19
$\begingroup$

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

$\endgroup$
43
$\begingroup$

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 $\sum\limits_x\frac1{a_x}$, $$ \sum_{x,y,z}\frac{1}{a_x\cdot(a_xa_y+a_xa_z+a_za_y)}=\frac13\left(\sum_x\frac{1}{a_x}\right)^3. $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That, sir, is very nice. $\endgroup$ – mez Jan 24 '13 at 19:23
  • 2
    $\begingroup$ Symmetry is essential here. $\endgroup$ – mrs Jan 24 '13 at 19:27
  • 2
    $\begingroup$ Awesome! +1 $\;$ $\endgroup$ – Stefan Hansen Jan 24 '13 at 19:40
  • 2
    $\begingroup$ Always try to look for symmetry when dealing with double, triple sums ... $\text{nice}^2$ (+1) $\endgroup$ – user 1591719 Jan 24 '13 at 21:54
  • $\begingroup$ wow nice answer ... $\endgroup$ – srijan May 15 '13 at 11:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.