Zeta Function and Dirichlet Series

This note has been used to help create the Riemann Zeta Function wiki

This blog entry is motivated in part by a beautiful question one of you posted here. It differs from the more standard presentations of the topic of Dirichlet series in two ways. First, I will ignore the deep connections between the zeta function and the distribution of prime numbers, and instead view Dirichlet series as a kind of an analogue of the more familiar generating functions technique: there are certain kinds of calculations that Dirichlet series can handle much better. Second, even though one should study Dirichlet series by viewing them as analytic functions of a complex variable, I will forgo the complex analytic perspective, so that the discussion that follows requires only a modest background in calculus. That being said, I hope that this post will motivate you to delve deeper into number theory and learn more about the zeta function.

The Riemann zeta function is defined by ζ(s)=n=11ns. \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}. It is easy to check (for instance, using the integral test) that this series converges for any real number s>1s>1 (and diverges for s1s\leq 1). This function is the most famous example of a Dirichlet series, where a general Dirichlet series is a function of the form f(s)=n=1anns f(s) = \sum_{n=1}^\infty \frac{a_n}{n^s} where {an}\{a_n\} is a fixed series of real (or complex) numbers. Of course, in order to treat this f(s)f(s) as an actual function (rather than a formal infinite sum), one has to restrict the domain to those values of ss for which the series converges.

We look at a few examples before discussing the general theory.

Worked Examples

1) Let's calculate the sum n=1τ(n)n2,\sum_{n=1}^\infty \frac{\tau(n)}{n^2}, where τ(n)\tau(n) is the number of positive integer divisors of nn. Note that unlike this example, proving convergence here is easy: for instance, one can check without difficulty that τ(n)=O(n)\tau(n)=O(\sqrt{n}) as nn\to\infty. It is well known that n=11n2=π26\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}. Form the square of the last series and regroup the terms as follows: (n=11n2)2=d,q=11d2q2=n=1(dq=n1n2) \left( \sum_{n=1}^\infty \frac{1}{n^2} \right)^2 = \sum_{d,q=1}^\infty \frac{1}{d^2 q^2} = \sum_{n=1}^\infty \left(\sum_{dq=n}\frac{1}{n^2}\right) Again, in this case, rearranging terms in any way you like is easy to justify because we are dealing with absolutely convergent series. Note that the definition of τ(n)\tau(n) can be rewritten as τ(n)=dq=n1\tau(n)=\sum_{dq=n}1 (the sum ranges over integers d,q1d,q\geq 1 whose product is nn), so we obtain n=1τ(n)n2=ζ(2)2=π436\sum_{n=1}^\infty \frac{\tau(n)}{n^2} = \zeta(2)^2 = \frac{\pi^4}{36}.


2) Our second example is the sum n=1(1)n1n\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}. It can be easily calculated using generating functions; however, I will explain the Dirichlet series approach because it brings up several important ideas that are useful in many other contexts. This series is easily shown to converge using the alternating series test. However, it does not converge absolutely. This suggests considering the function f(s)=n=1(1)n1nsf(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}; for any s>1s>1, the convergence is absolute. A general result about Dirichlet series (see below) guarantees that because the initial series converges, we have n=1(1)n1n=lims1f(s)\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} = \lim\limits_{s\downarrow 1} f(s), where the limit is taken as ss approaches 11 from above. To get a handle on f(s)f(s), we notice that it looks very similar to the zeta function, except that the signs in the sum that defines f(s)f(s) alternate (the terms corresponding to even nn appear with a minus sign). The Euler product formula (see below) naturally suggests that we consider the product (where pp ranges over prime numbers) (112s14s18s)p odd(1+1ps+1p2s+). \left( 1 - \frac{1}{2^s} - \frac{1}{4^s} - \frac{1}{8^s} - \ldots \right) \cdot \prod_{p\ \mbox{odd}} \left( 1 + \frac{1}{p^s} + \frac{1}{p^{2s}} + \ldots \right). Formally expanding the last product, we find that it is equal to f(s)f(s), and the validity of this formal calculation for s>1s>1 follows from the fact that all infinite series and products we have here are absolutely convergent when s>1s>1. The last formula makes it easy to relate f(s)f(s) to the zeta function itself: using the identity 112s14s=2(1+12s+14s+)=2112s 1 - \frac{1}{2^s} - \frac{1}{4^s} - \ldots = 2 - \left( 1 + \frac{1}{2^s} + \frac{1}{4^s} + \ldots \right) = 2 - \frac{1}{1-2^{-s}} and Euler's formula for ζ(s)\zeta(s), we get f(s)=(121s)ζ(s)f(s) = (1 - 2^{1-s})\cdot\zeta(s). It is known that lims1(s1)ζ(s)=1\lim\limits_{s\to 1}(s-1)\cdot\zeta(s)=1, which means that n=1(1)n1n=lims1f(s)=lims1121ss1. \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} = \lim\limits_{s\downarrow 1}f(s) = \lim\limits_{s\downarrow 1} \frac{1 - 2^{1-s}}{s-1}. It is an easy exercise to calculate the last limit and obtain the answer log(2)\log(2), which of course is also the value you would have gotten from the generating functions technique.


3) Using Dirichlet series for example #2 is clearly an overkill. However, exactly the same method can be used to handle this example, for which there is no easy approach using generating functions (to the best of my knowledge). The calculation I will sketch is very similar to the one found by one of you. As a first step, we verify that the series S=n=0(1)nτ(2n+1)2n+1 S = \sum_{n=0}^\infty \frac{(-1)^n \, \tau(2n+1)}{2n+1} converges (conditionally); this of course follows from this other calculation, but a somewhat more direct proof of convergence, which does not compute the value of the sum but can be generalized to many other examples, can be found here. Thus if we define f(s)=n=0(1)nτ(2n+1)(2n+1)s f(s) = \sum_{n=0}^\infty \frac{(-1)^n \, \tau(2n+1)}{(2n+1)^s} (this series converges absolutely for all s>1s>1 by the integral test), then the sum of the series SS is equal to the limit lims1f(s)\lim\limits_{s\downarrow 1}f(s). By analogy with the first example we looked at, it is easy to check that if s>1s>1, then f(s)=L1(s)2f(s)=L_1(s)^2, where L1(s)=n=0(1)n(2n+1)s. L_1(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}. Now if we plug s=1s=1 into the last formula, we get a series that converges conditionally by the alternating series test. Hence lims1L1(s)=n=0(1)n2n+1=π4, \lim\limits_{s\downarrow 1} L_1(s) = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = \frac{\pi}{4}, where the last equality follows by considering the Taylor series for arctan(x)\arctan(x). Therefore lims1f(s)=π216\lim\limits_{s\downarrow 1}f(s)=\frac{\pi^2}{16}. (I "cheated" a bit, because I did not explain how one could have figured out that L1(s)2L_1(s)^2 is the right thing to consider. The motivation for this step again comes from the Euler product formula, and is explained here.)

The methods used in the three examples given above can be combined and generalized in various ways to handle a large variety of situations. This week's Understanding Mathematics discussion invites you to try out several related problems to test your mastery of this material.

Now I will summarize the general principles used in all such situations.

