Prove $\binom{n}{1} - \frac{1}{2}\binom{n}{2} + \frac{1}{3}\binom{n}{3} + \ldots = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n}$

Asked Apr 14 '18 at 23:30

Active Apr 14 '18 at 23:30

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I thought of induction but that was unwieldy. Then considered using $(1+x)^n$. Didn't succeed.

Any hint? Thanks

asked Apr 14 '18 at 23:30

sku

2

Hint. The sum can be understood as the value of $$\int_{0}^{1}\frac{1-(1-x)^n}{x},dx = \int_{0}^{1}\frac{1-y^n}{1-y},dx$$ where $y=1-x$. – Sangchul Lee Apr 14 '18 at 23:37
1

I tried this:
$\int \frac{(1-x)^n}{x}dx = \ln x - x\binom {n}{1} + \frac{x^2}{2}\binom{n}{2} - \ldots $ but didnt know how to proceed with the LHS. Now with your hint, I can replace LHS (taking $\ln x$ the other side) with $1 + y+y^2+...$ and integrating this we will get the harmonic series... Cool. Thanks
– sku Apr 14 '18 at 23:53
No worries. And I found that the question has already been answered in this community, so let me provide the link as well. – Sangchul Lee Apr 14 '18 at 23:59

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