Triple the fun!

First, let's suppose (for the time being) that we only need to sum for values such that $i < j < k$ . That is, let's find the following sum: $\sum_{i=0}^{\infty} \sum_{j = i+1}^{\infty} \sum_{k = j+1}^{\infty} \frac{1}{3^i 3^j 3^k}$ $= \sum_{i=0}^{\infty} \frac{1}{3^i} \sum_{j = i+1}^{\infty} \frac{1}{3^j} \sum_{k = j+1}^{\infty} \frac{1}{3^k}$ $= \sum_{i=0}^{\infty} \frac{1}{3^i} \sum_{j = i+1}^{\infty} \frac{1}{3^j} \frac{1/3^{j+1}}{1 - 1/3}$ $= \frac{1}{2} \sum_{i=0}^{\infty} \frac{1}{3^i} \sum_{j = i+1}^{\infty} \frac{1}{9^{j}}$ $= \frac{1}{2} \sum_{i=0}^{\infty} \frac{1}{3^i} \frac{1/9^{i+1}}{1-1/9}$ $= \frac{1}{16} \sum_{i=0}^{\infty} \frac{1}{27^i}= \frac{1}{16} * \frac{27}{26} = \frac{27}{416}$

Now all we need to do is multiply this sum by 6, since there are exactly 6 different possibilities for how $i,j,k$ are ordered. Therefore, the total sum is $\frac{27}{416} * 6 = \frac{81}{208}$ . Therefore the answer is $81*208 = \boxed{16848}$ .

$\displaystyle \sum_{i = 0}^{\infty } \sum_{j = 0}^{\infty } \sum_{k = 0}^{\infty } = [\text { sum when i,j,k are independent }] - 3 \times [ \text{sum when any 2 of i,j,k are equal }] +2 \times [\text {sum when i = j = k}]$

= $\displaystyle \sum_{i =0}^{\infty } \sum_{j =0}^{\infty } \sum_{k =0}^{\infty } \frac{1}{3^{i} 3^{j} 3^{k}} - 3 \sum_{i=0}^{\infty } \sum_{k =0}^{\infty } \frac{1}{9^{i} 3^{k}} + 2\sum_{i =0}^{\infty} \frac{1}{27^{i}}$

= $\displaystyle \biggl(\frac{3}{2}\biggr)^{3} -3\biggl(\frac{9}{8}\biggr)\biggl(\frac{3}{2}\biggr) + 2\biggl(\frac{27}{26}\biggr)$

