Math Problem - Binomial Coefficients + Harmonic Numbers

Problem

Prove that

r = 1 \sum n (- 1)^{r - 1} (r n) H_{r} = \frac{1}{n}

Beginning with the LHS

r = 1 \sum n (- 1)^{r - 1} (r n) H_{r}

We apply the integral representation of the harmonic sum and interchange the integral and sum (because the sum is finite).

r = 1 \sum n (- 1)^{r - 1} (r n) H_{r} = r = 1 \sum n (- 1)^{r - 1} (r n) \int_{0}^{1} \frac{1 - x ^{r}}{1 - x} d x = \int_{0}^{1} \frac{1}{1 - x} r = 1 \sum n (- 1)^{r - 1} (r n) (1 - x^{r}) d x .

Splitting the inner sum:

r = 1 \sum n (- 1)^{r - 1} (r n) (1 - x^{r}) = r = 1 \sum n (- 1)^{r - 1} (r n) - r = 1 \sum n (- 1)^{r - 1} (r n) x^{r} .

Applying the binomial identities we get

r = 1 \sum n (- 1)^{r - 1} (r n) r = 1 \sum n (- 1)^{r - 1} (r n) x^{r} = 1, = 1 - (1 - x)^{n},

Substituting back

r = 1 \sum n (- 1)^{r - 1} (r n) H_{r} = \int_{0}^{1} \frac{( 1 - x ) ^{n}}{1 - x} d x = \int_{0}^{1} (1 - x)^{n - 1} d x = \frac{1}{n}

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