The value of the sum
$$600\sum_{a = 1}^\infty \sum_{b = 1}^\infty \sum_{c = 1}^\infty \frac{ab(3a + c)}{4^{a + b + c} (a + b)(b + c)(c + a)}$$
is ______
My attempt:
This looks very symmetric, except for the numerator. We can resolve it into two sums:
$$\begin{align}
S_1(a,b,c) &= 600\sum_{a = 1}^\infty \sum_{b = 1}^\infty \sum_{c = 1}^\infty\frac{abc}{4^{a+b+c}(a+b)(b+c)(c+a)} \\
S_2(a,b,c) &= 1800\sum_{a = 1}^\infty \sum_{b = 1}^\infty \sum_{c = 1}^\infty\frac{a^2b}{4^{a+b+c}(a+b)(b+c)(c+a)}
\end{align}$$
I don't know how to evaluate these two sums. The function does not 'separate' cleanly into different summations. I also read the Symmetric sums article on AoPS but coudn't get how to apply that concept here. Any hints/solutions are appreciated.
Best Answer
We should consider the sum $$S(n)=\sum_{a = 1}^n \sum_{b = 1}^n \sum_{c = 1}^n \frac{ab(3a + c)}{4^{a + b + c} (a + b)(b + c)(c + a)}$$
Exchange $a, b, c$ and we get: $$S(n)=\sum_{a = 1}^n \sum_{b = 1}^n \sum_{c = 1}^n \frac{ac(3a + b)}{4^{a + b + c} (a + b)(b + c)(c + a)}$$ and $$S(n)=\sum_{a = 1}^n \sum_{b = 1}^n \sum_{c = 1}^n \frac{ab(3b + c)}{4^{a + b + c} (a + b)(b + c)(c + a)}$$ etc.
We eventually get:
$6S(n)\\=\sum_{a = 1}^n \sum_{b = 1}^n \sum_{c = 1}^n \frac{ab(3a+c) + ac(3a+b)+ab(3b+c)+bc(3b+a)+ac(3c+b)+bc(3c+a)}{4^{a+b+c} (a+b)(b+c)(c+a)}\\=\sum_{a = 1}^n \sum_{b = 1}^n \sum_{c = 1}^n \frac{3(a+b)(b+c)(c+a)}{4^{a+b+c} (a+b)(b+c)(c+a)}\\=\sum_{a = 1}^n \sum_{b = 1}^n \sum_{c = 1}^n \frac{3}{4^{a+b+c}}\\=3 \sum_{a = 1}^{n} \frac{1}{4^a} \sum_{b = 1}^{n} \frac{1}{4^b} \sum_{c = 1}^{n} \frac{1}{4^c}$
When $n \longrightarrow \infty$, $$600\lim_{n \to \infty}S(n)=600\cdot\frac{1}{2} \left(\frac{\frac{1}{4}}{1-\frac{1}{4}}\right)^3=\frac{100}{9}$$
Edit :
A similar question was found here, in which we have to evaluate $$\sum_{a=1}^6\sum_{b=1}^6\sum_{c=1}^6\frac{ab(3a+c)}{2^a2^b2^c(a+b)(b+c)(c+a)}$$