= $\displaystyle \frac{81}{208}$

$81 \times 208 = \boxed{16848}$

clearly we can see that i , j , k are different

instead of solving it we can apply the method we use in

combinatorics , for example to find permutations of telephone

numbers in a telephone directory which do not end with 4

then we do this simple method that is total - numbers ending with 4

Similarly

here y = infinity

$total = ( \sum_{i=0}^y \frac{1}{3^{i}})^{3}$

when i = k not equal to j

here

$= ( \sum_{i=0}^y \frac{1}{3^{2i}})( \sum_{i=0}^y \frac{1}{3^{j}})$

$= \sum_{i=0}^y \frac{1}{3^{2i}}(\frac{3}{2} - \frac{1}{3^{i}}$

$= \frac{27}{16} - \frac{27}{26} = \frac{135}{8.26}$

Required sum is

$= \frac{27}{8} - \frac{27}{26} - 3. \frac{135}{8.26} = \frac{81}{208}$

m.n =16848

very good question Krishna Sharma , enjoyed very much,continue to give such more

A simple simplification will do like $\mathop{\sum_{i=0}^{\infty} \sum_{j=0}^{\infty} \sum_{k=0}^{\infty}}_{i \neq j \neq k} \frac {1}{3^i3^j3^k}\\ =\mathop{\sum_{i=0}^{\infty} \sum_{j=0}^{\infty}}_{i \neq j} \frac{1}{3^i3^j} \left( \sum_{k=0}^{\infty} \frac{1}{3^k} - \frac{1}{3^i} - \frac{1}{3^j} \right)\\ =\mathop{\sum_{i=0}^{\infty} \sum_{j=0}^{\infty}}_{i \neq j} \frac{1}{3^i3^j} \left( \frac{1}{1 - \frac{1}{3}} - \frac{1}{3^i} - \frac{1}{3^j} \right)\\ =\mathop{\sum_{i=0}^{\infty} \sum_{j=0}^{\infty}}_{i \neq j} \frac{1}{3^i3^j} \frac {3}{2} - \mathop{\sum_{i=0}^{\infty} \sum_{j=0}^{\infty}}_{i \neq j} \frac{1}{3^{2i}3^j} - \mathop{\sum_{i=0}^{\infty} \sum_{j=0}^{\infty}}_{i \neq j} \frac{1}{3^i3^{2j}}\\ =\frac{3}{2}\sum_{i=0}^{\infty} \frac{1}{3^i} \left( \sum_{j=0}^{\infty} \frac{1}{3^j} - \frac{1}{3^i} \right) - \sum_{i=0}^{\infty} \frac{1}{3^{2i}} \left( \sum_{j=0}^{\infty} \frac{1}{3^j} - \frac{1}{3^i} \right) - \sum_{i=0}^{\infty} \frac{1}{3^i} \left( \sum_{j=0}^{\infty} \frac{1}{3^{2j}} - \frac{1}{3^{2i}} \right)\\ =\frac{3}{2}\sum_{i=0}^{\infty} \frac{1}{3^i} \left( \frac{1}{1 - \frac{1}{3}} - \frac{1}{3^i} \right) - \sum_{i=0}^{\infty} \frac{1}{3^{2i}} \left( \frac{1}{1 - \frac{1}{3}} - \frac{1}{3^i} \right) - \sum_{i=0}^{\infty} \frac{1}{3^i} \left( \frac{1}{1 - \frac{1}{9}} - \frac{1}{3^{2i}} \right)\\ =\frac{3}{2}\sum_{i=0}^{\infty} \frac{1}{3^i} \left( \frac{3}{2} - \frac{1}{3^i} \right) - \sum_{i=0}^{\infty} \frac{1}{3^{2i}} \left( \frac{3}{2} - \frac{1}{3^i} \right) - \sum_{i=0}^{\infty} \frac{1}{3^i} \left( \frac{9}{8} - \frac{1}{3^{2i}} \right)\\ =\frac{3}{2}\sum_{i=0}^{\infty} \left( \frac{1}{3^i} \frac{3}{2} - \frac{1}{3^{2i}} \right) - \sum_{i=0}^{\infty} \left( \frac{1}{3^{2i}} \frac{3}{2} - \frac{1}{3^{3i}} \right) - \sum_{i=0}^{\infty} \left( \frac{1}{3^i} \frac{9}{8} - \frac{1}{3^{3i}} \right)\\ =\frac{3}{2} \left( \frac{1}{1 - \frac{1}{3}} \frac{3}{2} - \frac{1}{1 - \frac{1}{9}} \right) - \left( \frac{1}{1 - \frac{1}{9}} \frac{3}{2} - \frac{1}{1 - \frac{1}{27}} \right) - \left( \frac{1}{1 - \frac{1}{3}} \frac{9}{8} - \frac{1}{1 - \frac{1}{27}} \right)\\ =\frac{3}{2} \left( \frac{3}{2} \frac{3}{2} - \frac{9}{8} \right) - \left( \frac{9}{8} \frac{3}{2} - \frac{27}{26} \right) - \left( \frac{3}{2} \frac{9}{8} - \frac{27}{26} \right)\\ =\frac{27}{8} - \frac{3 * 27}{16} + \frac{2 * 27}{26}\\ =\frac{2 * 27}{16} - \frac{3 * 27}{16} + \frac{2 * 27}{26}\\ =\frac{2 * 27}{26} - \frac{27}{16}\\ =\frac{16 * 27 - 13 * 27}{2 * 13 * 8}\\ =\frac{3 * 27}{2 * 13 * 8}$

so the answer is $3 * 27 * 2 * 13 * 8 = 16848$