closed form expression for the summation $\sum_{i=0}^{\infty}\binom{i+k}{i}x^ii^t$

Question

I’m trying to find a closed form expression for this summation.

$\sum_{i=0}^{\infty}\binom{i+k}{i}x^ii^t$

Where k and t are given nonnegative integers. According to ( Checking Jaynes' formula 6.108 for $\sum\limits_{m=0}^\infty{m+a \choose m} m^nx^m$ ) the answer is

${(x\frac{d}{dx})}^t\frac{1}{{(1-x)}^{k+1}}$

And based on (What's the property of this series? Is it special? Coefficients of $\left(x\frac{d}{dx}\right)^n f(x) $) we can write

$\sum_{i=0}^{\infty}\binom{i+k}{i}x^ii^t=\left(x\frac{d}{dx}\right)^t\frac{1}{\left(1-x\right)^{k+1}}=\sum_{r=1}^{t}{S\left(t,r\right)x^r\frac{d^r}{dx^r}\frac{1}{\left(1-x\right)^{k+1}}} =\sum_{r=1}^{t}{S\left(t,r\right)x^r\frac{\left(k+1\right)\left(k+2\right)\ldots\left(k+r\right)}{\left(1-x\right)^{k+r+1}}} =\frac{1}{\left(1-x\right)^{k+1}}\sum_{r=1}^{t}{(k+1)(k+2)...(k+r)\ S\left(t,r\right)\ {(\frac{x}{1-x})}^r}$

Where $S(t,r)$ is the Stirling number of the second kind. This can be written in two ways. But none of them seems simpler to me.

$=\frac{1}{\left(1-x\right)^{k+1}}\sum_{r=1}^{t}{\frac{(k+r)!}{k!}\ S\left(t,r\right)\ {(\frac{x}{1-x})}^r}$

Or since it includes a falling factorial ($\left(k+r\right)_{\bar{r}}$), we can write

$=\frac{1}{\left(1-x\right)^{k+1}}\sum_{r=1}^{t}{\sum_{u=0}^{r}{S\left(t,r\right)\left(\frac{x}{1-x}\right)^rs\left(r,u\right)\left(k+r\right)^u}\ }$ $=\frac{1}{\left(1-x\right)^{k+1}}\sum_{r=1}^{t}{\sum_{u=1}^{r}{S\left(t,r\right)\left(\frac{x}{1-x}\right)^rs\left(r,u\right)\left(k+r\right)^u}\ }$ $=\frac{1}{\left(1-x\right)^{k+1}}\sum_{u=1}^{t}{\sum_{r=u}^{t}{S\left(t,r\right)s\left(r,u\right)\left(\frac{x}{1-x}\right)^r\left(k+r\right)^u}\ } $

Where $s(r,u)$ is the signed Stirling number of the first kind (Remember $\sum_{r=u}^{t}{S\left(t,r\right)s\left(r,u\right)=\delta_{tu}}$ but this one is more complicated). The question has not been solved yet (finding a closed form expression for this summation. $\sum_{i=0}^{\infty}\binom{i+k}{i}x^ii^t$). Can you help me?

(I wrote what I've tried since it may inspire you).

Answer 1

You can use Mathematica to verify the result for some cases:

f[inputT_] := 
 Sum[Binomial[i + k, i]*x^i*i^t /. {t -> inputT}, {i, 0, Infinity}]
f /@ Range[1, 5]

For $t=1$,

$$ \sum\limits_{i=0}^{\infty} \tbinom{i+k}{i}x^i i^1= \frac{(k+1) x (1-x)^{-k}}{(x-1)^2} $$

For $t=2$,

$$ \sum\limits_{i=0}^{\infty} \tbinom{i+k}{i}x^i i^2 = -\frac{(k+1) x (1-x)^{-k} (k x+x+1)}{(x-1)^3} $$

For $t=3$,

$$ \sum\limits_{i=0}^{\infty} \tbinom{i+k}{i}x^i i^3 =\frac{(k+1) x (1-x)^{-k} \left(k^2 x^2+2 k x^2+3 k x+x^2+4 x+1\right)}{(x-1)^4} $$

For $t=4$,

$$ \sum\limits_{i=0}^{\infty} \tbinom{i+k}{i}x^i i^4= \, _4F_3(2,2,2,k+2;1,1,1;x) \cdot (k+1) x $$

For $t=5$,

$$ \sum\limits_{i=0}^{\infty} \tbinom{i+k}{i}x^i i^5= \, _5F_4(2,2,2,2,k+2;1,1,1,1;x) \cdot (k+1) x $$

For $t=6$,

$$ \sum\limits_{i=0}^{\infty} \tbinom{i+k}{i}x^i i^6= \, _6F_5(2,2,2,2,2,k+2;1,1,1,1,1;x) \cdot (k+1) x $$

---

Here the involved hypergeometric function is HypergeometricPFQ.

