Calculate a multiple sum of inverse integers.

Question

The question is to calculate a following sum: \begin{equation} {\mathcal S}_p(n) :=\sum\limits_{1\le j_1 < j_2 < \dots <j_p \le n-1} \prod\limits_{q=1}^p \frac{1}{n-j_q} \end{equation} for $p=1,2,..$ and $n\ge 1$. From purely combinatorial reasoning we have: \begin{eqnarray} {\mathcal S}_1(n) &=& H_{n-1} \\ {\mathcal S}_2(n) &=& \frac{1}{2!} \left(H_{n-1}^2 - H^{(2)}_{n-1} \right) \\ {\mathcal S}_3(n) &=& \frac{1}{3!} \left(H_{n-1}^3 - 3 H_{n-1} H_{n-1}^{(2)} + 2 H_{n-1}^{(3)}\right) \\ {\mathcal S}_4(n) &=& \frac{1}{4!} \left(H^4_{n-1} - 6 H_{n-1}^2 H_{n-1}^{(2)} + 8 H_{n-1} H_{n-1}^{(3)} + 3 H_{n-1}^{(2)} H_{n-1}^{(2)} - 6 H_{n-1}^{(4)}\right) \\ {\mathcal S}_5(n) &=& \frac{1}{5!} \left(H_{n-1}^5 - 10 H_{n-1}^3 H_{n-1}^{(2)} + 20 H_{n-1}^2 H_{n-1}^{(3)} + 15 H_{n-1} ((H_{n-1}^{(2)})^2 - 2 H_{n-1}^{(4)}) - 20 H_{n-1}^{(2)} H_{n-1}^{(3)} + 24 H_{n-1}^{(5)}\right) \\ {\mathcal S}_6(n) &=& \frac{1}{6!} \left(H_{n-1}^6 - 15 H_{n-1}^4 H_{n-1}^{(2)} + 40 H_{n-1}^3 H_{n-1}^{(3)} + 45 H_{n-1}^2 ((H_{n-1}^{(2)})^2 - 2 H_{n-1}^{(4)}) - 24 H_{n-1} (5 H_{n-1}^{(2)} H_{n-1}^{(3)} - 6 H_{n-1}^{(5)}) + 5 (-3 (H_{n-1}^{(2)})^3 + 18 H_{n-1}^{(2)} H_{n-1}^{(4)} + 8 ((H_{n-1}^{(3)})^2 - 3 H_{n-1}^{(6)}) \right) \end{eqnarray} where $H_{n-1}^{(r)} := \sum\limits_{j=1}^{n-1} 1/j^r$ is the generalised Harmonic number.

Is it possible to find the result for generic $p\ge 1$?

Answer 1

Start with $$1 + \sum_{p=1}^{n-1} \mathcal{S}_p(n) z^p = \prod_{k=1}^{n-1}\left(1 + \frac{z}{n-k}\right) = \prod_{k=1}^{n-1}\left(1 + \frac{z}{k}\right) = \exp\left[\sum_{k=1}^{n-1}\log\left(1+\frac{z}{k}\right)\right] = \exp\left[\sum_{p=1}^\infty \frac{(-1)^{p-1}}{p} \left(\sum_{k=1}^{n-1}\frac{1}{k^p}\right)z^p\right] = \exp\left[\sum_{p=1}^\infty \frac{(-1)^{p-1} H_{n-1}^{(p)}}{p} z^p\right]$$ Taking logarithm and apply $z\frac{\partial}{\partial z}$ on both sides, we get

$$\sum_{p=1}^{n-1}p\mathcal{S}_p(n) z^p = \left(\sum_{p=1}^\infty (-1)^{p-1} H_{n-1}^{p} z^p\right) \left(1 + \sum_{p=1}^{n-1}\mathcal{S}_p(n) z^p\right) $$ Expanding both sides and compare coefficients of $z^p$, we obtain a chain of identities:

$$\begin{align} \mathcal{S}_1(n) &= H_{n-1}\\ 2\,\mathcal{S}_2(n) &= H_{n-1} \mathcal{S}_1(n) - H_{n-1}^{(2)}\\ 3\,\mathcal{S}_3(n) &= H_{n-1} \mathcal{S}_2(n) - H_{n-1}^{(2)}\mathcal{S}_1(n) + H_{n-1}^{(3)}\\ 4\,\mathcal{S}_4(n) &= H_{n-1} \mathcal{S}_3(n) - H_{n-1}^{(2)}\mathcal{S}_2(n) + H_{n-1}^{(3)}\mathcal{S}_1(n) - H_{n-1}^{(4)}\\ &\;\vdots\\ p\,\mathcal{S}_p(n) &= \left(\sum_{k=1}^{p-1}(-1)^{k-1} H_{n-1}^{(k)} \mathcal{S}_{p-k}(n)\right) + (-1)^{p-1} H_{n-1}^{(p)} \end{align}$$

