User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/existence

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/integrand

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/evaluation

User:GFauxPas/Sandbox/Zeta2/FourierSeries/

User:GFauxPas/Sandbox/Zeta2/Informal Proof

Theorem

 * $m^{\overline n} \equiv 0 \bmod n!$.

Proof
If $m = 1$, then $1^{\overline n} = n! \equiv 0 \bmod n!$.

Else WLOG consider $\left({m+1}\right)^{\overline n}$.

Writing:


 * $\left({m+1}\right)^{\overline n} = \left({m+1}\right)\left({m+2}\right)\ldots\left({m+n-1}\right)\left({m+n}\right)$

we see any prime occuring in the prime factorization of $\left({m+1}\right)^{\overline n}$ is given by the number of times it occurs in $\left({m+n}\right)!$, less those occuring in the prime factorization of $m!$.

Thus the power of each prime in $\left({m+1}\right)^{\overline n}$ is given by: