Is this some magical summation?

Question

I have been playing around with some probability calculations and somehow came to this expression $$\sum_{u_1 = 1}^n \sum_{u_2 = 1}^n \ldots \sum_{u_n = 1}^n \frac{1/u_1}{\sum_{j=1}^n 1/u_j} \prod_{i=1}^n \binom{n-1}{u_i-1}p^{u_i-1}(1-p)^{n-u_i}$$ which magically is always equal to $1/n$ for any $0< p < 1$. Is that obvious? I have been able to show that through numerical simulations by evaluating the expression numerically for different $n$ and $p$ but I am not able to establish that analytically.

Another question I have is what happens if only the outer sum for $u_1$ starts from 2? It doesn't seem to affect the overall result for large $n$ (especially as $p$ gets closer to 1). Again is there a way to show that?

Answer 1

Now I see the result follows directly by symmetry. In particular, there is nothing special about the index 1, hence the expression does not change if we change the fraction to $1/u_k$, i.e., $$\sum_{u_1 = 1}^n \sum_{u_2 = 1}^n \ldots \sum_{u_n = 1}^n \frac{1/u_k}{\sum_{j=1}^n 1/u_j} \prod_{i=1}^n \binom{n-1}{u_i-1}p^{u_i-1}(1-p)^{n-u_i}$$ for any $k \in \{1,\ldots,n\}$. Therefore, if I sum over $k$: $$\sum_{k=1}^n\sum_{u_1 = 1}^n \sum_{u_2 = 1}^n \ldots \sum_{u_n = 1}^n \frac{1/u_k}{\sum_{j=1}^n 1/u_j} \prod_{i=1}^n \binom{n-1}{u_i-1}p^{u_i-1}(1-p)^{n-u_i}$$ I get $n$ times the expression I am interested in. Changing the order of summations, the expression above reduces to: $$\sum_{u_1 = 1}^n \sum_{u_2 = 1}^n \ldots \sum_{u_n = 1}^n \prod_{i=1}^n \binom{n-1}{u_i-1}p^{u_i-1}(1-p)^{n-u_i} = 1$$ since the inner terms are binomial expansions of $(p+1-p)^{n-1}$. Hence, the expression of interest is equal to $1/n$.