Who's up to the challenge? 67

Let the summation be denoted by $\displaystyle S = \sum _{ n=0 }^{ \infty }{ \dfrac { { 2 }^{ n+1 }n!^{ 2 } }{ (2n+1)!(n+1)^{ 2 } } }$

We rewrite the summation as $\displaystyle S = \sum_{n=1}^{\infty} \frac{2^{n+1}}{n^3\binom{2n}{n}}$

By Gregory Series , $\displaystyle ytan^{-1}y = \sum_{m=1}^{\infty} \frac{(-1)^{m-1}y^{2m}}{2m-1}$

Substitute $\displaystyle y=\frac{x}{\sqrt{1-x^2}}$ ,

$\displaystyle \frac{x}{\sqrt{1-x^2}}\sin^{-1}x = \sum_{m=1}^{\infty} \frac{(-1)^{m-1}x^{2m}}{(2m-1)(1-x^2)^m}$

$\displaystyle = \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{2m-1} \sum_{k=0}^{\infty} \binom{m+k-1}{k} x^{2(k+m)}$

$\displaystyle \frac{x}{\sqrt{1-x^2}}\sin^{-1}x = \sum_{r=1}^{\infty}x^{2r} \sum_{m=1}^{r\Gamma(r)}{\Gamma(m)\Gamma(r-m+1)(2m-1)}$

where co-efficient of $x^{2r}$ is half of $\displaystyle \sum_{j=1}^{\infty} \frac{(2x)^{2j}}{j\binom{2j}{j}}$ i.e.

$\displaystyle r\binom{2r}{r} \sum_{v=0}^{r-1} (-1)^v \binom{r-1}{v}\frac{1}{2v+1} = 2^{2r-1}$

To verify rewrite as ,

$\displaystyle r\binom{2r}{r} \sum_{v=0}^{r-1} (-1)^v \binom{r-1}{v}\frac{1}{2v+1} = r\binom{2r}{r} \int_{0}^{1}\sum_{v=0}^{r-1} (-1)^v \binom{r-1}{v} y^{2v} dy = r\binom{2r}{r} \int_{0}^{1} (1-y^2)^{r-1}dy$

Hence , $\displaystyle r\binom{2r}{r} \sum_{v=0}^{r-1} (-1)^v \binom{r-1}{v}\frac{1}{2v+1} = r\binom{2r}{r} \int_{0}^{1} (sin\theta)^{2r-1} d\theta$

Using Walli's integral , $\displaystyle r\binom{2r}{r} \sum_{v=0}^{r-1} (-1)^v \binom{r-1}{v}\frac{1}{2v+1}=2^{2r-1}$

So we are verified with the co-efficient equating task and hence $\displaystyle \color{#20A900}{\sum_{j=1}^{\infty} \frac{(2x)^{2j}}{j\binom{2j}{j}} = \frac{2x}{\sqrt{1-x^2}}\sin^{-1}x}$

Dividing by $2x$ and integrating from $0$ to $x$ twice on the green sum we get,

$\displaystyle \sum_{m=1}^{\infty} \frac{(2x)^{2m}}{m^3\binom{2m}{m}}\ = 4\int_{0}^{x} \frac{(\sin^{-1}y)^2}{y} dy$

If we put $\displaystyle x=\frac{1}{\sqrt{2}}$ then the equation becomes,

$\displaystyle \frac{S}{2} = 4\int_{0}^{\frac{1}{\sqrt{2}}} \frac{(\sin^{-1}y)^2}{y} dy$ , where $S$ is our required sum

R.H.S is a representation of Log-Sine Integral which we can use as a standard result or even evaluate ,

Using IBP on RHS we get the second term as a Log-Sine integral $\displaystyle \int_{0}^{\frac{\pi}{2}} t\ln|\sin\frac{t}{2}| dt$ where we could make a substitution $\displaystyle t\to 2t$ to get upper limit $\pi$

We are missing a $'2'$ inside the Logarithm but that doesn't create any difference as we are not using any standard formulae for the evaluation , So let us consider a function ,

$\displaystyle F(a,b) = \int_{0}^{\pi} (\sin z)^{a}\cos bz dz = \frac{2^{-a}\pi\Gamma(a+1)\cos\frac{b\pi}{2}}{\Gamma(\frac{a+b+2}{2})\Gamma(\frac{a-b+2}{2})}$

So our integral equals $\displaystyle \frac{\partial^2 F(a,b)}{\partial a \partial b}]_{\text{a=b=1}}$ which can now be calculated using Trigamma & Catalan's constant.

Putting all the values successively which is lengthy and an easy exercise we get $\displaystyle \frac{S}{8} = \frac{\pi G}{4} -\frac{35}{64}\zeta(3) + \frac{\pi^2}{32}\ln2$

$\displaystyle \sum _{ n=0 }^{ \infty }{ \dfrac { { 2 }^{ n+1 }n!^{ 2 } }{ (2n+1)!(n+1)^{ 2 } } } = 2\pi G - \frac{35}{8}\zeta(3) +\frac{\pi^2}{4}\ln 2$

Thus making the answer $\boxed{2+35+8+3+2+4+2=56}$