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Show that if $s \ge 2,$ then $\sum\limits_{k \ge 1} \frac{(-1)^k}{k!} s_k = 0$ where $s_k = \sum\limits_{b_1+\dots+b_k=s-k} \prod\limits_i \frac{1}{b_i+1}$ and the sum is over all non-negative $b_i.$

I was working on a problem related to the probability of certain cycles appearing and managed to show that the result I needed was equivalent to $\frac{t}{n} = \sum\limits_{r \ge 1} \sum\limits_{a_1, \dots, a_r \ge 1} \frac{(-1)^{r+1}}{r!} \binom{t}{a_1, \dots, a_r} \prod\limits_{i=1}^r \frac{(a_i-1)!}{n^{a_i}}$ for all $n.$ The coefficient of $1/n$ is clearly $t,$ so I realized that it suffices to show the coefficient of $\frac{1}{n^s}$ is zero for all $s \ge 2.$ After rearranging and removing the chaff, the claim that these coefficients vanish is the equality in the title of this post.

After all of the work I've done on my problem, I would hate to have to start over from scratch. Hopefully, the question I posed has a nice and simple proof. You can rewrite $s_k$ as $$\int\limits_{[0,1]^k}\left[\sum_{b_1+\dots+b_k = s-k} x_1^{b_1} \cdots x_k^{b_k} \right] dx_1 \dots dx_k,$$ but I'm not sure whether this will help. You can go one step further and sneak $\frac{(-1)^k}{k!}$ in there: $$\frac{(-1)^k}{k!}s_k = \int\limits_{[0,1]^k}\left[\sum_{b_1+\dots+b_k = s-k} \prod\limits_i -x_i^{i(b_i+1)-1)} \right] dx_1 \dots dx_k.$$ But that still doesn't make combining all of the integrals any easier.

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  • $\begingroup$ I don't know if it helps, but $s_k$ can be rewritten as $\sum_{b_1 + b_2 + ... + b_k = s} \prod_i \frac{1}{b_i}$ where the $b_i$ are positive. $\endgroup$ Aug 14, 2020 at 20:34
  • $\begingroup$ @VarunVejalla It doesn't help because that's what I had earlier. It's usually easier to work with sums starting from $0$ than sums starting from $1,$ so I reindexed. $\endgroup$ Aug 15, 2020 at 6:08

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Let's change the notation and "reindex back": for integers $1\leqslant k\leqslant s$, we define $$A_{s,k}=\sum_{\substack{b_1,\ldots,b_k\geqslant 0\\b_1+\ldots+b_k=s-k}}\prod_{j=1}^k\frac{1}{b_j+1}=\sum_{\substack{n_1,\ldots,n_k>0\\n_1+\ldots+n_k=s}}\prod_{j=1}^k\frac{1}{n_j},\quad B_s=\sum_{k=1}^s\frac{(-1)^k}{k!}A_{s,k}.$$

Then (formally, or for $|z|<1$) \begin{align*} \sum_{s=1}^\infty B_s z^s &=\sum_{s=1}^\infty z^s\sum_{k=1}^s\frac{(-1)^k}{k!}A_{s,k} \\&=\sum_{k=1}^\infty\frac{(-1)^k}{k!}\sum_{s=k}^\infty A_{s,k}z^s \\&=\sum_{k=1}^\infty\frac{(-1)^k}{k!}\sum_{n_1,\ldots,n_k>0}\prod_{j=1}^k\frac{z^{n_j}}{n_j} \\&=\sum_{k=1}^\infty\frac{(-1)^k}{k!}\left(\sum_{n=1}^\infty\frac{z^n}{n}\right)^k \\&=\sum_{k=1}^\infty\frac{1}{k!}\big(\log(1-z)\big)^k=-z. \end{align*} Thus, $B_1=-1$ and $B_s=0$ for $s>1$.

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