Generalizing the sum for the binomial expansion of a number with negative exponent

Question

It is known that

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

but what about sums of the form $$ \sum_{n=t}^{\infty} \binom{z+n-1}{n}(-1)^{n}x^{n}? $$

Is there an explicit formula for them? Notice that shifting the iterator by applying $n=n+t$ will not help because then one will have $n+t$ as the second binomial argument which does not let one apply the sum formula from above. Furthermore, notice that $$ \binom{z+n-1+t}{n+t} = \binom{z+n-1}{n}\frac{z+n-1+1}{n+1}\frac{z+n-1+2}{n+2}...\frac{z+n-1+t}{n+t}. $$ This might be useful but I do not see how to apply it if it is so.

Answer 1

This sum is hypergeometric, so there's actually a computer program which will either find you a closed form in terms of (generalized) hypergeometric functions or will tell you no such closed form exists. See the book A=B (which is freely available at that link) for a description of this and similar algorithms.

The computer algebra system sage has this algorithm built in, so we can ask if a closed form exists. The answer is "yes", but it's nowhere near as nice as you might like. I suspect there's nothing better, though.

sage: sum(binom(z+n-1, n) * (-1)^n * x^n, n, t, oo)
(-1)^t*x^t*binomial(t + z - 1, t)*hypergeometric((1, t + z), (t + 1,), -x)

In latex, we see your sum is

$$(-x)^{t} {z + t - 1 \choose t} \,_2F_1\left(\begin{matrix} 1,t + z \\ t + 1 \end{matrix} ; -x \right)$$

where ${}_2 F_1$ is probably the best studied hypergeometric function (indeed, it's the original hypergometric function that the others are generalizing, and has been studied since at least Gauss). See here and here for starting points regarding the known identities, asymptotics, etc. for ${}_2 F_1$.

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I hope this helps ^_^

Answer 2

There is unfortunately no simple closed form that can be expressed purely with binomials. However, a "closed form" exists using the hypergeometric function.

Recall that the hypergeometric function ${}_2F_1\left(\alpha,\beta;\gamma; x\right)$ is defined as

$${}_2F_1\left(\alpha,\beta;\gamma;x\right)=\sum\limits_{k=0}^{+\infty}\frac {(\alpha)_k (\beta)_k}{(\gamma)_k}\frac {x^k}{k!}$$

Where $(\alpha)_k$ is the pochhammer symbol and can be expressed in terms of the gamma function as

$$(\alpha)_k=\frac {\Gamma(\alpha+k)}{\Gamma(k)}$$

To express our sum in terms of ${}_2F_1$, it's necessary to first apply a shift on the index. I've rewritten the sum with $k$ as the indexing variable, i.e.,

$$S\equiv\sum\limits_{k=n}^{+\infty}\binom {z+k-1}k (-x)^k$$

Replace $k$ with $k+n$ such that the new index starts from $k=0$. Next, express the binomial coefficient in terms of the gamma function since

$$\binom {n}{k}=\frac {\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}$$

The sum becomes

$$\begin{align*}S & =(-x)^n\frac {\Gamma(z+n)}{\Gamma(n+1)}\sum\limits_{k=0}^{+\infty}\color{red}{\frac {\Gamma(z+n+k)}{\Gamma(z+n)}}\color{blue}{\frac {\Gamma(n+1)}{\Gamma(n+k+1)}}\frac {\color{brown}{\Gamma(k+1)}}{\Gamma(z)}\frac {(-x)^k}{k!}\end{align*}$$

The colored components can all be expressed in terms of the pochhammer notation giving

$$\begin{align*}\sum\limits_{k=n}^{+\infty}\binom {z+k-1}k(-x)^k & =(-x)^n\binom {z+n-1}n\sum\limits_{k=0}^{+\infty}\frac {\color{red}{(z+n)_k}\color{brown}{(1)_k}}{\color{blue}{(n+1)_k}}\frac {(-x)^k}{k!}\\ & =(-x)^n\binom {z+n-1}n {}_2F_1(z+n,1;n+1;-x)\end{align*}$$