Formula for sum of combinations

Question

I am trying to find a closed-form formula or, at least, a different and more useful representation for the following sum of combinations:

\begin{equation} \sum_{i=1}^{n}\frac{n!}{i!(n-i)!}\times\frac{(-1)^i}{i} \end{equation}

Answer 1

We have that

$$\sum_{i=1}^{n}\frac{n!}{i!(n-i)!}\times\frac{(-1)^i}{i}=\sum_{i=1}^{n}\frac{(-1)^i}{i}\binom{n}{i}$$

then refer to

• Sum of Pascal's triangle column
• Let $n$ be a positive integer, Prove that $\sum_{k=1}^n\frac{ (-1)^{k-1}}{k}{n \choose k} = H_n$

Answer 2

Hint : sub this into the sum \begin{eqnarray*} \frac{1}{i}=\int_0^1 x^{i-1} dx. \end{eqnarray*} Now invert the order of the integral and the plum. Use the binomial theorem, do the obvious substitution, expand and integrate to get the result stated by Interstellar Probe.

Answer 3

Starting from $$(1-x)^n = 1 +\color{blue}{x}\sum_{i=\color{blue}{1}}^n\binom ni (-1)^ix^{\color{blue}{i-1}} \Leftrightarrow \frac{(1-x)^n-1}{x} = \sum_{i=1}^n\binom ni (-1)^ix^{i-1}$$

you get

$$\Rightarrow \frac{(1-x)^n-1}{x} = -\sum_{i=0}^{n-1}(1-x)^i = \sum_{i=1}^n\binom ni (-1)^ix^{i-1}$$

Now, integrate

$$ -\sum_{i=0}^{n-1}\int_0^1(1-x)^i dx = \sum_{i=1}^n\binom ni (-1)^i \int_0^1x^{i-1}dx$$

Hence, $$-\sum_{i=0}^{n-1}\frac{1}{i+1} = -H_n = \sum_{i=1}^n\binom ni \frac{(-1)^i}{i} $$