Combinatorics

Similar to the set up for the triangular numbers, "pyramid number," or \(P_n\), refers to the required number of dots for a \(n\) triangular base pyramid that has \(n\) dots per side on its base. And here is a fun fact for \(P_n\): $$\sum_{n=1}^{\infty}{\frac{1}{P_n}} = \frac{3}{2}$$ To prove this equality, we first may need some kind of formula for \(P_n\) so that we can do something with the LHS. As you may have found in homework, "Pyramid numbers" are a series of numbers right next to "triangular numbers" in Pascal's triangle. Therefore we know a general formula for \(P_n\): $$P_n = {n+2\choose 3} = \frac{(n+2)\cdot (n+1)\cdot n}{3!}$$ Now, we can rewrite the LHS and try to split the sum expression: $$\begin{eqnarray} \sum_{n=1}^{\infty}{\frac{1}{P_n}} &=& \sum_{n=1}^{\infty}{\frac{1}{(\frac{(n+2)\cdot (n+1)\cdot n}{3!})}}\\ &=&3!\sum_{n=1}^{\infty}{\frac{1}{(n+2)\cdot (n+1)\cdot n}}\\ &=&3!(\frac{1}{1\cdot 2 \cd