Rewrite the factorial parts of the summand as
[tex]\dfrac{(2n+1)!}{n!(n+2)!} = \dfrac{(2n+1)(2n!)}{(n+2)(n+1)(n!)^2} = \dfrac{2n+1}{(n+2)(n+1)} \dbinom{2n}n[/tex]
where [tex]\binom nk[/tex] is the binomial coefficient, and [tex]\binom{2n}n[/tex] are the so-called central binomial coefficients.
Expand the rational expression into partial fractions:
[tex]\dfrac{2n+1}{(n+2)(n+1)} = \dfrac3{n+2} - \dfrac1{n+1}[/tex]
Pull out a constant factor and collect the exponential terms.
[tex]\dfrac{(-1)^{n+1}}{4^{2n+3}} = -\dfrac1{64} \left(-\dfrac1{16}\right)^n[/tex]
The sum we want is now
[tex]\displaystyle \frac{13}8 - \frac1{64} \sum_{n=0}^\infty \binom{2n}n \left(\frac3{n+2} - \frac1{n+1}\right) \left(-\frac1{16}\right)^n[/tex]
Let f(x) and g(x) be functions with power series expansions
[tex]\displaystyle f(x) = \sum_{n=0}^\infty \binom{2n}n \frac{x^n}{n+1}[/tex]
[tex]\displaystyle g(x) = \sum_{n=0}^\infty \binom{2n}n \frac{x^n}{n+2}[/tex]
and recall the well-known binomial series
[tex]\displaystyle \dfrac1{\sqrt{1-4x}} = \sum_{n=0}^\infty \binom{2n}n x^n[/tex]
which converges for |x| < 1/4.
Integrating both sides yields
[tex]\displaystyle \int \frac{dx}{\sqrt{1-4x}} = \int \sum_{n=0}^\infty \binom{2n}n x^n \, dx[/tex]
[tex]\displaystyle -\frac12 \sqrt{1-4x} = C_1 + \sum_{n=0}^\infty \binom{2n}n \frac{x^{n+1}}{n+1}[/tex]
Taking x = 0 on both sides, it follows that C₁ = -1/2. We then see that
[tex]\displaystyle f(x) = \frac{1-\sqrt{1-4x}}{2x}[/tex]
Step back and multiply both sides of the binomial series identity by x, then integrate. This yields
[tex]\displaystyle \int \frac x{\sqrt{1-4x}} \, dx = \int \sum_{n=0}^\infty \binom{2n}n x^{n+1} \, dx[/tex]
[tex]\displaystyle -\frac1{12} \sqrt{1-4x} (1 + 2x) = C_2 + C_1 x + \sum_{n=0}^\infty \binom{2n}n \frac{x^{n+2}}{n+2}[/tex]
Taking x = 0 again points to C₂ = -1/12. Hence
[tex]\displaystyle g(x) = \frac{1 - \sqrt{1-4x}(1+2x)}{12x^2}[/tex]
Then the value of the sum we want is
[tex]\displaystyle \frac{13}8 - \frac1{64} \left(3g\left(-\frac1{16}\right) - f\left(\frac1{16}\right)\right) = \frac{1+\sqrt5}2 = \boxed{\phi}[/tex]
where ɸ ≈ 1.618 is the golden ratio.
