Prove that $\phi=\frac{13}{8} + \sum \limits_{n=0}^{\infty}\frac{(-1)^{(n+1)}(2n+1)!}{n!\,(n+2)!\,4^{(2n+3)}}$.

Question

Prove that $$\phi=\dfrac{13}{8} + \sum \limits_{n=0}^{\infty}\dfrac{(-1)^{(n+1)}(2n+1)!}{n!\,(n+2)!\,4^{(2n+3)}}$$,where $\phi$ is the golden ratio ($\approx1.6180$).

Evaluating the first few few partial sums makes it obvious that the expression approaches $\approx 1.61$ as the limit of the infinite sum is $\approx -0.007$. But is there any algebraic method to prove the same ?

Stirling's approximation seems to work a little but not much as the limit of the sum has to calculated and not the limit of the fraction. If there is no other known way than Stirling's, how should I properly implement it ? I have tried my best, but there does not seem any way out. As far as I think, only numerical method will help here.

Answer 1

Hint. One may recall that $$\begin{eqnarray*} 2\sum_{n=1}^\infty (-1)^n {2n\choose n}u^{n} &=& \frac{2}{\sqrt{1+4u}} - 2, \qquad |u|<1/4, \end{eqnarray*}$$ then integrating gives $$ 4\sum \limits_{n=0}^{\infty}\dfrac{(-1)^{(n+1)}(2n+1)!}{n!\,(n+2)!}\cdot \,u^{n+2}=\sqrt{1+4u}-2u-1 $$ and one may conclude with $u=\dfrac1{4^2}$.