# Relation between binomial sum and harmonic numbers

Prove that

$$\sum_{r=1}^n {n\choose r}(-1)^{r+1}\dfrac{1}{r}=\sum_{r=1}^n \dfrac{1}{r}$$

$$proof$$

Start by

$$\sum_{r=0}^n {n\choose r}x^r=(1+x)^n$$

Which can be converted to integration

$$\sum_{r=1}^n {n\choose r}\frac{(-1)^{r}}{r}=\int^{-1}_0 \frac{(x+1)^n-1}{x} dx$$

By substitution we have

$$\sum_{r=1}^n {n\choose r}\frac{(-1)^{r+1}}{r}=\int^{1}_0 \frac{t^n-1}{t-1} dt = H_n$$

Note the last step by expanding $(1-t)^{-1}$.

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