# How to evaluate the sum over a hyperplane

I have difficulties in evaluating the following expression:

$$\sum_{\small n_1+...+n_{k}=m-k}\; \prod_{i=1}^{k}\frac{1}{(n_i+1)(n_i+2)}$$

I have tried the function `RootSum`, but it doesn't work and gives the following warning:

````RootSum[Sum[Subscript[n, j], {j, 1, k}] - m + k, 
            Product[1/((Subscript[n, i] + 2) (Subscript[n, i] + 1)), {i, 1, k}]] 
````

"RootSum::pfn - m + k + $\sum\limits_{j=1}^{k} n_j$ is not a pure function".

I will really appreciate it if someone could give me some hints. Thanks!

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Could you please write down here the Mathematica expression you've tried to evaluate? – belisarius Jan 24 at 17:00
@belisarius: RootSum[Sum[Subscript[n, j], {j, 1, k}] - m + k,   Product[1/((Subscript[n, i] + 2) (Subscript[n, i] + 1)), {i, 1, k}]] – user5551 Jan 24 at 17:27
`RootSum` expects both arguments to be a pure function of a single variable. That's not the case here, and I don't think  you can state your problem that way. BTW Any restrictions on the $n_i$'s? Can we assume they are integer or so? For reals it doesn't seem to make sense at all. – Sjoerd C. de Vries Jan 24 at 18:18

## 1 Answer

The formula is a little vague because starting limits for the $n_i$ are not given. However, its form suggests it started out as

$$\sum_{(n_1, n_2, \ldots, n_k) \mid 1 \le n_i \wedge n_1+n_2+\cdots+n_k=m} \prod_{i=1}^k \frac{1}{n_i (n_i+1)}$$

for a fixed value of $k$. The relationship to the expression in the question is that we can replace each $n_i$ by $n_i-1$ and allow their values to start at $0$ rather than $1$; their sum is thereby reduced by $k \times 1 = k$ and we obtain precisely the sum in the question, assuming its $n_i$ are allowed to start at $0$ rather than $1$.

In any event, the sum appears to be over all ordered partitions $(n_1, n_2, \ldots, n_k)$ of $m$ consisting of exactly $k$ nonzero values or else it can be re-expressed in such a manner. The crux of the problem is to generate these partitions. This solution relies on the one-to-one-correspondence between any such partition and the set of its distinct partial sums $0, n_1, n_1+n_2, \ldots, n_1+n_2+\cdots+n_k=m$. The intermediate terms determine a $k-1$ element subset of $\{1,2,\ldots,m-1\}$ and, conversely, each such subset when ordered can be construed as such a sequence of partial sums from which all the $n_i$ can be recovered by taking successive differences. Mathematica has a function to produce all such subsets, naturally called `Subsets`.

All these key elements can be found in the following expression which comprises `Subsets` to generate the partition data (already sorted), `Append` and `Prepend` to tack on the final $m$ and initial $0$, `Differences` to recover all the $n_i$, `Product` to multiply the reciprocals, and finally `Total` to sum the terms:

````f[m_Integer, k_Integer] /; 0 < k <= m := 
 Product[1/(i (i + 1)), {i, #}] & /@ 
  (Differences[Append[Prepend[#, 0], m]] & /@ Subsets[Range[m - 1], {k - 1}]) // Total
````

For example, consider the case $m=5, k=2$. The ordered two-partitions of $5$ are $(n_1,n_2)$ = $(1,4), (2,3), (3,2), (4,1)$, introducing terms $\frac{1}{1 \cdot 2}\frac{1}{4 \cdot 5} = \frac{1}{40}$, $\frac{1}{2 \cdot 3}\frac{1}{3 \cdot 4} = \frac{1}{72}$, $\frac{1}{3 \cdot 4}\frac{1}{2 \cdot 3} = \frac{1}{72}$, and $\frac{1}{4 \cdot 5}\frac{1}{1 \cdot 2} = \frac{1}{40}$, respectively. They sum to $\frac{2}{40} + \frac{2}{72}= \frac{7}{90}$. And indeed,

````f[5,2]
````

$\frac{7}{90}$

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