How would you solve it: Part 4

$\large \displaystyle \sum_{n=0}^{\infty}\dfrac{(-1)^{n}}{n+1}{2n \choose n}^{-1} = \dfrac{\alpha}{\sqrt{\beta}}\ln^{\eta}{\phi} - \delta(\ln^{\gamma}{\phi})$ If the above summation is true for positive integers: $\alpha,\beta,\gamma,\delta,\eta$ and that $\beta$ is a prime, then compute: $\alpha+\beta+\gamma+\delta+\eta$ .
Note that: $\phi$ is the Golden ratio
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