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

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).

• I suspect you have it in simplest form. Also not sure what the downvote is for.
– anon
Aug 14, 2020 at 3:48

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.

• As you say, I can (and did) check up to $t=3$, but you and Mathematica lose me at $_4F_3(2,2,2,k+2;1,1,1;x)$. I gather it's the generalized hypergeometric function and involves an infinte sum using the Pochhammer symbol for rising factorials. Aug 25 at 2:54

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.

• Is $(x\frac d{dx})^t=x^t\frac{d^t}{dx^t}$ or what exactly? Aug 25 at 1:51
• $(x\frac{d}{dx})$ means take the derivative then multiply by $x$. $(x\frac{d}{dx})^t$ repeat this process $t$ times: derivative, multiply, derivative, multiply, ... Aug 25 at 2:31