Compute $\sum\limits_{a=1}^{\infty} \sum\limits_{b=1}^{\infty} \sum\limits_{c=1}^{\infty}\frac{ab(3a+c)}{4^{a+b+c}((a+b)(b+c)(a+c)}$ Compute the following sum:
$$
\sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty}\frac{ab(3a+c)}{4^{a+b+c}((a+b)(b+c)(a+c)}\
$$
I have no head or tail of how to even start this since the summation is interlinked.
 A: The basic idea is that, the final value of the sum doesn't depend on the dummy indices, i.e. whether you choose $(a,b,c)$, $(x,y,z)$ or some other triplet - it simply does not matter. Take advantage of this, and swap the indices $(a,b,c)$ around. How many permutations of these $3$ exist? Exactly $3! = 6$ permutations. I leave it to you to write them down. Now that you know that these $6$ different ways of writing the sum mean exactly the same thing - we're good to go. Denote the sum by $S$, and add all these (six) different representations of the same sum together. Get a nice expression for $6S$, and thus find $S$. See the magic unfold!
$$6S = \sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac{ab(3a+c) + ac(3a+b) + bc(3b+a) + ba(3b+c) + ca(3c+b) + cb(3c+a)}{4^{a+b+c} (a+b)(b+c)(c+a)}$$
$$6S = \sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac{3a^2b+3a^2c+3b^2c+3b^2a+3c^2a+3c^2b+6abc}{4^{a+b+c} (a+b)(b+c)(c+a)}$$
$$6S = \sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac{3(a+b)(b+c)(c+a)}{4^{a+b+c} (a+b)(b+c)(c+a)}$$
$$S = \frac12\sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac1{4^{a+b+c}} = \frac12\left(\sum_{a=1}^{\infty}\frac1{4^a}\right)^3 = \frac12\left(\frac13\right)^3 = \boxed{\frac1{54}}$$

If you want to practice this trick that I just mentioned, try to compute the following sum in a similar fashion (it has definitely been asked on this site before):
$$\sum_{m=1}^\infty\sum_{n=1}^\infty\dfrac{m^2n}{3^m(n3^m+m3^n)} = \frac{9}{32}$$
