Earlier a question in MSE:
Find $\sum_{r=1}^{3n-1}{ (-1)^{r-1}r\over{3n \choose r}}$, if $n$ is even
meant to ask the summation
$$S_n=\sum_{k=0}^{n} (-1)^k \frac{k}{{n \choose k}}~~~~~(1)$$
when is $n$ even. Due to the limitation of a procedure it could only be proved for even positive integers $n$. Here, we show that that the sum (1) can be written in closed form for both even and odd values of $n$. Let us use rhe integral representation of the reciprocal of the binomial coefficient as
$${n \choose k}^{-1} = (n+1) \int_{0}^{1}x^k (1-x)^{n-k} dx$$
Further, using $$\sum_{k=0}^{n}k z^k=\frac{z}{(1-z)^2}-\frac{z^{n+1}}{(1-z)^2}-\frac{n z^{n+1}}{1-z}$$
Then $$S_n=(n+1)\int_{0}^{1} \sum_{k=0}^{n} k \left(\frac{x}{x-1}\right)^k (1-x)^n dx= (n+1)\int_{0}^{1}[-x(1-x)^{n+1}+(-1)^n x^{n+1}(1-x)+(-1)^n n x^{n+1}]dx.$$
Use $\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx$ in the second integral
$$S_n=(n+1) \left(\int_{0}^{1} -x^{n+1}(1-x) dx+(-1)^n\int_{0}^{1} x^{n+1}(1-x) dx+(-1)^n n \int_{0}^{1} x^{n+1} dx \right).$$
$$\implies S_n=-(n+1)[1+(-1)^{n+1}] \int_{0}^{1} (x^{n+1}-x^{n+2}) dx+(-1)^n\frac{n(n+1)}{n+2}. $$
$$S_n=-[1+(-1)^{n+1}]\frac{n+1}{(n+2)(n+3)}+ (-1)^n \frac{n(n+1)}{n+2}~~~~(2)$$
The question is: what are other methods of obtainng this result (2).
Best Answer
It fits into a nice exponential generating function:
$$\begin{align*} \sum_{n=0}^\infty \sum_{k=0}^n (-1)^k \frac{k}{\binom{n}{k}} x^n &= \sum_{n=0}^\infty \left(\sum_{k=0}^n (-1)^k (k+1)! (n-k)! \right)\frac{x^n}{n!} \\ &= \frac{\frac{((x-3) x-2) x^2}{(x-1) (x+1)^2}+2 \log (1-x)+2 \log (x+1)}{x^3} \end{align*}$$ where the second line is what Mathematica spits out; presumably it's relatively straightforward to reproduce the result by hand. In any case, you're after the coefficient of $x^n$, which is a straightforward if mildly tedious exercise in partial fractions and Taylor series. Filling in the details and organizing it nicely is probably about the same length as your integral approach, though this has the benefit of using more generic generating function tools.