What is the sum of a binomial series where the bionomials terms are raised to a power?: $ \sum\limits_{k=0}^\infty \binom{n}{k}^dx^k$

Question

We know that:

$$ \sum_{k=0}^\infty \binom{n}{k}x^k = (1+x)^n $$

Is there a closed-form expression to

$$ \sum_{k=0}^\infty \binom{n}{k}^dx^k, $$ where $d \in \Bbb{R} $? Can you help me? Thanks!

Answer 1

Here will be an attempt at a closed form for $d\in\Bbb N$:

You can rewrite it in terms of the Pochhammer symbol:

$$\sum_{k=0}^\infty \binom nk ^d x^k=Γ(n+1)^d\sum_{k=0}^\infty \frac{x^k}{Γ(k+1)^dΓ(n-k+1)^d}= Γ(n+1)^d\sum_{k=0}^\infty \frac{x^k}{\prod_{j=1}^d Γ(k+1)\prod_{j=1}^d Γ(n-k+1)}= Γ(n+1)^d\sum_{k=0}^\infty \frac{x^k}{\prod_{j=1}^d (1)_{k}\prod_{j=1}^d (1)_{n-k}} $$

Which can probably be put into a Generalized Hypergeometric function form using the following definition:

$$\,_p\text F_q(a_1,…,a_p;b_1,…b_q;z)=\sum_{k=0}^\infty \frac{\prod_{j=1}^p(a_j)_k\,z^k}{\prod_{j=1}^q(b_j)_k\,k!}$$

Let’s try and find a pattern:

$d=3:$

$$Γ(n+1)^3\sum_{k=0}^\infty \frac{x^k}{Γ(k+1)^3Γ(n-k+1)^3}=\,_3\text F_2(-n,-n,-n,1,1,x)$$ $d=4:$

$$Γ(n+1)^4\sum_{k=0}^\infty \frac{x^k}{Γ(k+1)^4Γ(n-k+1)^4}=\,_4\text F_3(-n,-n,-n,-n,1,1,1,x)$$

$d=d$:

$$\sum_{k=0}^\infty \binom nk ^d x^k = Γ(n+1)^d\sum_{k=0}^\infty \frac{x^k}{Γ(k+1)^dΓ(n-k+1)^d} =\boxed{\,_d\text F_{d-1}(-n,…,-n;1,…,1;x)}$$

where there are a $d$ amount of $-n$s and a $d-1$ amount of $1$s. I challenge you to try the $d\not\in\Bbb N$ case. The formula is surprisingly simple for a hypergeometric function. Please correct me and give me feedback!

Here is an attempt to find a closed form for the following. Instead of writing out the full sum, let’s define:

$$S(n)\mathop=^\text{def}\sum_{k=0}^\infty\binom nk ^\frac12$$

It is linked to polynomial roots/ values of a trigonometric for $\frac \pi n$ because:

$$S(-1)=\lim_{n\to\infty} n+1+ni=\infty(1+i)$$$$S(0)=1$$$$S(1)=2$$$$S(2)=\sqrt 2+1=\text{Silver Ratio}$$$$ S(3)=1+2\sqrt 3$$$$S(4)=6+\sqrt 6$$$$S(5)=2+2\sqrt{5(3+2\sqrt2)}$$$$S(6)=2\sqrt6+2(\sqrt {15}+\sqrt 5+1)$$$$S(7)=2(1+\sqrt 7+\sqrt {21}+\sqrt 35)$$$$S(24)=2(1+2\sqrt 6+2\sqrt{69}+2\sqrt{506}+6\sqrt{9614}+4\sqrt{10626}+2\sqrt{33649}+7\sqrt{81719}+4\sqrt{156009}+\sqrt{490314}+\sqrt{676039})$$

and so on as the sum gets increasingly larger. Can you find a pattern form for $$S(n)\mathop=^\text{def}\sum_{k=0}^\infty\binom nk ^\frac12?$$ Notice how $S(7)=2(1+\sqrt 7+\sqrt {3\cdot 7}+\sqrt {5\cdot7})$ and try to rewrite the binomial coefficient.