# Symbolically implementing the following product

I have two products I hope to compare symbolically and I am hoping to use Mathematica for this. They are as follows: $$\prod_{1 \leq i < j \leq n} {(a_i + \cdots + a_{j-1}) + j - i \over j- i} \tag{1}$$ and $${\prod_{\ell = 1}^{n-1}\left(\prod_{j = \ell}^{n-1}\left[\sum_{i = 1}^j a_i + j - 1\right]\right) \over (n-1)!\cdot (n - 2)!\cdots 2!\cdot 1!} \tag{2}$$ However, I'm not very familiar with the Mathematica syntax required to do this. Regards

• Ponder on Product[(Sum[a[k], {k, i, j - 1}] + j - i)/(j - i), {j, 1, n}, {i, 1, j - 1}], and you might be able to figure how to write what you need. – J. M.'s discontentment Jun 18 '13 at 1:31
• @0x4A4D I have come up with the code (Product[(Sum[a[k], {k, 1, j}] + j - 1), {l, 1, n - 1}, {j, l, n - 1}])/(Product[(n - i)!, {i, 1, n - 1}]) for the second product, but I am confused about how to implement the first product with the $1 \leq i < j \leq n$. I have the following so far: Product[(Sum[a[k], {k, i, j - 1}] + j - i)/(j - i), {1 <= i < j <= n}] – Moderat Jun 18 '13 at 2:54
• Possibly using the Boole function? – Moderat Jun 18 '13 at 2:57
• Well, for the first: did you already ponder upon how I built the snippet in my first comment? – J. M.'s discontentment Jun 18 '13 at 3:00
• Ah! I just saw how your answer was actually a response to the first product. Many thanks. Does Mathematica have an == expression to test equivalence for two products such as these? – Moderat Jun 18 '13 at 3:02

Product[(Sum[a[k], {k, i, j - 1}] + j - i)/(j - i), {j, 1, n}, {i, 1,

Product[Sum[a[k], {k, 1, j}] + j - 1, {l, 1, n - 1}, {j, l, n - 1}]/