Monday Math 33

We know that the harmonic series $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$ diverges, but what about the alternating harmonic series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$ ? The key is the Taylor series for the natural logarithm, known as the Mercator series:
$\ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}x^n=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$ , which is valid for -1<x≤1. Setting x=1 tells us that the alternating harmonic series converges to $\ln2$ .

Now, recall the Riemann Zeta Function, which for $\Re(s)\gt1$ is given by
$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$ .
Suppose we define an analogous function with alternating terms:
$\eta(s)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}$ .
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:
$\begin{eqnarray}\zeta(s)-\eta(s)&=&\sum_{n=1}^{\infty}\frac{1}{n^s}-\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}\\&=&\sum_{n=1}^{\infty}\frac{(1-(-1)^{n-1})}{n^s}\end{eqnarray}$
We see that the odd n terms cancel, leaving only the even terms:
$\begin{eqnarray}\zeta(s)-\eta(s)&=&\sum_{n=1}^{\infty}\frac{(1-(-1)^{n-1})}{n^s}\\&=&\sum_{n=1}^{\infty}\frac{2}{(2n)^s}\\&=&\sum_{n=1}^{\infty}\frac{1}{2^{s-1}n^s}\\&=&2^{1-s}\sum_{n=1}^{\infty}\frac{1}{n^s}\\&=&2^{1-s}\zeta(s)\end{eqnarray}$ ,
and solving for eta,
$\eta(s)=(1-2^{1-s})\zeta(s)$
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:
$\eta(0)=\frac{1}{2}$
$\eta(1)=\ln2$
$\eta(2)=\frac{\pi^2}{12}$
$\eta(3)=\frac{3}{4}\zeta(3)$
$\eta(4)=\frac{7\pi^4}{720}$
$\eta(5)=\frac{15}{16}\zeta(5)$
$\eta(6)=\frac{31\pi^6}{30240}$
$\eta(7)=\frac{63}{64}\zeta(7)$

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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. You can leave a response, or trackback from your own site.

5 Responses to “Monday Math 33”

johnsmithinfinith Says:
August 19, 2008 at 9:52 am | Reply
Hi there. I just started blogging, myself and I’d wonder whether you could give me a couple of tips for starters;).
Why I’m asking you? Well, your blog has a lot of interesting math stuff and I liked it from the beginning. So, I’m kinda interested about blogging from your point of view, because my idea was to, also, blog about some mathematical topics.
Maybe you could take a brief look at my blog. Thanks!
Bye.

http://johnsmithinfinith.wordpress.com/
Monday Math 35 « Twisted One 151’s Weblog Says:
September 1, 2008 at 2:20 am | Reply
[...] , which gives us: . Plugging into our series, . We should recognize that last sum as the Dirichlet eta function. Thus, just as , we see [...]
Monday Math 74 « Twisted One 151’s Weblog Says:
June 1, 2009 at 12:14 am | Reply
[...] 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 [...]
Monday Math 133 « Twisted One 151's Weblog Says:
August 30, 2010 at 12:20 am | Reply
[...] 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 [...]
cody Says:
January 18, 2011 at 5:29 am | Reply
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.

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Monday Math 33

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