Formula for sum of combinations 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}
 A: 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 $\sum_{k=1}^n\frac{ (-1)^{k-1}}{k}{n \choose k} = H_n$
A: 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.
A: 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,
$$-\sum_{i=0}^{n-1}\frac{1}{i+1}  = -H_n = \sum_{i=1}^n\binom ni \frac{(-1)^i}{i} $$
