## Tuesday, 10 March 2009

### INVERSE POLYGONAL NUMBERS SERIES

Let there be $latex \displaystyle P(n,s)$ the n-th Polygonal Number [4] of s sides, then:

$latex \displaystyle P(n,s)=\frac{ (s-2)*n^{2}-(s-4)*n }{2}$

For s=3, we get the formula for Triangular Numbers

$latex \displaystyle P(n,3)=\frac{n^{2}+n}{2}=T_n$

and with s=4 then we get the Squares:

$latex \displaystyle P(n,4)=n^{2}$

Polygonal numbers hold for the next identity:

$latex \displaystyle P(n,s+1)=P(n,s)+ P(n-1,3)=P(n,s)+ T_{n-1}$

The sum of the inverse of the first k polygonal numbers with side s, is:

$latex \displaystyle S_{k}(s)=\sum_{n=1}^k{} \frac{1}{P(n,s)}$

And its infinite series sum:
$latex \displaystyle S_{\infty}(s)=\sum_{n=1}^{\infty{}} \frac{1}{P(n,s)}$

Than this infinite series is convergent, can be proved, using some convergence test:

Raabe's convergence Test:

$latex \displaystyle \rho=\displaystyle\lim_{n \to{+}\infty}{n*\left(\frac{P(n+1,s)}{P(n,s)}-1\right) }=2>1$

A series with a lower number of sides, upper bounds, series with higher sides: So the convergence in triangular numbers, implies the convergence of the remaining polygonal numbers series:

If $latex \displaystyle s_1>s_2 \longrightarrow{} \frac{1}{P(n,s_1)}<\frac{1}{P(n,s_2)} \longrightarrow{} S_{\infty}(s_1)$

The inverse series with Triangular numbers is a telescoping series [3]:

$latex \displaystyle S_{\infty}(3)=2\cdot\sum_{n=1}^{\infty{}} \frac{1}{n\cdot(n+1)}= 2$

The squares sum is the so called Basel Problem, [1] [2], related with the Riemann Zeta Function:

$latex \displaystyle S_{\infty}(4)=\sum_{n=1}^{\infty{}} \frac{1}{n^2}=\zeta(2)=\frac{\pi^2}{6}$

If $latex \displaystyle s\neq4$, then:

The series is a more general case of a telescoping series [3], related with the Harmonic Numbers.

$latex \displaystyle S_{\infty}(s)=2\cdot \sum_{n=1}^{\infty{}} \frac{1}{(s-2)\cdot n^{2}-(s-4)*n}=\frac{2}{s-2}\cdot\sum_{n=1}^{\infty{}}\frac{1}{n\cdot(n+\frac{4-s}{s-2})}$

$latex \displaystyle \sum_{n=1}^{\infty{}} \frac{1}{n*(n+k)}=\frac{H_k}{k}$

$latex \displaystyle S_{\infty}(s)= \frac{2}{4-s}\cdot H_{\frac{4-s}{s-2}}$

Example: In the hexagonal numbers case:

$latex \displaystyle S_{\infty}(6)= -H_{-\frac{1}{2}} \approx 1.38629$

Archives:

References:

[1]-Estimating Basel Problem@ MAA Online How Euler did it, Ed.Sandifer
[2]-Basel Problem @ Wikipedia Basel Problem
[3]-Telescoping Series @ Wikipedia Telescoping Series
[4]-Polygonal Numbers @ Wikipedia Polygonal Number