$\sum\nolimits_{i=0}^{\infty }{\sum\nolimits_{j=0}^{\infty }{\sum\nolimits_{k=0}^{\infty }{{{3}^{-\left( i+j+k \right)}}}}}$

Question

Some said to use Residue theorem (Special case for series )

S=$\sum\nolimits_{i=0}^{\infty }{\sum\nolimits_{j=0}^{\infty }{\sum\nolimits_{k=0}^{\infty }{{{3}^{-\left( i+j+k \right)}}}}}$ for $i\ne j\ne k$

My question is there any shortcut ??

Answer 1

$$\sum\nolimits_{i=0}^{\infty }{\sum\nolimits_{j=0}^{\infty }{\sum\nolimits_{k=0}^{\infty }{{{3}^{-\left( i+j+k \right)}}}}} = \sum\nolimits_{i=0}^{\infty }{3}^{-i}{\sum\nolimits_{j=0}^{\infty }{3}^{-j}{\sum\nolimits_{k=0}^{\infty }{3}^{-k}}}\\ =\left(\sum_{i=0}^\infty \frac{1}{3^i}\right)^3 = \left(\frac{1}{1-1/3}\right)^3$$

Answer 2

Since you want all the variables to be distinct, we can sum it over all values of i,j,k then subtract the sum that occurs when they are not distinct, and apply PIE(Principle of Inclusion/Exclusion).

Can you take it from there?

Answer 3

You can avoid inclusion-exclusion by considering $S$ as a nested sum: $$S=6\sum_{i=0}^\infty 3^{-i}\sum_{j=i+1}^\infty 3^{-j}\sum_{k=j+1}^\infty 3^{-k}\ ,$$ whereby the factor $6$ compensates for the assumption $i<j<k$. The innermost sum has the value $$3^{-j}\sum_{k'=1}^\infty 3^{-k'}={1\over2} 3^{-j}\ .$$ The next-inner sum then becomes $${1\over2}\sum_{j=i+1}^\infty3^{-2j}={1\over2}3^{-2i}\sum_{j'=1}^\infty 3^{-2j'}={1\over2}\cdot{1\over8}\>3^{-2i}\ .$$ In this way we finally obtain $$S=6\cdot{1\over2}\cdot{1\over8}\sum_{i=0}^\infty 3^{-3i}=6\cdot{1\over2}\cdot{1\over8}\cdot{27\over26}={81\over208}\ .$$