## How to find the sum? [closed]

The problem:

How to find this sum?

$$\sum_{a=0}^{\infty}\frac{1}{(\frac{(a(p-1)+b)!}{p^{q_a}} \mod p) \times p^q}$$

where:

$p \in Primes$

$b \in \mathbb{N}\quad$, $0 \leq b \leq p-2$, but not defined.

$q_a$ is the greatest power of $p$ who divides the term $(a(p-1)+b)!$

Details:

$b$ refers to the congruence modulus $p-1$, so $0 \leq b \leq p-2$.

$a(p-1)+b$ for differents $a$'s and $b$'s we can express all natural numbers,

so the summation looks like the sequence of $e$ but a bit modified.

$q_a$ can also be expressed as: $q_a=\lfloor{\frac{(a(p-1)+b)}{p}}\rfloor+\lfloor{\frac{(a(p-1)+b)}{p^2}}\rfloor+\lfloor{\frac{(a(p-1)+b)}{p^3}}\rfloor+\ldots$

$\frac{(a(p-1)+b)!}{p^{q_a}} \bmod p$ is just the remainder of the division by $p$.

Expandind the sum we have:

$\frac{1}{(\frac{b!}{p^{q_0}} \mod p)p^{q_0}}+\frac{1}{(\frac{(p-1+b)!}{p^{q_1}} \mod p)p^{q_1}}+\frac{1}{(\frac{(2p-2+b)!}{p^{q_2}} \mod p)p^{q_2}}+\frac{1}{(\frac{(3p-3+b)!}{p^{q_3}} \mod p)p^{q_3}}+\ldots$

just to clarify, to $p=2$ and $b=0$ we have:

$\frac{1}{(0! \mod 2)2^0}+\frac{1}{(1! \mod 2)2^0}+\frac{1}{(\frac{2!}{2^1} \mod 2)2^1}+\frac{1}{(\frac{3!}{2^1} \mod 2)2^1}++\frac{1}{(\frac{4!}{2^3} \mod 2)2^3}+\frac{1}{(\frac{5!}{2^3} \mod 2)2^3}+\frac{1}{(\frac{6!}{2^4} \mod 2)2^4}\ldots$

The motivation to find this sum is analyze some properties of the prime numbers using the expansion of $e$.

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This was posted to math.stackexchange.com/questions/82204/… a few minutes before you posted it here. – Yemon Choi Nov 15 2011 at 1:41
To explicit, cross-posting is bad form, because answers are not linked to questions at the other site, and this is unhelpful to other users. Also, some of the detail you provided is not of research-level (e.g. people here don't need telling what 'mod p' means), and 'analyze some properties of the prime numbers using the expansion of e.' is not exactly detailed motivation. People study analytic number theory/additive combinatorics/arakelov geometry/etc/ for this reason. – David Roberts Nov 15 2011 at 1:57
Although I think this question could be a lot better, would the closers please give a reason? I've started a meta thread meta.mathoverflow.net/discussion/1207/… – David Roberts Nov 15 2011 at 2:07
Do you have a reason to believe this sum is meaningful? When we ask for motivation, we mean that you should describe how such a sum would help you in your study of properties of primes. At any rate, anyone here who knows an answer to your question is welcome to give a response at math.stackexchange.com, because there is a corresponding question there. – S. Carnahan Nov 15 2011 at 2:24
I think close isn't the better way to solve this. The question isn't "too localizable" (even in Math.SE) and have mathematician interest as in FAQ. About the motivation, the truth is I didn't have a theory yet and I'm asking some help (solving this question) to go ahead. Otherwise, where in FAQ says put questions in two different sites aren't allowed? – GarouDan Nov 15 2011 at 19:19