TETRAHEDRAL NUMBERS RECIPROCALS SUM

25 12 2009

TETRAHEDRAL NUMBERS SERIES:

This post follows with the exercises on special numbers reciprocals related series, after the blog entries about Square Pyramidal Numbers and Polygonal Numbers . In fact, this example it is not very much interesting, but I wanted to write it before to deal with more difficult problems.

\displaystyle T_{n}=\frac{n(n+1)(n+2)}{6}=\binom{n+2}{3}

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

If we split the main fraction into others:

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

Solving the linear system of equations it gives:

\displaystyle A=\frac{1}{2} \; ; B=-1 \; ; C=\frac{1}{2};

This three series can be summed easily with the aid of the Harmonic Numbers:

\displaystyle \sum_{k=1}^{n}{\frac{1}{k}}=1+\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n}=H_n

\displaystyle \sum_{k=1}^{n}{\frac{1}{k+1}}=\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n}+\frac{1}{n+1}=H_n-1+\frac{1}{n+1}

\displaystyle \sum_{k=1}^{n}{\frac{1}{k+2}}=\frac{1}{3}+\frac{1}{4}+ \cdots +\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}=H_n-1-\frac{1}{2} + \frac{1}{n+1}+\frac{1}{n+2}

If we sustitute everything in the expression for the reciprocals sum:

\displaystyle \frac{S(n)}{6}=\frac{n}{n+1}-\frac{1}{2}-\frac{1}{4} +\frac{1}{2(n+1)}+\frac{1}{2(n+2)}

In the previous step we can see what does exactly means to be a “telescoping series“, the term H_n, has vanished and there is no need to handle the Euler Mascheroni Gamma and the Digamma Function:

H_{n}=\gamma + \psi_{0}(n+1)

Then the formula for the n-th partial sum is:

\displaystyle S(n)=\frac{3n(3+n)}{2(1+n)(2+n)}

And taking the limit we get:

\displaystyle S(\infty)=\lim_{n \leftarrow \infty}{S(n)}=\frac{3}{2}


References:[1]-Tetrahedral Number at- Wikipedia
[2]-Weisstein, Eric W. “Tetrahedral Number.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/TetrahedralNumber.html
[3] A000292-Tetrahedral (or pyramidal) numbers: C(n+2,3) = n(n+1)(n+2)/6. The On-Line Encyclopedia of Integer Sequences!


Advertisements

Actions

Information




%d bloggers like this: