Prove series identity (Alternating reciprocal factorial sum)

In summary, the alternating series identity with ascending and descending reciprocal factorials can be proven using induction and the telescoping property of binomial coefficients. This identity has connections to the Gamma and Poisson distributions and can be useful in probability calculations.
  • #1
uart
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This alternating series indentity with ascending and descending reciprocal factorials has me stumped.
[tex] \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)} [/tex]
Or more compactly,
[tex] \sum_{r=0}^{n} ( \frac{(-1)^r}{(k+r)! \, (n-r)!} ) = \frac{1}{(k-1)! \, n! \, (k+n)} [/tex]

BTW. (Assume n,k integers, with [itex]n \ge 0[/itex] and [itex]k \ge 1[/itex].)

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 [itex]1 - poisson\_cdf(k-1,\lambda) = gamma\_cdf(x,k,\lambda) [/itex] 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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  • #2
Identities of that kind often yield to induction. In this case you'd need two inductions, one on k and one on n.
I'd take the base case of k=1, n=0 and observe the identity's truth for that - both sides equalling 1.
Then I'd try proving that if the identity holds for k=k', n=n' then it holds for k=k',n=n'+1 and also for k=k'+1,n=n'. The result would follow by induction.

Sometimes the nature of the induction step proof requires us to have already proven more than one base case, eg not only k=1,n=0 but also k=1,n=1 and k=1,n=2. But if such a requirement exists it will become apparent when trying to prove one or other of the induction steps. In that case, you'd need to prove the additional base cases to complete the proof.
 
  • #3
Equation to be proved is transformed to
[tex](-)^k\sum_{r=0}^n \ _{n+k}C_{r+k} (-)^{r+k}=\ _{n+k-1}C_n[/tex]
Terms under summation are interpreted as coefficients of ##x^{n-r}## in expansion of ##(x-1)^{n+k}##. I hope such a transformation could be helpful.
 
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  • #4
Thanks all. Yes I did try induction first up, but I didn't get anywhere with it.

That's a great approach anuttarasammyak, looks very promising.

And using the combinations property that,
[tex] _nC_r = \, _{n-1}C_r + \, _{n-1}C_{r-1}[/tex]
with the end condition that [itex] _nC_0 = \, _{n-1}C_0[/itex], I'm pretty sure I can make that whole LHS "telescope".
 
  • #5
Just sketching out a proof based on anuttarasammyak's insights.
[tex] \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)} [/tex]
Multiplying both sides by [itex](k+n)![/itex] puts the proposed identity into the following form.
[tex] _{n+k}C_n -\, _{n+k}C_{n-1} +\, _{n+k}C_{n-2} - \cdots (-1)^n \,\, _{n+k}C_{n-n}= \, _{n+k-1}C_n [/tex]
Using the relationship that, [itex] \, _nC_k = \, _{n-1}C_k + \, _{n-1}C_{k-1}[/itex] (for n,k >=1), the LSH can be written as,
[tex] [ _{n+k-1}C_n + \, _{n+k-1}C_{n-1}] - [ _{n+k-1}C_{n-1} + \, _{n+k-1}C_{n-2}] + \cdots (-1)^{n-1} [ _{n+k-1}C_1 + \, _{n+k-1}C_0] + (-1)^n [ _{n+k}C_0 ] [/tex]
Here the second binomial coefficient in each square bracketed term cancels with the first binomial coeff in the following bracketed term, so the series telescopes leaving just the very first coeff of, [itex]\, _{n+k-1}C_n[/itex], hence LHS = RHS.
 
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1. What is a series identity?

A series identity is a mathematical equation that relates the terms of a series to each other. It is used to prove that a series is convergent or to find the sum of a series.

2. How do you prove a series identity?

To prove a series identity, you must manipulate the terms of the series using algebraic or mathematical operations to show that the left side of the equation is equal to the right side. This can involve using known identities or properties of series, such as the associative or distributive properties.

3. What is an alternating series?

An alternating series is a series in which the signs of the terms alternate between positive and negative. For example, 1 - 2 + 3 - 4 + 5 - 6 + ... is an alternating series.

4. What is a reciprocal factorial sum?

A reciprocal factorial sum is a series in which the terms are the reciprocals of factorials, such as 1/1! + 1/2! + 1/3! + ... This type of series is often used in mathematical proofs and can be challenging to evaluate.

5. How is the alternating reciprocal factorial sum identity proved?

The alternating reciprocal factorial sum identity, also known as the alternating harmonic series, can be proved using the alternating series test. This test states that if a series is alternating and the absolute value of the terms decreases as n increases, then the series is convergent. By applying this test to the alternating reciprocal factorial sum, we can prove its identity.

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