We know that the harmonic series diverges, but what about the alternating harmonic series ? The key is the Taylor series for the natural logarithm, known as the Mercator series:

, which is valid for -1<*x*≤1. Setting *x*=1 tells us that the alternating harmonic series converges to .

Now, recall the Riemann Zeta Function, which for is given by

.

Suppose we define an analogous function with alternating terms:

.

This series does not have a pole at *s*=1, and in fact, can be defined via analytic continuation to be defined over the entire complex plane. This function is called the Dirichlet eta function.

Now, let us consider the difference of the Riemann zeta and Dirichlet eta functions:

We see that the odd *n* terms cancel, leaving only the even terms:

,

and solving for eta,

which allows us to find exact values for the Dirichlet eta function at positive even integers (and find values for positive odd integers in terms of the zeta function of those integers).

The values for the first few integers are:

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Tags: Alternating Harmonic Series, Dirichlet Eta Function, Math, Mercator Series, Monday Math, Riemann Zeta Function, Taylor Series

This entry was posted on August 18, 2008 at 12:01 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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August 19, 2008 at 9:52 am |

Hi there. I just started blogging, myself and I’d wonder whether you could give me a couple of tips for starters;).

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Maybe you could take a brief look at my blog. Thanks!

Bye.

http://johnsmithinfinith.wordpress.com/

September 1, 2008 at 2:20 am |

[…] , which gives us: . Plugging into our series, . We should recognize that last sum as the Dirichlet eta function. Thus, just as , we see […]

June 1, 2009 at 12:14 am |

[…] The Maclaurin series for is given by the Mercator series: for -1<x≤1. Thus where is the Dirichlet eta function. We also found here that , so […]

August 30, 2010 at 12:20 am |

[…] Riemann zeta function has a pole at s=1, while the term in the brackets is zero at s=1. Recall from here that the Dirichlet eta function η(s), which is defined for all complex numbers s, is related to […]

January 18, 2011 at 5:29 am |

Hi TwistedOne:

Do you happen to know what

SUM([(-1)^(n-1)*zeta(2n)]/n), n=1..infinity

converges to?.

I have tried finding its sum, but to no avail.