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.
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4You 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
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1You have one in the case $i = 0$. – ncmathsadist Aug 30 '18 at 11:02
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2@ncmathsadist I think for $i=0$ the sum is $\frac{2^n}{n!}$ – Henry Aug 30 '18 at 11:02
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1@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
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1@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
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1WA gives the sum as $\frac1{n!},_2F_1(1,i-n;i+1;-1)$, where $_2F_1(a,b;c;x)$ is the hypergeometric function. (Link) I've got a feeling you're unlikely to find any simpler representation of the formula than that, except for specific values of $n,i$. – Jam Aug 30 '18 at 11:35
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Also, I'd note that expressing the sum in sigma notation, as a beta function or as a hypergeometric function may be redundant since it's just different ways of expressing the same thing without actually simplifying it. – Jam Aug 30 '18 at 11:43
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Sorry there's a typo in my first comment, it should be $\frac{\binom{n}{i}}{n!},_2F_1(\ldots)$. – Jam Aug 30 '18 at 11:48
1 Answers
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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.
References and Further Reading
Despite having no closed form, the sum has been studied before and can be approximated and bounded:
