Hint for proving $H_n=\sum_{i=1}^n\binom{n}{i}\frac{(-1)^{i+1}}i$, where $H_i=\sum_{k=1}^i\frac1k$

Question

I have been trying to prove the following identity: $$H_n = \sum_{i=1}^{n} \binom{n}{i}\frac{(-1)^{i+1}}{i}$$ where $$ H_i = \sum_{k=1}^{i} \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{i} $$ is the $i$-th harmonic number.

I have been trying to apply various other identites, but to no avail. What I want to get is but a hint on how to solve it, not the solution. Please take into consideration the fact that I wish to prove this identity by means of elementary methods, i.e. without calculus, linear algebra or anything college-level.

I should be exceedingly thankful for any help.

Answer 1

\begin{align*} H_n&=\sum_{k=1}^n \dfrac{1}{k}= \sum_{k=0}^{n-1} \int_0^1 x^kdx\\ &=\int_{0}^{1}\dfrac{(x-1+1)^n-1}{x-1}dx=\int_{0}^{1}\dfrac{dx}{1-x}+\sum_{k=0}^n\binom{n}{k}\int_{0}^{1}\dfrac{(x-1)^{k}1^{n-k}}{x-1}dx\\ &=\sum_{k=1}^n\binom{n}{k}\int_{0}^{1}(x-1)^{k-1}dx =\sum_{k=1}^n\binom{n}{k}\cdot \dfrac{(-1)^{k-1}}{k} \end{align*}

Answer 2

The thread linked by Brian Moehring has a lot of good answers to this (though not hints), but I’d like to add one more that might help provide some insight into why the formula works.

In the diagram below, you can see that $H_5$ is equal to the area of the union of all the rectangles (times some constant). If you use the inclusion exclusion principle to find this area, you can get to the formula your in your question.