Race of the wealthy gamblers: How do I get this closed form?

Let $a_t \triangleq \left( \frac{\begin{pmatrix} 2t+1 \\ t \end{pmatrix}}{2^{2t+1}} \right)^2 \frac{1}{t+2}$. We need $S_p \triangleq \sum\limits_{t=0}^p a_t$ in closed form.

First, note that $$\frac{a_{t+1}}{a_t} = \frac{t+2}{t+3} . \frac{1}{2^4} . \left( \frac{ \begin{pmatrix} 2t+3 \\ t+1 \end{pmatrix} }{ \begin{pmatrix} 2t+1 \\ t \end{pmatrix} } \right)^2 = \frac{t+2}{t+3} . \frac{1}{2^4} . \left( \frac{ (2t+3)(2t+2) }{ (t+1)(t+2) } \right)^2 = \frac{(t+3/2)^2}{(t+2)(t+3)} \tag 1$$

Using $(1)$ repeatedly starting with $a_0 = 1/8$, we get:

$$a_t = \frac{1}{8} . \frac{(3/2)^2}{2.3} . \frac{(5/2)^2}{3.4} \ldots \frac{(t+1/2)^2}{(t+1).(t+2)} \text{ for $t\geq 1$} \tag 2$$

Using $(2)$, we have

$$S_0 = a_0 = \frac{1}{8} = \frac{9}{8} - 1 = \bbox[yellow]{\frac{1}{8}.3^2 - 1}$$ $$S_1 = a_1 + a_0 = a_1 + S_0 = \frac{1}{8}.\frac{(3/2)^2}{2.3} + \frac{3^2}{8} - 1 $$ $$ = \bbox[yellow]{\frac{1}{8}.\frac{(3/2)^2}{2.3}. 5^2 -1}$$ $$S_2 = a_2 + S_1 = \frac{1}{8} . \frac{(3/2)^2}{2.3} . \frac{(5/2)^2}{3.4} + \frac{5^2}{8}.\frac{(3/2)^2}{2.3} -1 $$ $$= \bbox[yellow]{\frac{1}{8} . \frac{(3/2)^2}{2.3} . \frac{(5/2)^2}{3.4}. 7^2 - 1}$$ The general pattern we observe is (can be proved formally via induction) $$ \begin{align} S_p &= \frac{1}{8}. \frac{(3/2)^2}{2.3}. \frac{(5/2)^2}{3.4} \ldots \frac{((2p+1)/2)^2}{(p+1)(p+2)}.(2p+3)^2 -1 \\ &= \frac{1}{16}. \frac{(3/2)^2}{3^2}. \frac{(5/2)^2}{4^2} \ldots \frac{((2p+1)/2)^2}{(p+2)^2}.(2p+3)^2.(p+2) -1 \\ &= \frac{1}{2^{2p+4}}. \frac{3^2}{3^2}. \frac{5^2}{4^2} \ldots \frac{(2p+1)^2}{(p+2)^2}.(2p+3)^2.(p+2) -1 \\ &= \frac{p+2}{2^{2p+4}} \left[ \frac{3.5.7 \ldots (2p+1)(2p+3)}{2.3.4 \ldots (p+1)(p+2)}.2\right]^2 -1 \\ &= \frac{p+2}{2^{2p+4}} \left[ \frac{3.5.7 \ldots (2p+1)(2p+3)}{(p+2)!}.\frac{(p+1)! 2^{p+1}}{(p+1)! 2^{p+1}}.2\right]^2 -1 \\ &= \frac{p+2}{2^{2p+4}} \left[ \frac{(2p+3)!}{(p+2)!(p+1)!}.\frac{1}{ 2^{p}}\right]^2 -1 \\ &= \frac{p+2}{2^{4p+4}} \begin{pmatrix} 2p+3 \\ p+1 \end{pmatrix}^2 -1 \\ \end{align} $$ which is what we want.

This answer is based upon the Gosper algorithm. It can also be used to solve structural similar identities (see (4) below). We follow closely Summa Summarum by M.E. Larsen.

We set \begin{align*} \color{blue}{A(p):=\sum_{t=0}^p\frac{1}{t+2}\binom{2t+1}{t}^2\frac{1}{2^{4t+2}}}\tag{1} \end{align*}

We rewrite $A(p)$ as follows \begin{align*} A(p)&=\sum_{t=0}^pa_t=a_0+a_1+a_2+\cdots+a_p\\ &=a_0+a_0\frac{a_1}{a_0}+a_0\frac{a_1a_2}{a_0a_1}+\cdots+a_0\frac{a_1a_2\cdots a_p}{a_0a_1\cdots a_{p-1}}\\ &=a_0\sum_{t=0}^p\prod_{j=0}^{t-1}\frac{a_{j+1}}{a_j}\tag{2} \end{align*}

We obtain $a_0=\frac{1}{8}$ and \begin{align*} \frac{a_{j+1}}{a_j}&=\frac{j+2}{j+3}\cdot\frac{\binom{2j+3}{j+1}^2}{\binom{2j+1}{j}^2}\cdot\frac{2^{4j+6}}{2^{4j+2}}=\frac{(2j+3)^2}{4(j+2)(j+3)}\\ &=\frac{\left(-\frac{3}{2}-j\right)^2}{(-2-j)(-3-j)} \end{align*}

In the following we use the falling factorial notation $x^{\underline{t}}=x(x-1)(x-2)\cdots(x-t+1)$.

