Question:

The sum $S = \sum_{r=1}^{n}(-1)^{r-1}\left(1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{r}\right)^nC_r$ is converted using integrals: each harmonic sum equals $\int_0^1 (1+x+x^2+...+x^{r-1})dx$. After exchanging sum and integral, the inner sum becomes $\sum_r ^nC_r(-1)^{r-1}(1-x^r)/(1-x)$. This simplifies through careful algebraic manipulation using the binomial expansion $(1-x)^n$, ultimately yielding $\frac{1}{n}$.