# Triple sigma($\sum\sum\sum$) question

I was solving "multiple sigma" questions, while I stumbled upon this:

Find $$\sum_{i=0}^n\sum_{j=0}^n\sum_{k=0}^n \frac{1}{3^i3^j3^k} \ ,i\ne j \ne k$$

I'm used to solve double sigma problems (like $$\sum_{1\le i) using triangles like these:

However I'm not sure about how I'm going to construct a cube and splitting it into tetrahedrons(if at all possible)!! Else how do I approach this?

P.S.: The variables are dependent

The answer is $$\boxed{\frac{81}{208}}$$ when $$n\to\infty$$

• You can just do some casework: take the sum where all indices are independent, then subtract the extra terms (2 indices the same, or all 3 indices the same) Sep 20, 2020 at 16:35
• I tried doing something like that but was really confused about what to add/subtract Sep 20, 2020 at 16:36
• If 2 indices the same, then you have a sum of the form $\sum_{i=0}^n\sum_{j=0}^n \frac{1}{3^{2i}3^j}$ (there are 3 pairs like this), if 3 indices the same, you have $\sum_{i=0}^n\frac{1}{3^{3i}}$ Sep 20, 2020 at 16:38
• Yes, I understand. But again, which ones should I add/subtract? I couldn't get any clear idea(probably a difficult case of inclusion/exclusion?) Sep 20, 2020 at 16:42
• Should this be the limit as $n \to \infty$? Sep 21, 2020 at 2:40

## 2 Answers

The requitred sum is $$S=(\sum_{0k=0}^{\infty} 3^{-k})^3-3 \sum_{i=0}^{\infty} 9^{-i} \sum_{j=0} 3^{-j} +2 \sum_{k=0}^{\infty} 3^{-3k}$$ $$\implies S=\left(\frac{1}{1-1/3}\right)^3-3\frac{1}{1-1/9}.\frac{1}{1-1/3}+2\frac{1}{1-1/27}=\frac{27}{8}-\frac{81}{16}+\frac{27}{13}=\frac{81}{208}$$

You may see my answer in this post for the explanation:

$\sum_i \sum_j \sum_k (1 \le i \le j \le k \le n) ~~f_i f_j f_k$ in terms of $\sum f_i~, \sum f^2 _i ~\mbox{and}~ \sum f^3 _i$

• Why is the $i=j=k$ sum added twice? Sep 20, 2020 at 17:32
• You may see: my answer in math.stackexchange.com/questions/3274832/… Sep 20, 2020 at 17:40
• It was indeed helpful. So principle of inclusion/exclusion has got nothing to do with this? Sep 20, 2020 at 17:51
• No, exclusion/inclusion it is actually there. The first part is the free sum then three sums where two of (i,j,k) are equal are subtracted (excluded). But since in these THREE cases, $i=j=k$ features 3 times. So we have over subtracted 2 times extra. So double of (i=j=k) should be added (included) Sep 21, 2020 at 4:44
• I see....thanks for the help Sep 21, 2020 at 4:46

I'm going to construct a cube and splitting it into tetrahedrons(if at all possible)!!

It is not necessary to construct a cube. Let $$A = \{0,1,2,...,n\}$$. The required combinations (which you mentioned in the square) is basically Cartesian product $$A×A$$. Extending this further $$A×A×A = (A×A)×A$$. So write down elements of $$A×A$$ (which you found by your square) in row and elements of $$A$$ in column. Combine elements in the rectangle to arrive at the required combinations.

EDIT 2.0 (Approach towards solution)

While the given problem can also be solved using Principle of Inclusion and Exclusion, I am going to solve it using the 'square'/'rectangle' method that I have mentioned earlier. Write $$\{ (0,0),(0,1),...,(0,n),(1,0),(1,1),...,(1,n),...,(n,0),(n,1),...,(n,n) \}$$ in the horizontal direction and $$\{ 0,1,2,...,x,...,n \}$$ in the vertical direction. This rectangle can be split into squares formed by Cartesian Product of $$\{ (x,0),(x,1),...,(x,n) \}$$ and $$\{0,1,...,n\}$$ for $$0\leq x \leq n$$. This is shown below:

for $$i \neq j \neq k$$, we have to find for total $$-$$(minus) yellow portion. Total $$=3^{-x} (\sum_{i=o}^n{3^{-i}})^2$$ [solve row-wise and add up all the row-wise results to arrive at this result]. Yellow portion = diagonal + (x,x) column + x row - 2(x,x,x)[since I have deducted (x,x,x) 3 times but want it to be deducted only once]. Diagonal $$= 3^{-x} \sum_{i=0}^n{3^{-2i}}$$. (x,x) Column $$= 3^{-2x} \sum_{i=0}^n{3^{-i}}$$. x row $$= 3^{-2x} \sum_{i=0}^n{3^{-i}}$$ .So, Yellow portion $$= 3^{-x} \sum_{i=0}^n{3^{-2i}} + 2 \times 3^{-2x} \sum_{i=0}^n{3^{-i}} -2 \times 3^{-3x}$$. Therefore required answer is $$\sum_{x=0}^n{(3^{-x} (\sum_{i=o}^n{3^{-i}})^2-3^{-x} \sum_{i=0}^n{3^{-2i}} -2 \times 3^{-2x} \sum_{i=0}^n{3^{-i}} + 2 \times 3^{-3x})}$$ $$= \frac{27}{8}-\frac{27}{16}-2\times \frac{27}{16} + 2\times \frac{27}{26} = \boxed{\frac{81}{208}}$$

Note: We can solve for $$i by reducing the original square into a smaller version $$\{ (x,x+1),...,(x,n) \} \times \{x+1,...,n\}$$ (see the diagram). The logic here is area of green portion = (area of reduced square -(minus) area of diagonal of reduced square)/2

• I suppose the symmetry would be a problem... Feb 3, 2022 at 17:06
• @DatBoi I made an edit regarding Symmetry. Feb 4, 2022 at 6:23
• I really appreciate your efforts!! It does seem very tedious but I'll try looking for other possible symmetries to exploit! Feb 4, 2022 at 6:40
• I found a better symmetry that is why I made another edit. The key is to split rectangles into squares. Feb 4, 2022 at 12:28
• Can you please clarify what $x$ here is? Feb 4, 2022 at 13:01