<s>[INST] Confused about infinite sum $\sum\limits_{a,b,c}\frac{a+b+c+abc} {2^a(2^{a+b}+2^{b+c}+2^{a+c})}$ $$ \displaystyle \sum^{\infty}_{a=0} \displaystyle \sum^{\infty}_{b=0} \displaystyle \sum^{\infty}_{c=0}\dfrac{a+b+c+abc} {2^a(2^{a+b}+2^{b+c}+2^{a+c})}= \ ? \ $$ 
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.
 [/INST] 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
$$\sum_{a=0}^{\infty} \frac{a}{2^{a}}=
\lim_{n\to \infty} \sum_{k=0}^{n} \frac{k}{2^{k}}=
\lim_{n\to \infty} \left(2-\frac{n}{2^{n}}-\frac{1}{2^{n-1}} \right)=2$$
</s>