$_p F_q(a ; b ; z)$ has series expansion $$\sum\limits_{k=0}^{\infty}\left(a_1\right)_k \ldots\left(a_p\right)_k /\left(b_1\right)_k \ldots\left(b_q\right)_k z^k / k !$$

where $(a)_k$ is the Pochhammer symbol.

Answer 2

I found this question because I am reading Jaynes Chapter 6 and encountered 6.108. Using your variables, it says $$\sum_{i=0}^{\infty}\binom{i+k}{i}x^ii^t = (x\frac{d}{dx})^t \frac{1}{{(1-x)}^{k+1}}$$ I do not believe a simple closed form expression exists for arbitrary non-negative integers. Of course, for t = 0, it says $$\sum_{i=0}^{\infty}\binom{i+k}{i}x^i = \frac{1}{{(1-x)}^{k+1}}$$ which is the negative binomial expansion of $(1-x)^{-(k+1)}$. This is what Jaynes uses in the rest of Chapter 6. He doesn't need $t>0$.

I haven't seen where Jaynes uses $t > 0$, but Blitzstein and Hwang (p. 173) use $t > 0$ but $k =0$ when finding the moments of a random variable $Y$ that has the geometric probability distribution with 'failure' probability $x$, $0 \le x \le 1$: $P(Y=i) = (1-x)x^i$. $$E(Y) = (1-x)\sum_{i=1}^{\infty}ix^i$$ Ignoring the $(1-x)$ out front, they use $$\sum_{i=1}^\infty ix^{i} = \left(x\frac{d}{dx}\right)\frac{1}{1-x} = \frac{x}{(1-x)^2}$$ and $E(Y) = x/(1-x)$. $$E(Y^2) = (1-x)\sum_{i=0}^{\infty}i^2x^i$$ Again, ignoring the $(1-x)$ out front, they use $$\sum_{i=1}^{\infty}i^2x^i = \left(x\frac{d}{dx}\right)^2\frac{1}{1-x} = \frac{x(1+x)}{(1-x)^3}$$ and $E(Y^2) = \frac{x(1+x)}{(1-x)^2}$.

It appears that this process of differentiating and then replenishing the supply of $x$'s is standard to calculate sums of the form $$\sum_{i=1}^{\infty}i^tx^i = \left(x\frac{d}{dx}\right)^t\frac{1}{1-x}$$ One could continue for $t = 3, 4, ...$, but still $k=0$. $$\sum_{i=1}^{\infty}i^3x^i = \left(x\frac{d}{dx}\right)^3\frac{1}{1-x} = \frac{x(x^2 + 4x +1)}{(1-x)^4}$$ $$\sum_{i=1}^{\infty}i^4x^i = \left(x\frac{d}{dx}\right)^4\frac{1}{1-x} = \frac{x(x^3 + 11x^2 + 11x +1)}{(1-x)^5}$$

Jaynes's general summation formula 6.108 combines the above with the negative binomial expansion of $(1-x)^{-(k+1)}$ to get $$\sum_{i=0}^{\infty}\binom{i+k}{i}x^ii^t = (x\frac{d}{dx})^t \frac{1}{{(1-x)}^{k+1}}$$

For $k\ge 0$ and $t=1$, the closed form expression is

$$\sum_{i=1}^\infty \binom{i+k}{i} ix^{i} = \left(x\frac{d}{dx}\right)\frac{1}{(1-x)^{k+1}} = \frac{x(k+1)}{(1-x)^{k+2}}$$

For $k\ge 0$ and $t=2$, the closed form expression is

$$\sum_{i=1}^\infty \binom{i+k}{i} i^2x^{i} = \left(x\frac{d}{dx}\right)\frac{x(k+1)}{(1-x)^{k+2}} = \frac{x(k+1)(kx+x+1)}{(1-x)^{k+3}}$$

For $t=3$, it's

$$\sum_{i=1}^\infty \binom{i+k}{i} i^3x^{i} = \left(x\frac{d}{dx}\right)\frac{x(k+1)(kx+x+1)}{(1-x)^{k+3}} = \frac{x(k+1)\left((k+1)^2x^2+(3k+4)x+1)\right)}{(1-x)^{k+4}}$$

For $t=4$, it's $$\sum_{i=1}^\infty \binom{i+k}{i} i^4x^{i} = \frac{x(k+1) \left((k+1)^3x^3+(6k^2+16k +11)x^2 + (7k+11)x + 1)\right)}{(1-x)^{k+5}}$$

Here, I have provided closed form expressions for all integers $k \ge 0$ and t = 0, 1, 2, 3, or 4.