# 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 $\displaystyle\sum_{x=0}^\infty\sum_{y=0}^\infty\sum_{z=0}^\infty\frac{1}{2^x(2^{x+y}+2^{x+z}+2^{z+y})}$ ?

By symmetry, the sum $S$ of this triple series $$S=\sum_{x,y,z}\frac{1}{2^x\cdot(2^{x+y}+2^{x+z}+2^{z+y})}$$ is also $$S=\sum_{x,y,z}\frac{1}{2^\color{red}{y}\cdot(2^{x+y}+2^{x+z}+2^{z+y})}=\sum_{x,y,z}\frac{1}{2^\color{red}{z}\cdot(2^{x+y}+2^{x+z}+2^{z+y})}.$$ 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 $\sum\limits_x\frac1{a_x}$, $$\sum_{x,y,z}\frac{1}{a_x\cdot(a_xa_y+a_xa_z+a_za_y)}=\frac13\left(\sum_x\frac{1}{a_x}\right)^3.$$
• Awesome! +1 $\;$ Jan 24, 2013 at 19:40
• Always try to look for symmetry when dealing with double, triple sums ... $\text{nice}^2$ (+1) Jan 24, 2013 at 21:54