# Series of binomial coefficient denominators

I'm not sure how to evaluate the following :

$$\sum_{k=i}^n \frac{1}{k!(n-k)!}$$

Where $i,n \in \mathbb{N}, n > i$ are given.

I don't have any working for this, I just looked it at and don't have any idea how to go about evaluating it.

• You can at most write it as $\frac1{n!}\sum_{k=i}^n\binom{n}{k}$ but there is no closed form for it. – drhab Aug 30 '18 at 11:00
• You have one in the case $i = 0$. – ncmathsadist Aug 30 '18 at 11:02
• @ncmathsadist I think for $i=0$ the sum is $\frac{2^n}{n!}$ – Henry Aug 30 '18 at 11:02
• @drhab - there may be a form related to the regularized incomplete beta function though this is may not be helpful – Henry Aug 30 '18 at 11:06
• @Henry I see. It somehow shifts from sums to integrals (which can always be labeled with some name). – drhab Aug 30 '18 at 11:14

### No Closed Form

As pointed out by @drhab, your sum is equivalent to $$\frac{1}{n!}\sum_{k=i}^n\binom{n}{k}$$. Given that $$\sum_{k=0}^n\binom{n}{k}=2^n$$, your sum can be rearranged to $$\frac{2^n}{n!}-\frac1{n!}\sum_{k=0}^{i-1}\binom{n}{k}$$

Hence, the partial sum of binomial coefficients, $$\sum_{k=0}^{i-1}\binom{n}{k}$$, is the heart of your problem. The sum can be expressed in various other ways (such as with hypergeometric or beta functions) and, as pointed out by @Henry, it has some nice expressions for specific $$n,i$$. But unfortunately, it has no closed form in terms of the sum of a fixed number of hypergeometric terms (Petrovsek, 1996. Theorem 5.6.3, pp. 88, 94, 102; 2, p. 6). But then again, it could possibly have a different type of closed form.

Despite having no closed form, the sum has been studied before and can be approximated and bounded:

1. M. Petkovˇsek, G. S. Wilf and D. Zeilberger, 'A=B' (1996) (purchase eBook) (PDF)

2. M. Boardman, 'The Egg-Drop Numbers' (2004) (link)