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This alternating series indentity with ascending and descending reciprocal factorials has me stumped.
\frac{1}{k! \, n!} + \frac{-1}{(k+1)! \, (n-1)!} + \frac{1}{(k+2)! \, (n-2)!} \cdots \frac{(-1)^n}{(k+n)! \, (0)!} = \frac{1}{(k-1)! \, n! \, (k+n)}
Or more compactly,
\sum_{r=0}^{n} ( \frac{(-1)^r}{(k+r)! \, (n-r)!} ) = \frac{1}{(k-1)! \, n! \, (k+n)}
BTW. (Assume n,k integers, with n \ge 0 and k \ge 1.)
In this particular instance (for me), this series arises from a connection between the Gamma (Erlang) distribution and the Poisson distribution, where for certain parameters they represent the same probability scenario.
Specifically 1 - poisson\_cdf(k-1,\lambda) = gamma\_cdf(x,k,\lambda) at x=1. I know this is true, but if I try to prove it by doing a series expansion of each of those CDFs and equating coefficients, then I end up with the alternating series that has me stumped.
Has anyone seen that series before or know of any insights in how best to prove it?
\frac{1}{k! \, n!} + \frac{-1}{(k+1)! \, (n-1)!} + \frac{1}{(k+2)! \, (n-2)!} \cdots \frac{(-1)^n}{(k+n)! \, (0)!} = \frac{1}{(k-1)! \, n! \, (k+n)}
Or more compactly,
\sum_{r=0}^{n} ( \frac{(-1)^r}{(k+r)! \, (n-r)!} ) = \frac{1}{(k-1)! \, n! \, (k+n)}
BTW. (Assume n,k integers, with n \ge 0 and k \ge 1.)
In this particular instance (for me), this series arises from a connection between the Gamma (Erlang) distribution and the Poisson distribution, where for certain parameters they represent the same probability scenario.
Specifically 1 - poisson\_cdf(k-1,\lambda) = gamma\_cdf(x,k,\lambda) at x=1. I know this is true, but if I try to prove it by doing a series expansion of each of those CDFs and equating coefficients, then I end up with the alternating series that has me stumped.
Has anyone seen that series before or know of any insights in how best to prove it?
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