Can be written as $\frac{A}{B}$ , where $A$ and $B$ are not coprime?!

$\displaystyle \sum_{m = 1}^{\infty} \displaystyle \sum_{n=1}^{\infty} \frac{1}{A^m B^n} = \displaystyle \sum_{m = 1}^{\infty} \frac{1}{A^m} \displaystyle \sum_{n = 1}^{\infty} \frac{1}{B^n}$

We know that $\frac{1}{A}, \frac{1}{B} < 1$ because the sum would be infinite otherwise and thus it would not be able to expressed as $\frac{A}{B}$ . So using the geometric series formula

$(\frac{\frac{1}{A}}{1 - \frac{1}{A}})(\frac{\frac{1}{B}}{1 - \frac{1}{B}}) = (\frac{1}{A-1})(\frac{1}{B-1}) = \frac{A}{B}$ .

Cross multiplying, we get $A(A-1)(B-1) = B$ and then $B = \frac{A^2 - A}{A^2 - A - 1} = 1 + \frac{1}{A^2 - A - 1}$

Since $B$ is an integer, then $A^2 - A - 1 = 1$ , so $A = 2$ and $B = 2$ .

Thus, $A + B = 4$ .

This can be represented as the product of two geometric series like this. $\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\dfrac{1}{A^iB^j} =\sum_{i=1}^{\infty}\dfrac{1}{A^i}\times\sum_{j=1}^{\infty}\dfrac{1}{B^j}$

We can take the sum of these series individually. Let the first sequence be $X$ and the second one be $Y.$ We are looking for $x\times Y.$

$X$ is a geometric sequence with first term $\frac{1}{A}$ and ratio $\frac{1}{A}.$ The sum of the terms is equal to $\dfrac{\frac{1}{A}}{1-\frac{1}{A}}=\dfrac{\frac{1}{A}}{\frac{A-1}{A}}={\dfrac{\textbf{1}}{\textbf{A-1}}}$

Similarly, $Y$ is a geometric sequence with first term $\frac{1}{B}$ and ratio $\frac{1}{B}.$ The sum of the terms is equal to $\dfrac{\frac{1}{B}}{1-\frac{1}{B}}=\dfrac{\frac{1}{B}}{\frac{B-1}{B}}={\dfrac{\textbf{1}}{\textbf{B-1}}}$

$X\times Y=\frac{1}{(A-1)(B-1)}.$ We have found what the sum of the reciprocals of $A^mB^n$ is. Now we need to solve for $A$ and $B$ in this equation. $\dfrac{1}{(A-1)(B-1)}=\dfrac{A}{B}$ We find that $B=\frac{A^2-A}{A^2-A-1}\text{***}$

Solving this equation yields $\{A,B\}=\{2,2\}$ as solutions, so $A+B=\boxed{4}$

* We don't have to worry about a non-unique solution because this function is asymptotic at $A=1$ and $B=1$ (in the A-B coordinate plane, which is exactly the same as the Cartesian plane). $A$ and $B$ have to be greater than or equal to $2$ because if either of them were $1$ then the total sum would be infinite. Any $A>2$ would produce a non-integer $B\text{-value}$ in the range $1<B<2.$ A graph of this function is here .