"Pyramid Numbers" Fun Fact!
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:
$$\begin{eqnarray}
\sum_{n=1}^{\infty}{\frac{1}{P_n}} &=& \sum_{n=1}^{\infty}{\frac{1}{(\frac{(n+2)\cdot (n+1)\cdot n}{3!})}}\\
&=&3!\sum_{n=1}^{\infty}{\frac{1}{(n+2)\cdot (n+1)\cdot n}}\\
&=&3!(\frac{1}{1\cdot 2 \cdot 3}+\frac{1}{2\cdot 3 \cdot 4}+\frac{1}{3\cdot 4 \cdot 5}+...)
\end{eqnarray}$$
Is there anyway that we can simplify this expansion of fractions? The answer is yes and that involves a trick to split a fraction into a difference between two other smaller fractions. For example, \(\frac{1}{1\cdot 2 \cdot 3}\) is the same as \(\frac{1}{2}\cdot (\frac{1}{1\cdot 2} - \frac{1}{2\cdot 3})\), and \(\frac{1}{2\cdot 3 \cdot 4}\) is the same as \(\frac{1}{2}\cdot (\frac{1}{2\cdot 3} - \frac{1}{3\cdot 4})\). And actually the proof of this property is obvious using algebra:
$$\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} = \frac{n+2-n}{n(n+1)(n+2)} = \frac{2}{n(n+1)(n+2)}$$
$$ \therefore \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \cdot (\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)})$$
Using this trick we can continue simplifying the expansion on the LHS:
$$\begin{eqnarray}
\sum_{n=1}^{\infty}{\frac{1}{P_n}}&=&3!(\frac{1}{1\cdot 2 \cdot 3}+\frac{1}{2\cdot 3 \cdot 4}+\frac{1}{3\cdot 4 \cdot 5}+...)\\
&=&\frac{3!}{2}(\frac{1}{1\cdot 2} - \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3} - \frac{1}{3\cdot 4}+\frac{1}{3\cdot 4} - \frac{1}{4\cdot 5}+...)\\
&=&\frac{3!}{2}\cdot \frac{1}{1\cdot 2}\\
&=&\frac{3 \cdot 2 \cdot 1}{2 \cdot 2} = \frac{3}{2}\\
\end{eqnarray}$$
We get LHS = RHS!
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:
$$\begin{eqnarray}
\sum_{n=1}^{\infty}{\frac{1}{P_n}} &=& \sum_{n=1}^{\infty}{\frac{1}{(\frac{(n+2)\cdot (n+1)\cdot n}{3!})}}\\
&=&3!\sum_{n=1}^{\infty}{\frac{1}{(n+2)\cdot (n+1)\cdot n}}\\
&=&3!(\frac{1}{1\cdot 2 \cdot 3}+\frac{1}{2\cdot 3 \cdot 4}+\frac{1}{3\cdot 4 \cdot 5}+...)
\end{eqnarray}$$
Is there anyway that we can simplify this expansion of fractions? The answer is yes and that involves a trick to split a fraction into a difference between two other smaller fractions. For example, \(\frac{1}{1\cdot 2 \cdot 3}\) is the same as \(\frac{1}{2}\cdot (\frac{1}{1\cdot 2} - \frac{1}{2\cdot 3})\), and \(\frac{1}{2\cdot 3 \cdot 4}\) is the same as \(\frac{1}{2}\cdot (\frac{1}{2\cdot 3} - \frac{1}{3\cdot 4})\). And actually the proof of this property is obvious using algebra:
$$\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} = \frac{n+2-n}{n(n+1)(n+2)} = \frac{2}{n(n+1)(n+2)}$$
$$ \therefore \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \cdot (\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)})$$
Using this trick we can continue simplifying the expansion on the LHS:
$$\begin{eqnarray}
\sum_{n=1}^{\infty}{\frac{1}{P_n}}&=&3!(\frac{1}{1\cdot 2 \cdot 3}+\frac{1}{2\cdot 3 \cdot 4}+\frac{1}{3\cdot 4 \cdot 5}+...)\\
&=&\frac{3!}{2}(\frac{1}{1\cdot 2} - \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3} - \frac{1}{3\cdot 4}+\frac{1}{3\cdot 4} - \frac{1}{4\cdot 5}+...)\\
&=&\frac{3!}{2}\cdot \frac{1}{1\cdot 2}\\
&=&\frac{3 \cdot 2 \cdot 1}{2 \cdot 2} = \frac{3}{2}\\
\end{eqnarray}$$
We get LHS = RHS!
Nice!!! Telescoping sums are delightful!
ReplyDeleteVery cool! This seems like it could be considered a figurative number set. Basically that is just different sets of numbers you get when generalizing triangular numbers to different types of shapes such as tetrahedral numbers. I'm sure you could use a similar proof like you used above to prove different shapes!
ReplyDelete