A particular sum involving product of binomial coefficients

Question

I am encountering a particular sum involving binomial coefficients, and I am looking for a possible closed-form solution. Here is the sum: suppose we are given two real numbers $a \in (0,1)$ and $b \in (0,1)$, and a positive integer $N \in \mathbb{N}$. The sum is \begin{equation} f(N,a,b) = \sum_{i=1}^N \cfrac{\binom{a}{2i-1}}{2i} \binom{2i}{i} b^i. \end{equation}

Here, since $a \in (0,1)$, we have $\binom{a}{2i-1}=\frac{a(a-1)\cdot\cdot\cdot(a-(2i-1)+1)}{(2i-1)(2i-2)\cdot\cdot\cdot2 \cdot 1}$. Thanks a lot for your help.

Answer 1

Hint. Note that $$ \frac{\binom{a}{2i-1}}{2i}\binom{2i}{i}=\frac{a!(2i)!}{2i (2i-1)! (a-2i+1)!i!i!}=\frac{(a+1)!}{i!i! (a-2i+1)!(a+1)} $$ and also $$ (1+1+b)^{a+1}=\sum_{0\le i,j\le a+1}\frac{(a+1)!}{i!j!(a+1-i-j)!}b^{j}. $$

Answer 2

See the link to WolframAlpha WolframAlpha

and the screen copy :

(after correction of a mistake in the binomial coefficient)

The limit for $N$ tending to infinity is :

This agrees with the answer given by Claude Leibovici

The result in case $N=$infinity can also be expressed as a Legendre function :

Answer 3

Another "nice" formula $$\frac{\, _2F_1\left(\frac{1}{2} (-a-1),-\frac{a}{2};1;4 b\right)-1}{a+1}-\frac{\Gamma (a+1) b^{n+1} \, _3\tilde{F}_2\left(1,-\frac{a}{2}+n+\frac{1}{2},-\frac{a}{2}+n+1;n+2,n+2;4 b\right)}{\Gamma (a-2 n)}$$

I have not been able to find a limit for infite $N$. Sorry for that. Enjoy !