coefficient is itself a sum of harmonic series. Any hints or ideas would be appreciated.

Answer:

It simplifies to $\dfrac{1}{n}$