$$ \displaystyle \sum^{\infty}_{a=0} \displaystyle \sum^{\infty}_{b=0} \displaystyle \sum^{\infty}_{c=0}\dfrac{a+b+c+abc} {2^a(2^{a+b}+2^{b+c}+2^{a+c})}= \ ? \ $$

I calculated its value as $\frac{32}{3}$ but I'm not sure whether I'm right or wrong.

This is what I did :

Considering $a$,$b$ and $c$ as identical and independent,

$$ S= \displaystyle \sum^{\infty}_{a=0} \displaystyle \sum^{\infty}_{b=0} \displaystyle \sum^{\infty}_{c=0}\dfrac{a+b+c+abc} {2^a(2^{a+b}+2^{b+c}+2^{a+c})} \\ 3S= \displaystyle \sum^{\infty}_{a=0} \displaystyle \sum^{\infty}_{b=0} \displaystyle \sum^{\infty}_{c=0}\dfrac{a+b+c+abc} {(2^{a+b}+2^{b+c}+2^{a+c})}×\left[ \frac{1}{2^a}+\frac{1}{2^b}+\frac{1}{2^c} \right] \\ = \displaystyle \sum^{\infty}_{a=0} \displaystyle \sum^{\infty}_{b=0} \displaystyle \sum^{\infty}_{c=0}\dfrac{a+b+c+abc}{2^a2^b2^c } \\ = \displaystyle \sum^{\infty}_{a=0}\frac{a}{2^a} \displaystyle \sum^{\infty}_{b=0}\frac{1}{2^b} \displaystyle \sum^{\infty}_{c=0}\frac{1}{2^c}+ \displaystyle \sum^{\infty}_{a=0}\frac{1}{2^a} \displaystyle \sum^{\infty}_{b=0}\frac{b}{2^b} \displaystyle \sum^{\infty}_{c=0}\frac{1}{2^c}+ \displaystyle \sum^{\infty}_{a=0}\frac{1}{2^a} \displaystyle \sum^{\infty}_{b=0}\frac{1}{2^b} \displaystyle \sum^{\infty}_{c=0}\frac{c}{2^c}+ \displaystyle \sum^{\infty}_{a=0}\frac{a}{2^a} \displaystyle \sum^{\infty}_{b=0}\frac{b}{2^b} \displaystyle \sum^{\infty}_{c=0}\frac{c}{2^c} \\ = 2^3+2^3+2^3+2^3 =32 \\ \Rightarrow S=\frac{32}{3} $$

Please tell me if I'm right.

  • 4
    $\begingroup$ What could be the problem with this? :-) $\endgroup$ – Did Feb 24 '16 at 14:52

Seems OK.

You've used $$\frac{1}{xy+yz+zx} \left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)= \frac{1}{xyz}$$ and $$\sum_{a=0}^{\infty} \frac{a}{2^{a}}= \lim_{n\to \infty} \sum_{k=0}^{n} \frac{k}{2^{k}}= \lim_{n\to \infty} \left(2-\frac{n}{2^{n}}-\frac{1}{2^{n-1}} \right)=2$$

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