How to find the sum? - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-06-19T12:26:58Z http://mathoverflow.net/feeds/question/80942 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80942/how-to-find-the-sum How to find the sum? GarouDan 2011-11-15T01:27:41Z 2011-11-15T01:27:41Z <p><strong>The problem:</strong></p> <p>How to find this sum?</p> <blockquote> <p>$$\sum_{a=0}^{\infty}\frac{1}{(\frac{(a(p-1)+b)!}{p^{q_a}} \mod p) \times p^q}$$</p> </blockquote> <p>where:</p> <p>$p \in Primes$</p> <p>$b \in \mathbb{N}\quad$, $0 \leq b \leq p-2$, but not defined.</p> <p>$q_a$ is the greatest power of $p$ who divides the term $(a(p-1)+b)!$</p> <p><strong>Details:</strong></p> <p>$b$ refers to the congruence modulus $p-1$, so $0 \leq b \leq p-2$.</p> <p>$a(p-1)+b$ for differents $a$'s and $b$'s we can express all natural numbers,</p> <p>so the summation looks like the sequence of $e$ but a bit modified.</p> <p>$q_a$ can also be expressed as: $q_a=\lfloor{\frac{(a(p-1)+b)}{p}}\rfloor+\lfloor{\frac{(a(p-1)+b)}{p^2}}\rfloor+\lfloor{\frac{(a(p-1)+b)}{p^3}}\rfloor+\ldots$</p> <p>$\frac{(a(p-1)+b)!}{p^{q_a}} \bmod p$ is just the remainder of the division by $p$.</p> <p>Expandind the sum we have:</p> <p>$\frac{1}{(\frac{b!}{p^{q_0}} \mod p)p^{q_0}}+\frac{1}{(\frac{(p-1+b)!}{p^{q_1}} \mod p)p^{q_1}}+\frac{1}{(\frac{(2p-2+b)!}{p^{q_2}} \mod p)p^{q_2}}+\frac{1}{(\frac{(3p-3+b)!}{p^{q_3}} \mod p)p^{q_3}}+\ldots$</p> <p>just to clarify, to $p=2$ and $b=0$ we have:</p> <p>$\frac{1}{(0! \mod 2)2^0}+\frac{1}{(1! \mod 2)2^0}+\frac{1}{(\frac{2!}{2^1} \mod 2)2^1}+\frac{1}{(\frac{3!}{2^1} \mod 2)2^1}++\frac{1}{(\frac{4!}{2^3} \mod 2)2^3}+\frac{1}{(\frac{5!}{2^3} \mod 2)2^3}+\frac{1}{(\frac{6!}{2^4} \mod 2)2^4}\ldots$</p> <p>The motivation to find this sum is analyze some properties of the prime numbers using the expansion of $e$.</p>