  1. Euler product formula. Whenever you deal with a Dirichlet series whose terms depend on divisibility properties of integers (such as example #3, which involves the divisor counting function), you can bet that the Euler product formula will be of use. Here is a quick derivation of this formula. By the fundamental theorem of arithmetic, every natural number nn can be written uniquely as a product p1k1prkrp_1^{k_1}\ldots p_r^{k_r}, where pip_i are pairwise distinct primes and ki1k_i\geq 1 are integers (we allow the "empty product" r=0r=0, which corresponds to n=1n=1). We then have ns=p1sk1prskrn^{-s} = p_1^{-sk_1}\ldots p_r^{-sk_r}, which leads to the expansion ζ(s)=n=11ns=p(1+1ps+1p2s+), \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \prod_p \left( 1 + \frac{1}{p^s} + \frac{1}{p^{2s}} + \ldots \right), where the product on the right ranges over all primes pp. As always, this sort of formal manipulation is valid when s>1s>1 because then we have absolute convergence for all infinite sums and products in sight. Recognizing a geometric series on the right hand side of the last equality, we obtain ζ(s)=p(1ps)1\zeta(s) = \prod_p (1-p^{-s})^{-1}, which is Euler's expansion. As we saw in example #2, it is useful to consider variations of this idea. For example, think about what will happen if you omit one particular prime from Euler's product formula. What will the corresponding Dirichlet series look like?

  2. Passing to the limit. It is worth remembering the following general result, which we already used in a few places. Consider a general Dirichlet series f(s)=n=1anns, f(s) = \sum_{n=1}^\infty \frac{a_n}{n^s}, and suppose that it converges for one particular value s=s0Rs=s_0\in\mathbb{R}. Then the series converges for every ss0s\geq s_0, and, moreover, defines a continuous function on the interval [s0,+)[s_0,+\infty). In particular, f(s0)=limss0f(s)f(s_0)=\lim\limits_{s\downarrow s_0}f(s).

  3. Multiplying Dirichlet series. The idea of relating a complicated Dirichlet series to the square of a simpler one, which we used in examples #1 and #3, can be obviously generalized to the product of two Dirichlet series: if f(s)=n=1anns   and   g(s)=n=1bnns, f(s) = \sum_{n=1}^\infty \frac{a_n}{n^s} \ \ \ \mbox{and}\ \ \ g(s) = \sum_{n=1}^\infty \frac{b_n}{n^s}, then by formally expanding and regrouping these terms, we obtain f(s)g(s)=n=1(dq=nadbq)ns, f(s) \cdot g(s) = \sum_{n=1}^\infty \left( \sum_{dq=n} a_d b_q \right) \cdot n^{-s}, where the inner sum ranges over all integers d,q1d,q\geq 1 with dq=ndq=n. Once again, this formal manipulation is always valid provided that the series defining f(s)f(s) and g(s)g(s) converge absolutely.

  4. Testing for convergence. Testing a Dirichlet series for absolute convergence is typically quite straightforward, and in most of the cases you are likely to encounter in practice, it can be done using the integral test. Here is one statement that is easy to remember and prove: if a sequence of real (or complex) numbers {an}\{a_n\} satisfies the asymptotic bound an=O(nr)|a_n|=O(n^r) as nn\to\infty, for some fixed rRr\in\mathbb{R}, then the corresponding Dirichlet series converges absolutely for all s>r+1s>r+1 (it may or may not converge for s=r+1s=r+1). Testing a Dirichlet series for conditional convergence could be very difficult. You should always first try one of the general tests for conditional convergence (such as the alternating series test), but if those do not apply, you may have to do quite a bit of work using things like summation by parts and more delicate asymptotic estimates. A fairly general pattern for using this method can be extracted from this example.

  5. Some special values. Two formulas worth remembering off the top of your head are these: lims1(s1)ζ(s)=1\lim\limits_{s\to 1}(s-1)\cdot\zeta(s)=1 and ζ(2)=π26\zeta(2)=\frac{\pi^2}{6}. Quite a bit more is known about the values of the zeta function at certain points.

Note by Calvin Lin
7 years, 2 months ago

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Loved this post!

Aditya Kumar - 5 years, 6 months ago

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I love this post!

Swapnil Das - 5 years, 6 months ago

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using this function you can prove that 1+2+3+4+.... all the way to infinite equals -1/12

Aubrey Bull - 6 years, 2 months ago

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Hmm.. but I think it is the Ramanujan Summation of the Series. As this is a divergent series, it has no well defined value.

Swapnil Das - 5 years, 6 months ago

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U r wrong!

Aditya Kumar - 5 years, 6 months ago

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