Triple sum

First observe that

$\dfrac{1}{5^{i+j+k}} = \dfrac{1}{5^i 5^j 5^k}$

Now we break down the sum into three separate sums as follows

$\begin{aligned} \displaystyle \bullet & \ S = \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \dfrac{1}{5^i 5^j 5^k} (i \neq j \neq k) = \sum_{i=0}^{\infty} \dfrac{1}{5^i} \cdot S_1 \\ \displaystyle \bullet & \ S_1 = \sum_{j=0, \ j \neq i}^{\infty} \dfrac{1}{5^j} \cdot S_2 \\ \displaystyle \bullet & \ S_2 = \sum_{k=0, \ k \neq i, \ k \neq j}^{\infty} \dfrac{1}{5^k} \end{aligned}$

Starting with $S_2$

$\begin{aligned} \displaystyle S_2 &= \sum_{k=0, \ k \neq i, \ k \neq j}^{\infty} \dfrac{1}{5^k} \\ \displaystyle &= \sum_{k=0}^{\infty} \dfrac{1}{5^k} - \dfrac{1}{5^i} - \dfrac{1}{5^j} \\ \displaystyle &= \dfrac 54 - \dfrac{1}{5^i} - \dfrac{1}{5^j} \end{aligned}$

Now

$\begin{aligned} \displaystyle S_1 &= \sum_{j=0, \ j \neq i}^{\infty} \dfrac{1}{5^j} \cdot S_2 \\ \displaystyle &= \sum_{j=0, \ j \neq i}^{\infty} \dfrac{1}{5^j} \left( \dfrac 54 - \dfrac{1}{5^i} - \dfrac{1}{5^j} \right) \\ \displaystyle &= \left( \dfrac 54 - \dfrac{1}{5^i} \right) \sum_{j=0, \ j \neq i}^{\infty} \dfrac{1}{5^j} - \sum_{j=0, \ j \neq i}^{\infty} \dfrac{1}{{25}^j} \\ \displaystyle &= {\left( \dfrac 54 - \dfrac{1}{5^i} \right)}^2 - \left( \dfrac {25}{24} - \dfrac{1}{{25}^i} \right) \\ \displaystyle &= \dfrac {25}{48} - \dfrac 52 \cdot \dfrac{1}{5^i} + 2 \cdot \dfrac{1}{{25}^i} \end{aligned}$

Hence

$\begin{aligned} \displaystyle S &= \sum_{i=0}^{\infty} \dfrac{1}{5^i} \cdot S_1 \\ \displaystyle &= \sum_{i=0}^{\infty} \dfrac{1}{5^i} \left( \dfrac{25}{48} - \dfrac 52 \cdot \dfrac{1}{5^i} + 2 \cdot \dfrac{1}{{25}^i} \right) \\ \displaystyle &= \dfrac{25}{48} \sum_{i=0}^{\infty} \dfrac{1}{5^i} - \dfrac 52 \sum_{i=0}^{\infty} \dfrac{1}{{25}^i} + 2 \sum_{i=0}^{\infty} \dfrac{1}{{125}^i} \\ \displaystyle &= \dfrac{25}{48} \times \dfrac 54 - \dfrac 52 \times \dfrac{25}{24} + 2 \times \dfrac{125}{124} \\ \displaystyle &= \dfrac{125}{1984} \end{aligned}$

Thus, $\dfrac ab = \dfrac{125}{1984} \implies a+b = \boxed{2109}$ .