Consider the following setting. Let $k = 1, \ldots, n$ and define $$y_k= \Phi^{-1}\left(\frac{k}{n+1}\right),$$ where $\Phi$ is the inverse of the CDF of a standard normal.
I noticed numerically that $$ \sum_{k =1}^{n} y_k = 0$$ But I cannot prove it. Any idea for a proof?
Hint: if $n$ is even, you can group the terms using Gauss summation:
$$y_1 + \cdots + y_n = (y_1+y_n) + (y_2+y_{n-1}) + \cdots.$$
Can you say anything about these pairs? Try using the symmetry of the inverse CDF:
If $n$ is odd, you can first show that the $(n+1)/2$-th element is zero, and use the same trick as above to establish your result.