Let $a$ and $b$ be relatively prime positive integers such that \[\dfrac ab=\dfrac1{2^1}+\dfrac2{3^2}+\dfrac3{2^3}+\dfrac4{3^4}+\dfrac5{2^5}+\dfrac6{3^6}+\cdots,\] where the numerators always increase by $1$, and the denominators alternate between powers of $2$ and $3$, with exponents also increasing by $1$ for each subsequent term.  Compute $a+b$.
1. Let's denote the given series by \( S \):
   \[
   S = \frac{1}{2^1} + \frac{2}{3^2} + \frac{3}{2^3} + \frac{4}{3^4} + \frac{5}{2^5} + \frac{6}{3^6} + \cdots
   \]

2. We can split \( S \) into two separate series, one involving powers of 2 and the other involving powers of 3:
   \[
   S_2 = \frac{1}{2^1} + \frac{3}{2^3} + \frac{5}{2^5} + \cdots
   \]
   \[
   S_3 = \frac{2}{3^2} + \frac{4}{3^4} + \frac{6}{3^6} + \cdots
   \]

3. First, consider the series \( S_2 \):
   \[
   S_2 = \frac{1}{2^1} + \frac{3}{2^3} + \frac{5}{2^5} + \cdots
   \]

4. To find a pattern, let's multiply \( S_2 \) by \(\frac{1}{4}\):
   \[
   \frac{S_2}{4} = \frac{1}{2^3} + \frac{3}{2^5} + \frac{5}{2^7} + \cdots
   \]

5. Subtracting \(\frac{S_2}{4}\) from \(S_2\):
   \[
   S_2 - \frac{S_2}{4} = \left( \frac{1}{2^1} + \frac{3}{2^3} + \frac{5}{2^5} + \cdots \right) - \left( \frac{1}{2^3} + \frac{3}{2^5} + \frac{5}{2^7} + \cdots \right)
   \]
   \[
   \frac{3S_2}{4} = \frac{1}{2^1} + \frac{2}{2^3} + \frac{2}{2^5} + \cdots
   \]

6. Notice that the remaining series is a geometric series:
   \[
   \frac{3S_2}{4} = \frac{1}{2} + \frac{2}{2^3} + \frac{2}{2^5} + \cdots
   \]

7. The geometric series sum formula is:
   \[
   \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}
   \]
   Here, \( a = \frac{1}{2} \) and \( r = \frac{1}{4} \):
   \[
   \frac{3S_2}{4} = \frac{\frac{1}{2}}{1 - \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3}
   \]

8. Solving for \( S_2 \):
   \[
   S_2 = \frac{2}{3} \cdot \frac{4}{3} = \frac{8}{9}
   \]

9. Now, consider the series \( S_3 \):
   \[
   S_3 = \frac{2}{3^2} + \frac{4}{3^4} + \frac{6}{3^6} + \cdots
   \]

10. To find a pattern, let's multiply \( S_3 \) by \(\frac{1}{9}\):
    \[
    \frac{S_3}{9} = \frac{2}{3^4} + \frac{4}{3^6} + \frac{6}{3^8} + \cdots
    \]

11. Subtracting \(\frac{S_3}{9}\) from \(S_3\):
    \[
    S_3 - \frac{S_3}{9} = \left( \frac{2}{3^2} + \frac{4}{3^4} + \frac{6}{3^6} + \cdots \right) - \left( \frac{2}{3^4} + \frac{4}{3^6} + \frac{6}{3^8} + \cdots \right)
    \]
    \[
    \frac{8S_3}{9} = \frac{2}{3^2} + \frac{2}{3^4} + \frac{2}{3^6} + \cdots
    \]

12. Notice that the remaining series is a geometric series:
    \[
    \frac{8S_3}{9} = \frac{2}{3^2} + \frac{2}{3^4} + \frac{2}{3^6} + \cdots
    \]

13. The geometric series sum formula is:
    \[
    \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}
    \]
    Here, \( a = \frac{2}{9} \) and \( r = \frac{1}{9} \):
    \[
    \frac{8S_3}{9} = \frac{\frac{2}{9}}{1 - \frac{1}{9}} = \frac{\frac{2}{9}}{\frac{8}{9}} = \frac{2}{9} \cdot \frac{9}{8} = \frac{2}{8} = \frac{1}{4}
    \]

14. Solving for \( S_3 \):
    \[
    S_3 = \frac{1}{4} \cdot \frac{9}{8} = \frac{9}{32}
    \]

15. Adding \( S_2 \) and \( S_3 \):
    \[
    S = S_2 + S_3 = \frac{8}{9} + \frac{9}{32}
    \]

16. Finding a common denominator:
    \[
    \frac{8}{9} = \frac{8 \cdot 32}{9 \cdot 32} = \frac{256}{288}
    \]
    \[
    \frac{9}{32} = \frac{9 \cdot 9}{32 \cdot 9} = \frac{81}{288}
    \]

17. Adding the fractions:
    \[
    S = \frac{256}{288} + \frac{81}{288} = \frac{337}{288}
    \]

18. Since \( a \) and \( b \) are relatively prime, we have \( a = 337 \) and \( b = 288 \).

19. Therefore, \( a + b = 337 + 288 = 625 \).

The final answer is \(\boxed{625}\).