This is the Newton's identities associated with the set of numbers $\;\displaystyle \frac{1}{n-j}\;$ for $1 \le j \le n-1$. One can use it to obtain the expression of $\mathcal{S}_p$ for higher $p$ recursively. Let $h_p = H_{n-1}^{(p)}$, following are some more $\mathcal{S}_p$ computed by this approach. $$\begin{array}{rcl} 6!\mathcal{S}_{6}(n) &=& -120{h}_{6}+144{h}_{1}{h}_{5}+90{h}_{2}{h}_{4}-90{h}_{1}^{2}{h}_{4}+40{h}_{3}^{2}-120{h}_{1}{h}_{2}{h}_{3}\\ &&+40{h}_{1}^{3}{h}_{3}-15{h}_{2}^{3}+45{h}_{1}^{2}{h}_{2}^{2}-15{h}_{1}^{4}{h}_{2}+{h}_{1}^{6}\\ 7!\mathcal{S}_{7}(n) &=& 720{h}_{7}-840{h}_{1}{h}_{6}-504{h}_{2}{h}_{5}+504{h}_{1}^{2}{h}_{5}-420{h}_{3}{h}_{4}+630{h}_{1}{h}_{2}{h}_{4}\\ &&-210{h}_{1}^{3}{h}_{4}+280{h}_{1}{h}_{3}^{2}+210{h}_{2}^{2}{h}_{3}-420{h}_{1}^{2}{h}_{2}{h}_{3}\\ &&+70{h}_{1}^{4}{h}_{3}-105{h}_{1}{h}_{2}^{3}+105{h}_{1}^{3}{h}_{2}^{2}-21{h}_{1}^{5}{h}_{2}+{h}_{1}^{7}\\ 8!\mathcal{S}_{8}(n) &=& -5040{h}_{8}+5760{h}_{1}{h}_{7}+3360{h}_{2}{h}_{6}-3360{h}_{1}^{2}{h}_{6}+2688{h}_{3}{h}_{5}\\ &&-4032{h}_{1}{h}_{2}{h}_{5}+1344{h}_{1}^{3}{h}_{5}+1260{h}_{4}^{2}-3360{h}_{1}{h}_{3}{h}_{4}\\ &&-1260{h}_{2}^{2}{h}_{4}+2520{h}_{1}^{2}{h}_{2}{h}_{4}-420{h}_{1}^{4}{h}_{4}-1120{h}_{2}{h}_{3}^{2}\\ &&+1120{h}_{1}^{2}{h}_{3}^{2}+1680{h}_{1}{h}_{2}^{2}{h}_{3}-1120{h}_{1}^{3}{h}_{2}{h}_{3}+112{h}_{1}^{5}{h}_{3}\\ &&+105{h}_{2}^{4}-420{h}_{1}^{2}{h}_{2}^{3}+210{h}_{1}^{4}{h}_{2}^{2}-28{h}_{1}^{6}{h}_{2}+{h}_{1}^{8}\\ \end{array}$$

Answer 2

Suppose we seek to evaluate

$$S_n(m) = \sum_{1\le j_1 \lt j_2 \lt\cdots\lt j_n \le m-1} \prod_{q=1}^n \frac{1}{m-j_q}.$$

By way of enrichment let me point out that we can express this sum in terms of the cycle index $Z(P_n)$ of the unlabeled set operator $\mathfrak{P}_{=n}$ by applying the Polya Enumeration Theorem.

Recall the recurrence by Lovasz for the cycle index $Z(P_n)$ of the set operator $\mathfrak{P}_{=n}$ on $n$ slots, which is $$Z(P_n) = \frac{1}{n} \sum_{l=1}^n (-1)^{l-1} a_l Z(P_{n-l}) \quad\text{where}\quad Z(P_0) = 1.$$ This recurrence lets us calculate the cycle index $Z(P_n)$ very easily.

The sum is then given by $$Z(P_n)(Q_1+Q_2+\cdots+Q_{m-1})$$ evaluated at $Q_k = \frac{1}{m-k}.$

The Polya enumeration rule says to substitute as follows: $$a_l = Q_1^l + Q_2^l + \cdots + Q_{m-1}^l$$

in other words $$a_l = H_{m-1}^{(l)}$$

for a final answer of $${\large \bbox[5px,border:2px solid #00A000]{ \left.Z(P_n)\right|_{a_l = H_{m-1}^{(l)}}}}$$

For example we have $$Z(P_5) = \frac{1}{5!} \left({a_{{1}}}^{5}-10\,a_{{2}}{a_{{1}}}^{3}+20\,a_{{3}}{a_{{1}}}^{2} +15\,a_{{1}}{a_{{2}}}^{2}-30\,a_{{4}}a_{{1}} -20\,a_{{2}}a_{{3}}+24\,a_{{5}}\right)$$ and the substitution should now be clear.

Answer 3

Expanding the right hand side of the identity given by achille hui we get ``a compact'' expression for the sum: \begin{eqnarray} {\mathcal S}_p(n) &=& \sum\limits_{m=1}^p \frac{(-1)^{-m+p} }{m!} \sum\limits_{p_1+p_2+\dots+p_m=p} \prod\limits_{q=1}^m \frac{H_{n-1}^{(p_q)}}{p_q} \\ &=& \frac{1}{1!} \frac{(-1)^{p-1}}{p} H^{(p)}_{n-1} + \frac{(-1)^{p-2}}{2!} \sum\limits_{p_1=1}^{p-1} \frac{1}{p_1 (p-p_1)} H_{n-1}^{(p_1)} H_{n-1}^{(p-p_1)} + \dots + \frac{1}{p!} H_{n-1}^p \end{eqnarray}