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

UPDATE:

If you can prove the following identity:

$$\sum\limits_{t=0}^p \left(\frac{{2t+1 \choose t}}{2^{2t+1}}\right)^2 \frac{1}{t+2} = -1 + \frac{(2+p)}{2^{4p+4}}{2p+3 \choose p+1}^2$$

Then this is good enough to solve this question and get my gratitude as well as the 50 point bounty. I got this identity from Mathematica. See my answer below for details.

My question relates to an infinite summation and it's very elegant closed form. The expression is the solution to a nice problem which I'll get into as well. Here is the summation:

Let: $$a_t = \left({2t+1 \choose t} - {2t+1 \choose t-1}\right)\frac{1}{2^{2+2t}}$$

and, $$b_t = \left({2t+2 \choose t} - {2t+2 \choose t-1}\right)\frac{1}{2^{3+2t}}$$

For the very first terms of these sequences, ${n \choose -1} = 0$ for all $n$.

And the summation:

$$S = \sum_{t=0}^{\infty} \left(1-\sum_{l=0}^{t-1}b_l\right) a_t = 1- \sum_{t=0}^{\infty} \left(\sum_{l=0}^{t-1}b_l\right) a_t = 7-\frac{20}{\pi} \tag{1}$$

I know the expression above is correct (verified with a python program), but have no idea how to prove this and would like to at least see how I might approach it.

Now, why do I care about this summation? It is the solution to the following problem:

Imagine two wealthy gamblers start tossing their own separate fair coins, winning 1\$ on heads and losing 1\$ on tails. Both start at 0\$ and have infinite bank balances. The first one wants to get to 2\$ and the second one wants to get to 3\$. What is the probability that the first one will reach his goal before the second one?

One way to solve this is to consider the probability that a gambler reaches exactly $k$ dollars for the first time on toss $n$. If he has $t$ tails, then he needs $t+k$ heads. So, $n=2t+k$ (note if k=2\$, he can only reach the target on even tosses and if k=3\$, he can only reach it on odd tosses). This probability turns out to be:

$$\left({k+2t-1 \choose t} - {k+2t-1 \choose t-1}\right) \frac{1}{2^{k+t}} \frac{1}{2^t}$$

Now, let $A_n$ be the probability that the 2\$ targeting gambler wins on toss $n$ and $A$ be the probability that he wins. Then we have $A = \bigcup\limits_{n=1}^{\infty}A_n$ and so, $P(A) = \sum\limits_{n=0}^\infty P(A_n)$. Now, for the 2\$ targeting gambler to win on the $n$th toss, two things should happen:

1. He should reach his target on the $n$th toss for some even $n$.
2. His competitor, the 3\$ gambler should not reach his target on any toss upto $n-1$ (since he can only reach his target on odd tosses).

Putting all of this together, you can see that the probability that the 2\$ gambler wins is given by equation (1) above. I have put together some python code that approximates $S$ by going upto a large number of tosses. A Reddit user pointed out the closed form for which he used a slightly different but related approach and Mathematica. Now, how do I prove that the summation above has the closed form mentioned $(7-\frac{20}{\pi})$?

EDIT:

Here is a short python snippet that demonstrates the summation in equation (1) above.

a_t = np.array([(comb(2*t+1,t)-comb(2*t+1,t-1))/2**(2*t+2) for t in range(500)])
b_t = np.array([(comb(2*t+2,t)-comb(2*t+2,t-1))/2**(2*t+3) for t in range(500)])
b_sum = 1-np.concatenate(([0],np.cumsum(b_t)))
s = sum(a_t*b_sum[:500])
print(7-20/np.pi-s)

Also, here is the Mathematica snippet that shows the result (thanks to @SteveKass for helping with that):

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*}