## SQUARE PYRAMIDAL NUMBERS RECIPROCALS SUM

11 04 2009

This post continues with the work on some special, sets of integers, related series:

The square pyramidal numbers expression can be found on [1] and it is:

$\displaystyle P(n)=\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$

The Square pyramidal numbers reciprocal finite serie is:

$\displaystyle S(n)=\sum_{k=1}^n{\frac{6}{k(k+1)(2k+1)}}$

Splitting the main fraction into others:

$\displaystyle \frac{S(n)}{6}=\sum_{k=1}^n{\frac{1}{k(k+1)(2k+1)}}=\sum_{k=1}^n{\frac{A}{k}+\frac{B}{k+1}+\frac{C}{2k+1}}$

Solving the equations gives:

$\displaystyle A=1 ;\; B=1;\; C=-4;$

Substituing the series (ii), and the expression for the Harmonic Numbers (i):

$\displaystyle \frac{S(n)}{6}= \sum_{k=1}^n {\frac{1}{k}}+\sum_{k=1}^n{\frac{1}{k+1}}-4\sum_{k=1}^n{\frac{1}{2k+1}}=2H_n-\frac{n}{n+1}-4\sum_{k=1}^n{\frac{1}{2k+1}}$

Taking into account the expressions: (i),(iv) and (vi), for the digamma function:

$\displaystyle \frac{S(n)}{6}= 2\gamma+2\psi_{0}(n+1)-\frac{n}{n+1}-4\sum_{k=1}^n{\frac{1}{2k+1}}$

$\displaystyle S(n)=6[ 2\gamma+2\psi_{0}(n+1) -\frac{n}{n+1}-2[\psi_{0}(n+\frac{3}{2})+\gamma+2\log{2}-2]]$

$\displaystyle S(n)=6[2\psi_{0}(n+1)-2\psi_{0}(n+\frac{3}{2})-\frac{n}{n+1} -4\log{2}+4]$

Calculating the limit, and using (vii):

$\displaystyle S(\infty)= \lim_{x \to{+}\infty}{S(n)}= 6(3-4\log{2})=18-24\log{2}\approx 1.3644676665...$

(see reference [4])

NOTES:

(i) Harmonic Numbers and Digamma Function at integer values:

$\displaystyle H_{n}=\sum_{k=1}^n {\frac{1}{k}}=\gamma+\psi_{0}(n+1)$

(ii) Changing series into Harmonic Numbers:

$\displaystyle \sum_{k=1}^n{\frac{1}{k+1}} =\sum_{k=1}^n{\frac{1}{k}}-1+\frac{1}{n+1}=H_{n}-\frac{n}{n+1}$

(iii) Changing one series into another:

$\displaystyle \sum_{k=1}^n{\frac{1}{2k-1}} -\sum_{k=1}^n{\frac{1}{2k+1}}=1-\frac{1}{2n+1}$

(iv) Digamma function at half-integer values:

$\displaystyle \psi_{0}(n+\frac{1}{2})= -\gamma-2\log{2}+2\sum_{k=1}^n{\frac{1}{2k-1}}$

(v) A Digamma function property:

$\displaystyle \psi_{0}(z+1)= \psi_{0}(z)+\frac{1}{z}$

(vi) Another expression for Digamma function at half-integer values:

Using (iii), (iv) and (v):

$\displaystyle \psi_{0}(n+\frac{3}{2}) = \psi_{0}(n+\frac{1}{2}) +\frac{1}{n+\frac{1}{2}}=\psi_{0}(n+\frac{1}{2})+\frac{2}{2n+1}= \\ = -\gamma-2\log{2} +2(1-\frac{1}{2n+1}+\sum_{k=1}^n{\frac{1}{2k+1}} )= \\= -\gamma-2\log{2}+2+2\sum_{k=1}^n{\frac{1}{2k+1}}$

(vii) Limit for Digamma function at half-integer values:

$\displaystyle L_{1}(\infty)=\lim_{x \to{+}\infty}{L_{1}(n)}= \lim_{x \to{+}\infty}{[\psi_{0}(n+1)-\psi_{0}(n+\frac{3}{2})] }$

$\displaystyle L_{1}(n)=\psi_{0}(n+1)-\psi_{0}(n+\frac{3}{2})=2\log{2} + \sum_{k=1}^n {\frac{1}{k}}-2 \sum_{k=1}^n {\frac{1}{2k-1}}$

$\displaystyle L_{1}(n)=2\log{2}-\sum_{k=1}^n {\frac{1}{k(2k-1)}}$

The last serie limit can be derived from the Mercator-Mengoli infinite series for $\displaystyle \log{2}\;$. See [3].
This proof is interesting enough for another entry on the blog.

$\displaystyle \log{2}=\sum_{k=1}^\infty {\frac{1}{2k(2k-1)}}$

So:

$\displaystyle L_{1}(\infty)=0$

References:

[1]-Square Pyramidal Numbers @ Wikipedia Square Pyramidal Numbers
[2]-Weisstein, Eric W. “Digamma Function.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/DigammaFunction.html
[3]-Collection of formulae for log 2-Numbers Constants and Computation-Xavier Gourdon and Pascal Sebah.
[4]-A159354-Decimal expansion of 18-24*log(2) The On-Line Encyclopedia of Integer Sequences!
[5]-A000330-Square pyramidal numbers The On-Line Encyclopedia of Integer Sequences!

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