# Beyond infinity and more

**Calculus**Level 5

If sum of the infinite series \[\dfrac12 + \dfrac28 + \dfrac{3}{16} + \dfrac{5}{32} + \dfrac{8}{64}+ \dfrac{13}{128} + \cdots = \dfrac{A}{B},\]

where \(\gcd(A,B)=1\), then find \(A+B\).

**Assumptions and Clarifications**

- \( \dfrac{1}{4} \) between \( \dfrac12 \) and \( \dfrac28 \) is deliberately omitted.
- With the exception of the missing \(\frac{1}{4}\) term, the numbers in the numerator for the Fibonacci sequence and the numbers in the denominator are terms of a G.P.