Combination and sum

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Source: The Jikei University School of Medicine entrance exam 2006

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#1 h Jan 29, 2006, 3:23 PM • 2 Y

Y by Adventure10, Mango247

Let $n$ be positive integer.Prove the following equality.

$\sum_{r=1}^n (-1)^r\ _n C_r\ \frac{1}{r}=-\sum_{r=1}^n \frac{1}{r}$

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#2 h Apr 28, 2025, 2:44 PM

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$\sum_{r=1}^{n} C_{r}^{n} \frac{(-1)^{r}}{r} = -\sum_{r=1}^{n}\frac{1}{r}$ ;

or $\sum_{r=1}^{n} C_{r}^{n} (-1)^{r}\int_{0}^{1}x^{r-1} dx = -\sum_{r=1}^{n}\int_{0}^{1}y^{r-1} dy$ ;

or $\int_{0}^{1} \frac{1}{x}\left[\sum_{r=1}^{n} C_{r}^{n} (-x)^{r}\right] dx = -\int_{0}^{1} \sum_{r=1}^{n}y^{r-1} dy$ ;

or $\int_{0}^{1} \frac{1}{x} \cdot [(1-x)^{n}-1] dx = -\int_{0}^{1} \frac{1-y^{n}}{1-y} dy$ ;

or $\int_{0}^{1} \frac{(1-x)^{n}-1}{x}dx = \int_{0}^{1} \frac{y^{n}-1}{1-y} dy$ (i).

Now, let $y=1-z, dy=-dz$ to ultimately obtain:

$\int_{0}^{1} \frac{(1-x)^{n}-1}{x}dx = -\int_{1}^{0} \frac{(1-z)^{n}-1}{1-(1-z)} dz$ ;

or $\int_{0}^{1} \frac{(1-x)^{n}-1}{x}dx = \int_{0}^{1} \frac{(1-z)^{n}-1}{z} dz$ .

Hence, $\sum_{r=1}^n (-1)^r C_{r}^{n}\ \frac{1}{r}=-\sum_{r=1}^n \frac{1}{r}$ holds for all $n \in \mathbb{N}$ .

QED

This post has been edited 2 times. Last edited by Mathzeus1024, Apr 29, 2025, 10:18 AM

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