19 03 2009

The final result, in the preceeding post, can not be derived from a telescoping series [3], if \displaystyle k is not integer (See comments at reference [1]).

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

This lack of generality, can be avoided, if we consider a more general definition for the Harmonic Numbers [4], extended to the complex plane, using the function:

\displaystyle H_z=\gamma+\psi_0(z+1)

Where \displaystyle \psi_0 \; is the so called digamma function, and \displaystyle \;\gamma\; is the Euler-Mascheroni constant.

If you take a look at the expresion (15), in the reference [2] : We can find that one asymptotic expansion for the digamma function is:

\displaystyle \psi_0(k+1)=-\gamma+\sum_{n=1}^\infty{\frac{k}{n(n+k)}}

This is why the Polygonal Numbers Series sum is working:

\displaystyle H_k=\gamma-\gamma+ \sum_{n=1}^\infty{\frac{k}{n(n+k)}}

\displaystyle \frac{H_k}{k}=\sum_{n=1}^\infty{\frac{1}{n(n+k)}}=\frac{\gamma+\psi_0(k+1)}{k}

And the polygonal numbers infinite sum, can be expressed (if \displaystyle \;s\neq4\;) as:

\displaystyle S_{\infty}(s)=\frac{2}{4-s}*(\gamma+\psi_{0}\left(\frac{2}{s-2}\right))

This expresion works for all \displaystyle s>2, as well as for all nonreal \displaystyle s, It also works for all \displaystyle s<2, except if \displaystyle s<2, and \displaystyle s is \displaystyle \;\;0, 1, 4/3, 6/4, 8/5, 10/6, ... \;, because \displaystyle \;\psi_0\; is not defined for negative integers (See reference) [1]


[1]-Charles R Greathouse IV – Comments @ My Math Forum Inverse Polygonal Series
[2]-Weisstein, Eric W. “Digamma Function.” From MathWorld–A Wolfram Web Resource.
[3]-Telescoping Series @ Wikipedia Telescoping Series
[4]-Sondow, Jonathan and Weisstein, Eric W. “Harmonic Number.” From MathWorld–A Wolfram Web Resource.
[5]-Photo Martin Gardner, Mathematical Games, Scientific American, 211(5):126-133, taken from




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