# How to compute the series $\sum\limits_{x=0}^\infty\sum\limits_{y=0}^\infty\sum\limits_{z=0}^\infty\frac{1}{2^x(2^{x+y}+2^{x+z}+2^{z+y})}$

How to compute the series $\displaystyle\sum_{x=0}^\infty\sum_{y=0}^\infty\sum_{z=0}^\infty\frac{1}{2^x(2^{x+y}+2^{x+z}+2^{z+y})}$ ?

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By symmetry, the sum $S$ of this triple series is also $$S=\sum_{x,y,z}\frac{1}{2^y(2^{x+y}+2^{x+z}+2^{z+y})}=\sum_{x,y,z}\frac{1}{2^z(2^{x+y}+2^{x+z}+2^{z+y})}.$$ Furthermore, $$\frac1{2^x}+\frac1{2^y}+\frac1{2^z}=\frac{2^{x+y}+2^{x+z}+2^{z+y}}{2^{x+y+z}}.$$ Hence, summing these three equivalent formulas for $S$, one gets $$3S=\sum_{x,y,z}\frac1{2^{x+y+z}}=\left(\sum_{x}\frac1{2^x}\right)^3,$$ and, finally, $$S=\frac13\cdot2^3=\frac83.$$ More generally, for every absolutely convergent series $\sum\limits_x\frac1{a_x}$, $$\sum_x\sum_y\sum_z\frac{1}{a_x(a_xa_y+a_xa_z+a_za_y)}=\frac13\left(\sum_x\frac{1}{a_x}\right)^3.$$
Awesome! +1 $\;$ –  Stefan Hansen Jan 24 '13 at 19:40
Always try to look for symmetry when dealing with double, triple sums ... $\text{nice}^2$ (+1) –  Chris's sis Jan 24 '13 at 21:54