# It seems rational but is truly irrational

Calculus Level 4

$\displaystyle \sum_{n = 0}^{\infty} \dfrac {(4n)! (1103 + 26390n)}{(n!)^4 396^{4n}}$

has a value such that the closest integer to it is $$x$$. If the actual value of it is $$\dfrac {a}{b \sqrt{c} \cdot \pi}$$ where the fraction is in the lowest terms and $$c$$ is prime, find the value of $$x + a + b + c$$?

×