Sequence Killer!

$\begin{aligned} \sum_{i \ne j \ne k}^\infty \frac 1{3^i 3^j 3^k} & = 6 \sum_{i < j < k}^\infty \frac 1{3^i 3^j 3^k} & \small \blue{\text{See Note.}} \\ & = 6 \sum_{i=0}^\infty \frac 1{3^i} \sum_{j=i+1}^\infty \frac 1{3^j} \sum_{k=j+1}^\infty \frac 1{3^k} \\ & = 6 \sum_{i=0}^\infty \frac 1{3^i} \sum_{j=i+1}^\infty \frac 1{3^j} \times \frac 1{3^{j+1}} \times \frac 1{1-\frac 13} \\ & = 6 \sum_{i=0}^\infty \frac 1{3^i} \sum_{j=i+1}^\infty \frac 1{3^{2j}} \times \frac 12 \\ & = 3 \sum_{i=0}^\infty \frac 1{3^i} \times \frac 1{9^{i+1}} \times \frac 1{1-\frac 19} \\ & = \frac 38 \sum_{i=0}^\infty \frac 1{27^i} \\ & = \frac 38 \times \frac {27}{26} = \frac {81}{208} \end{aligned}$

Therefore $m+n = 81+208 = \boxed{289}$ .

Generalization: $\displaystyle \sum_{i\ne j\ne k \ge 0}^\infty \frac 1{n^{i+j+k}} = \frac {6n^3}{(n^3-1)(n^2-1)(n-1)}$

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Note: Consider the summation $S(i<j<k) = \sum_{i < j < k}^\infty \frac 1{3^i3^j3^k}$ . $S_{i<j<k}$ sums for all $i$ but only $j >i$ and $k > j$ . Swapping $j$ with $k$ and do the same summation and add to the first one, $S(i<j<k) + S(i<\red{k<j}) = 2 S(i<j<k)$ , we get the sum of $i\ne j \ne k$ terms for all $i$ . Let's denote it as $S(i\ne j \ne k \mid i)$ . Since we can interchange $i$ , $j$ , and $k$ , we have $S(i\ne j \ne k) = 3 S(i\ne j \ne k \mid i) = 6 S(i<j<k)$

We'll use three ideas to solve this. First, sum of a geometric series: for $|r|<1$ , $\sum_{i=0}^{\infty} r^i=\frac{1}{1-r}$

Secondly, the fact that, for sequences $(a_i)$ and $(b_i)$ with convergent series $\sum_{i,j=0}^{\infty} a_i b_j = \left( \sum_{i,j=0}^{\infty} a_i \right) \cdot \left( \sum_{i,j=0}^{\infty} b_j \right)$

(this can be seen by multiplying out the product term by term, and for full rigour, using partial sums)

Thirdly, we'll need to use the inclusion-exclusion principle.

The geometric series sums we'll need are $\begin{aligned} \sum_{i=0}^{\infty} \frac{1}{3^i}&=\frac{3}{2} \\ \sum_{i=0}^{\infty} \frac{1}{9^i}&=\frac{9}{8} \\ \sum_{i=0}^{\infty} \frac{1}{27^i}=\frac{27}{26} \end{aligned}$

Now we'll use these with the second identity above to define and find some sums related to our target sum. First, call the target itself $S_{\text{distinct}} \sum_{i,j,k=0, \; (i \neq j,i \neq k,j \neq k)}^{\infty} \frac{1}{3^i 3^j 3^k}$

Now we'll work out the sum when all indices are permitted (ie they don't have to be distinct): $S_{\text{all}} \sum_{i,j,k=0}^{\infty} \frac{1}{3^i 3^j 3^k} = \left( \sum_{i=0}^{\infty} \frac{1}{3^i} \right)^3 = \frac{27}{8}$

Next the sum when two of the indices (say $j$ and $k$ ) are the same, and the third can take any value: $S_{\text{two same}} \sum_{i,j,k=0, \; j=k}^{\infty} \frac{1}{3^i 3^j 3^k} = \left( \sum_{i=0}^{\infty} \frac{1}{3^i} \right)\left( \sum_{j=0}^{\infty} \frac{1}{9^j} \right) = \frac{27}{16}$

Finally the case when all three indices are equal, ie $i=j=k$ : $S_{\text{three same}} \sum_{i,j,k=0, \; i=j=k}^{\infty} \frac{1}{3^i 3^j 3^k} = \sum_{i=0}^{\infty} \frac{1}{27^i} = \frac{27}{26}$

Lastly, we use inclusion-exclusion as follows: to find $S_{\text{distinct}}$ , start from $S_{\text{all}}$ . Now remove the cases when two of the indices are equal (note there are three of these corresponding to $i=j$ , $i=k$ and $j=k$ respectively). However, each of these sums also includes the case when all three indices are the same; so we need to add back in two copies of this sum. In other words, $S_{\text{distinct}} = S_{\text{all}}-3S_{\text{two equal}}+2S_{\text{three equal}}=\frac{27}{8}-3\cdot \frac{27}{16} + 2 \cdot \frac{27}{26} = \frac{81}{208}$

so that - at last - the answer is $\boxed{289}$ .