Forgot password? New user? Sign up
Existing user? Log in
n!an−1=an\dfrac { n! }{ { a }_{ n-1 } } ={ a }_{ n }an−1n!=an
Consider a recurrence relation above for n=1,2,3,…n = 1,2,3,\ldotsn=1,2,3,… and a0=1a_0 = 1a0=1. If the value of the series ∑n=0∞1a2n \displaystyle \sum_{n=0}^\infty \dfrac1{a_{2n} } n=0∑∞a2n1 can be expressed as eα/β e^{\alpha /\beta} eα/β, where α\alphaα and β\betaβ are coprime positive integers, find α+β\alpha+\betaα+β.
Clarification: e≈2.71828e \approx 2.71828e≈2.71828 denotes the Euler's number.
Problem Loading...
Note Loading...
Set Loading...