Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

The Riemann Zeta Function: Extended Confusion about an Analytic Continuation

Date: 11/11/2011 at 08:33:44
From: Kesavan.
Subject: Negative even numbers of Riemann zeta function

How do you calculate the value of Zeta(-2)? How do you prove it equals
zero? and similarly for every negative value?

When I substitute -2 in place of s in Riemann Zeta function, it becomes

   1 + 4 + 9 + 16 + ...

This is clearly unbounded and tending to infinity. But several articles
mention this equals zero.

I am dying to know where I went haywire and shall not breathe my last
before cracking this conundrum.

Thanks for your help.


Date: 11/11/2011 at 18:26:48
From: Doctor Vogler
Subject: Re: Negative even numbers of Riemann zeta function

Hi Kesavan,

Thanks for writing to Dr. Math. 

The Riemann Zeta Function is defined to be the complex analytic
continuation of the infinite sum you are clearly already familiar with,

   zeta(s) = sum  n^(-s)

In fact, the series for zeta(s) only converges when the real part of s is
strictly bigger than 1.

Indeed, the most common misunderstanding of the Riemann Zeta Function is
forgetting that it is defined as the *analytic continuation* of that
infinite series. The RZF is defined even where the series does not
converge. Forgetting this, and saying that the RZF equals that sum, would
be like saying that the function ...

  f(x) = 1/(1 - x)

... is equal to the sum of the geometric series ...

  g(x) = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ...

... and that therefore it couldn't possibly be negative when x is
positive. Right? So how can you say that f(2) = -1?

Well, f and g are equal when -1 < x < 1, but g does not converge when 
|x| >= 1, whereas f is still defined there (except at x = 1).

Similarly, the Riemann Zeta Function is equal to the infinite series you
know and love when the real part of s is greater than 1. At other values
of s, you'll have to use a different formula.

So how do you extend the domain of a function defined by an infinite sum
that doesn't converge everywhere? That is where analytic continuation
comes in. The technique of expanding the domain is called "analytic
continuation" and is a subject from complex analysis. If you are
unfamiliar with this subject, I would recommend a course or book, but I'll
say a little about complex analysis here.

It turns out that "analytic" is a very strong condition for a function to
have. "Analytic" means that it has all of its derivatives (first
derivative, second derivative, third derivative, etc.) and that it equals
its Taylor Series. But you can prove that if two analytic functions are
equal to one another on any open set, or even any open interval, then they
are equal everywhere that they are both defined. That means that if you
know an analytic function on a small interval, then there is only one way
to extend it to a wider domain and still have an analytic function. And
that is what we call its "analytic continuation."

For example, consider the function I defined above:

   g(x) = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ...

This is defined for all complex numbers x whenever the absolute value (or
"modulus" or "norm") of x is less than 1. But this function has an
analytic continuation which is defined for all complex numbers apart from
x = 1. I suspect that you won't be surprised to learn that the analytic
continuation is equal to

   f(x) = 1/(1 - x).

The usual way to determine the analytic continuation of a function is to
find some formula which is equal to your infinite sum in some region where
both are defined, but which is itself defined in a larger region. For
example, the MathWorld page on the RZF has some examples of such formulas
for the RZF; look at equation (16) and below:


The "functional equation" for the RZF (which usually refers to what the
MathWorld site calls equation (13), relating zeta(1 - s) to zeta(s)) is
the usual way to prove that the RZF has zeros at the negative even
integers. You might also look at the Wikipedia page on the RZF, which
includes a section on the functional equation:


It's easy to show that the power series equal to the RZF is positive
everywhere when the real part of s is greater than 1. This means that the
RZF has no zeros or poles in that region. This fact and the functional
equation are sufficient to show that there are no poles, but only the
"trivial" zeros in the region where the real part of s is negative. It
takes more work to show what happens where the real part of s is equal to
1 or 0 (there is a pole at s = 1, but no zeros on those lines), and it is
still an unproven conjecture (The Riemann Hypothesis) that all of the
other zeros are on the line where Re(s) = 1/2.

If you still want more, you might also be interested in a few other
related answers from our archives, such as these:




If you have any questions about this or need more help, please write back
and show me what you have been able to do, and I will try to offer further

- Doctor Vogler, The Math Forum

Date: 11/12/2011 at 10:23:15
From: Kesavan.
Subject: Thank you (Negative even numbers of Riemann zeta function)

Dear Dr. Vogler,

Hats off and encomium to the brightest explanation you have elucidated to
this simpleton, as I -- in my limited knowledge of English lexicon --
could not find words enough to appreciate and acknowledge your assistance.

The moment you started with a very fine example of functions 1/(1 - x) and
its counterpart 1 + x + x^2 + x^3 + x^4 + ..., my doubts and
uncertainties vanished into thin air, bidden adieu, or should I say were
cleverly dispelled, just like the morning sun's rays clears off the mist
enveloping the leaves of the tree. Kindly accept my kudos, adulations and

The function, and its infinite series cannot be equal for all values.
Nicely argued and well presented. The infinite series is convergent only
when the real part of 's' is greater than one. Substituting -2 is itself
wrong due to the wrong assumption of both being equal. Besides, the series
is not convergent for -2. So substitution here is wrong. Now I understand.

The beauty of mathematics could be appreciated only when it is understood

Once again thanking you for your time, labor and interest,

Associated Topics:
College Analysis

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994-2013 The Math Forum