Solving the Summation Cases

Question

Let $n$ be a positive integer. Prove that

$\displaystyle \sum_{k=1}^{n} \dfrac{(-1)^{k-1}} {k} \binom{n} {k} = 1 +\dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n}$

I and my friend discussed this two days ago. In this case, we prove that it goes to $\displaystyle \sum_{k=1}^{n} \dfrac{1}{k}$ (right-hand side expression in summation form), but unfortunately we went nothing. One thing that really evaporates the difficulty is when you need to apply the binomial expressions, related to summation lower bound and upper bound, to prove it, however I suspect that we may lack of knowledge knowing the identity/theorem which maybe useful to approach this problem. So, do you have any idea for this one?

Answer 1

There's a beautiful proof of this identity from the integral representation:

$$H_n = \int_0^1 \frac{1-x^n}{1-x}dx$$

This is easy to confirm because:

$$\frac{1-x^n}{1-x} = 1 + x + \dots + x^{n-1}$$

Then,

\begin{align*} H_n &= \int_0^1 \frac{1-x^n}{1-x}dx\\ &= \int_0^1 \frac{1-(1-u)^n}{u}du\\ &= \int_0^1 \frac{1-\sum_{k=0}^n\binom n k (-u)^k}{u}du\\ &= \int_0^1 \left(-\sum_{k=1}^n(-1)^k\binom n k u^{k-1}\right)du\\ &= -\sum_{k=1}^n (-1)^k\binom n k\int_0^1 u^{k-1}du\\ &= -\sum_{k=1}^n (-1)^k\binom n k \frac{1}{k}\\ \end{align*}

Answer 2

The following variation is Example 3 in section 1.2 of John Riordan's classic Combinatorial Identities.

Consider for $n=1,2,\ldots$ \begin{align*} f_n&=\sum_{k=1}^n(-1)^{k+1}\binom{n}{k}\frac{1}{k}\\ &=\sum_{k=1}^n(-1)^{k+1}\left[\binom{n-1}{k}+\binom{n-1}{k-1}\right]\frac{1}{k}\\ &=f_{n-1}-\frac{1}{n}\sum_{k=1}^n(-1)^k\binom{n}{k}\\ &=f_{n-1}-\frac{1}{n}\left[(1-1)^n-1\right]\\ &=f_{n-1}+\frac{1}{n}\\ &=H_n \end{align*} since $f_1=1$.

Answer 3

We can use binomial expansion. Consider series:

$ (\frac{1}{2})^{n-1}=(1-\frac{1}{2})^{n-1}=∑ ^n_{k=1} (-1)^{k-1}(\frac{1}{2})^{k-1}\binom{n}{k}$

But;..$∑ ^n_{k=1}\binom{n}{k}=(1+1)^{n-1}=2^{n-1} =∑ ^n_{k=1}2^{k-1}$

Therefore term by term multiplication of series on RHS and series$∑ ^n_{k=1} \frac{1}{n}$ is:

$ ∑ ^n_{k=1} (-1)^{k-1}(\frac{1}{2})^{k-1} . 2^{k-1}\frac{1}{n}\binom{n}{k}= ∑ ^n_{k=1} (-1)^{k-1}\frac{1}{n}\binom{n}{k}=2^n \times\frac{1}{2^n}∑ ^n_{k=1} \frac{1}{n}=∑ ^n_{k=1} \frac{1}{n}$