# A question on binomial theorem

If $$C_0$$, $$C_1$$, $$C_2$$,...$$C_n$$ are the coefficients in the expansion of $$(1+x)^n$$, where $$n$$ is a positive integer, show that $$C_1- {C_2\over 2} +{C_3\over 3}-...+{(-1)^{n-1} C_n\over n}=1+ {1\over 2}+ {1\over 3}+...+{1\over n}$$

Evaluate $$\int_0^1\frac{1-(1-x)^n}xdx$$ in two different ways.

First way: $$1-(1-x)^n=1-\sum_{k=0}^n(-1)^k\binom nkx^k=\sum_{k=1}^n(-1)^{k+1}\binom nkx^k$$ so $$\int_0^1\frac{1-(1-x)^n}xdx=\sum_{k=1}^n(-1)^{k+1}\binom nk\int_0^1x^{k-1}dx=\sum_{k=1}^n(-1)^{k+1}\binom nk\frac1k$$ $$=\frac{\binom n1}1-\frac{\binom n2}2+\frac{\binom n3}3-\cdots+(-1)^{n+1}\frac{\binom nn}n.$$

Second way: Substituting $$u=1-x$$, $$\int_0^1\frac{1-(1-x)^n}xdx=\int_0^1\frac{1-u^n}{1-u}du=\int_0^1(1+u+u^2+\cdots+u^{n-1})du$$$$=\frac11+\frac12+\frac13+\cdots+\frac1n.$$

In other words, you are saying that $$$$\sum_{k=1}^n \dfrac{\left(-1\right)^{k-1}}{k} \dbinom{n}{k} = \dfrac{1}{1} + \dfrac{1}{2} + \cdots + \dfrac{1}{n}$$$$ (because your $$C_k$$ are precisely the binomial coefficients $$\dbinom{n}{k}$$).

This is a fairly known identity. The one place I remember seeing a proof (because I wrote it) is Exercise 3.19 in my Notes on the combinatorial fundamentals of algebra, version of 10th of January 2019. I suspect it's been on math.stackexchange a few times already.

(If you're looking for a hint: Induction on $$n$$.)