$\sum_{j=1}^n \dfrac{(-1)^j\binom{n}{j-1}}{\sum_{1\le k\le j} k}$ closed form

Question

Find $\sum_{j=1}^n \dfrac{(-1)^j\binom{n}{j-1}}{\sum_{1\le k\le j} k}$

I am not able to get a single idea that can kill the problem. Some hints or solution?

Answer 1

Hint 1: First compute the sum in the denominator. Spoiler below.

$\sum_{1 \le k \le j} k = \frac{j(j+1)}{2}$

Hint 2: After completing hint 1, put the terms involving $j$ into a single binomial coefficient. Spoiler below.

$\frac{\binom{n}{j-1}}{\frac{j(j+1)}{2}} = \frac{\binom{n+2}{j+1}}{\frac{(n+2)(n+1)}{2}}$

Hint 3: For the sum over $j$, take out the constant factors and applying the Binomial Theorem, which says $(1 + x)^m = \sum_{j=0}^m \binom{m}{j} x^j$.

Answer 2

$$\sum_{j=1}^{n}\frac{2(-1)^j}{j(j+1)}\binom{n}{j-1}=\frac{2}{n+1}\sum_{j=1}^{n}\frac{(-1)^j}{j+1}\binom{n+1}{j}=\frac{2}{n+1}\int_{0}^{1}\sum_{j=1}^{n}\binom{n+1}{j}(-x)^j\,dx $$ hence the LHS equals $\color{red}{\large\frac{2((-1)^n-1-n)}{(n+1)(n+2)}}$.