From (1) and (2) we get \begin{align*} A(p)&=\frac{1}{8}\sum_{t=0}^{p}\prod_{j=0}^{t-1}\frac{\left(-\frac{3}{2}-j\right)^2}{(-2-j)(-3-j)}\\ &=\frac{1}{8}\sum_{t=0}^p\frac{\left(-\frac{3}{2}\right)^{\underline{t}}\left(-\frac{3}{2}\right)^{\underline{t}}}{(-2)^{\underline{t}}(-3)^{\underline{t}}}\tag{3} \end{align*}

We consider Corollary 6.2 of Summa Summarum which states for $a,b,c,d\in\mathbb{C}$ with $a+b=c+d$ \begin{align*} \sum_{t=0}^p\frac{a^{\underline{t}}b^{\underline{t}}}{(c-1)^{\underline{t}}(d-1)^{\underline{t}}} =\frac{1}{(a-c)(b-c)}\left(\frac{a^{\underline{p+1}}b^{\underline{p+1}}}{(c-1)^{\underline{p}}(d-1)^{\underline{p}}}-cd\right)\tag{4} \end{align*}

We can apply Corollary 6.2 since in (3) we have $a=b=-\frac{3}{2}, c=-1,d=-2$ so that $a+b=c+d$. We get \begin{align*} \color{blue}{A(p)}&=\frac{1}{8}\cdot\frac{1}{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)} \left(\frac{\left(-\frac{3}{2}\right)^{\underline{p+1}}\left(-\frac{3}{2}\right)^{\underline{p+1}}}{(-2)^{\underline{p}}(-3)^{\underline{p}}}-(-1)(-2)\right)\\ &=\frac{1}{2}\cdot\frac{\left(-\frac{3}{2}\right)^{\underline{p+1}}\left(-\frac{3}{2}\right)^{\underline{p+1}}}{(-2)^{\underline{p}}(-3)^{\underline{p}}}-1\tag{5}\\ &=\frac{1}{2}\left(\frac{(2p+3)!}{2^{2p+2}(p+1)!}\right)^2\cdot\frac{1}{(p+1)!\frac{1}{2}(p+2)!}-1\\ &\,\,\color{blue}{=\frac{p+2}{2^{4p+4}}\binom{2p+3}{p+1}^2-1} \end{align*} and the claim follows.

Comment:

• In (5) we use \begin{align*} \left(-\frac{3}{2}\right)^{\underline{p+1}}&=\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots\left(-\frac{3}{2}-p\right)\\ &=\frac{(-1)^{p+1}}{2^{p+1}}(2p+3)!!=\frac{(-1)^{p+1}}{2^{p+1}}\cdot\frac{(2p+3)!}{(2p+2)!!}=\frac{(-1)^{p+1}(2p+3)!}{2^{p+1}2^{p+1}(p+1)!}\\ &=\frac{(-1)^{p+1}(2p+3)!}{2^{2p+2}(p+1)!}\\ (-2)^{\underline{p}}&=(-2)(-3)\cdots(-2-(p+1))=(-1)^p(p+1)!\\ (-3)^{\underline{p}}&=(-3)(-4)\cdots(-3-(p+1))=(-1)^p\frac{1}{2}(p+2)! \end{align*}

As Steve Kass commented, there is a problem somewhere.

Using a CAS, what I obtained after simplifications is $$\sum_{l=1}^{t-1}b_l=\frac{7}{8}+\frac{ (t+3) (3 t+4) }{3 (t+1)\,2^{2 (t+1)}}\left(\binom{2 t+2}{t-1}-\binom{2 t+2}{t}\right)=\frac{7}{8}-\frac{(3 t+4)\, \Gamma \left(t+\frac{3}{2}\right)}{\sqrt{\pi } \,\,\Gamma (t+3)}$$ The partial sum $$S_p = \sum_{t=1}^{p} \left(\sum_{l=1}^{t-1}b_l\right) a_t$$ is evaluated (the result is very nasty but "almost" explicit ) but, given by the same CAS, $$S_\infty =\frac{20}{\pi }-\frac{195}{32}$$ is obtained without any problem (how ? this is the question).

Edit

After your two major changes (summations starting at $0$ and the formula for $S$), the results become quite different.

$$\sum_{l=0}^{t-1}b_l=1-\frac{(3 t+4) \Gamma \left(t+\frac{3}{2}\right)}{\sqrt{\pi } \,\,\Gamma (t+3)}$$ $$S_p = \sum_{t=0}^{p} \left(1-\sum_{l=0}^{t-1}b_l\right) a_t=7-\frac{4 (5 p+12) \,\Gamma \left(p+\frac{5}{2}\right)^2}{\pi \, \Gamma (p+3)^2}$$ Using Stirling approximation for lage values of $p$, we have $$S_p=7-\frac{20}\pi\left(1+\frac{3}{20 p}-\frac{59}{160 p^2}+\frac{573}{640 p^3}-\frac{21977}{10240 p^4}+O\left(\frac{1}{p^5}\right)\right)$$ $$S_\infty =7-\frac{20}{\pi }$$