Sequences and series (3)

We break down the sum into three separate sums as follows

$\begin{aligned} \displaystyle \bullet & \ S = \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \dfrac{1}{3^i 3^j 3^k} (i \neq j \neq k) = \sum_{i=0}^{\infty} \dfrac{1}{3^i} \cdot S_1 \\ \displaystyle \bullet & \ S_1 = \sum_{j=0, \ j \neq i}^{\infty} \dfrac{1}{3^j} \cdot S_2 \\ \displaystyle \bullet & \ S_2 = \sum_{k=0, \ k \neq i, \ k \neq j}^{\infty} \dfrac{1}{3^k} \end{aligned}$

Starting with $S_2$

$\begin{aligned} \displaystyle S_2 &= \sum_{k=0, \ k \neq i, \ k \neq j}^{\infty} \dfrac{1}{3^k} \\ \displaystyle &= \sum_{k=0}^{\infty} \dfrac{1}{3^k} - \dfrac{1}{3^i} - \dfrac{1}{3^j} \\ \displaystyle &= \dfrac 32 - \dfrac{1}{3^i} - \dfrac{1}{3^j} \end{aligned}$

Now

$\begin{aligned} \displaystyle S_1 &= \sum_{j=0, \ j \neq i}^{\infty} \dfrac{1}{3^j} \cdot S_2 \\ \displaystyle &= \sum_{j=0, \ j \neq i}^{\infty} \dfrac{1}{3^j} \left( \dfrac 32 - \dfrac{1}{3^i} - \dfrac{1}{3^j} \right) \\ \displaystyle &= \left( \dfrac 32 - \dfrac{1}{3^i} \right) \sum_{j=0, \ j \neq i}^{\infty} \dfrac{1}{3^j} - \sum_{j=0, \ j \neq i}^{\infty} \dfrac{1}{9^j} \\ \displaystyle &= {\left( \dfrac 32 - \dfrac{1}{3^i} \right)}^2 - \left( \dfrac 98 - \dfrac{1}{9^i} \right) \\ \displaystyle &= \dfrac 98 - 3 \cdot \dfrac{1}{3^i} + 2 \cdot \dfrac{1}{9^i} \end{aligned}$

Hence

$\begin{aligned} \displaystyle S &= \sum_{i=0}^{\infty} \dfrac{1}{3^i} \cdot S_1 \\ \displaystyle &= \sum_{i=0}^{\infty} \dfrac{1}{3^i} \left( \dfrac 98 - 3 \cdot \dfrac{1}{3^i} + 2 \cdot \dfrac{1}{9^i} \right) \\ \displaystyle &= \dfrac 98 \sum_{i=0}^{\infty} \dfrac{1}{3^i} - 3 \sum_{i=0}^{\infty} \dfrac{1}{9^i} + 2 \sum_{i=0}^{\infty} \dfrac{1}{{27}^i} \\ \displaystyle &= \dfrac 98 \times \dfrac 32 - 3 \times \dfrac 98 + 2 \times \dfrac{27}{26} \\ \displaystyle &= \dfrac{81}{208} \end{aligned}$

Thus, $\dfrac ab = \dfrac{81}{208} \implies a+b = \boxed{289}$ .