Can this be proved using definite integrals

Question

It's a problem from a high school math book that I've been unable to solve:

Prove using definite integrals that, $${n \choose 1}-\frac{1}{2}{n \choose 2}+\frac{1}{3}{n \choose 3}-...+(-1)^{n-1}\frac{1}{n}{n \choose n}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$

I've tried using binomial expansions but to no avail. I cannot see any other approach to it using definite integrals. Although I am not sure if this is true. So, if one cannot prove it, can you please at least give me some numerical evidence of its correctness. A proof will be much better though.

Shouldn't the term $(-1)^{n-1}{n \choose n}$ on the l.h.s. be $(-1)^{n-1}\frac{1}{n}{n \choose n}$? Just asking, in case of a typo... — Pierpaolo Vivo, Jan 09 '16 at 14:28
This question is not quite a duplicate of that other question, since this question says to use definite integrals and the other one doesn't, but the first answer to the other question is a duplicate since it does use definite integrals. Therefore I am voting to close this as a duplicate. — Rory Daulton, Jan 09 '16 at 15:31

Empy2 · Accepted Answer · 2016-01-09T14:34:31.930

2

The right-hand side is $$\int_0^11+x+x^2+...+x^{n-1}dx=\int_0^1\frac{1-x^n}{1-x}dx$$ The left-hand side is a similar integral, with $x$ replaced by $1-x$.

edited Jan 09 '16 at 14:34

answered Jan 09 '16 at 14:29

Empy2

50,853

1

how do you avoid the singularity? – Jack Tiger Lam Jan 09 '16 at 14:30
There is none because the denominator divides into the numerator. The other way, the denominator is $x$, and the constant term in the numerator is zero. – Empy2 Jan 09 '16 at 14:32
When you substitute $u=1-x$, you change the limits, so it is $-\int_1^0\frac{1-(1-u)^n}udu$. Expand the numerator, and it looks a lot like your left-hand side. – Empy2 Jan 09 '16 at 14:39
Yeah... Now I see... – MathIsNice1729 Jan 09 '16 at 14:50

Can this be proved using definite integrals

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