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## Senior thesis

I am writing my senior thesis (I am an undergrad math major at UCSB) on Dirichlet Series, which are, in the classical sense, series of the form

$$\xi (s)=\sum_{n=1}^{\infty}\frac{a_n}{n^s}$$

where $$a_n,s\in\mathbb{C}$$ and $$a_n$$ is multiplicative, hence

$$\forall n,m\in\mathbb{N}, \, a_{nm}=a_{n}a_{m}$$

I have begun this bit on analytic continuation for such series, here it goes:

$$\xi (s)+\sum_{n=1}^{\infty}(-1)^{n}\frac{a_n}{n^s}=\sum_{n=1}^{\infty}\frac{a_n}{n^s}+\sum_{n=1}^{\i nfty}(-1)^{n}\frac{a_n}{n^s}=2\sum_{n=1}^{\infty}\frac{a_{2n}}{(2n)^s}=2^{1-s}a_2\sum_{n=1}^{\infty}\frac{a_n}{n^s}$$

so that

$$\xi (s)=(1-a_22^{1-s})^{-1}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{a_n}{n^s}$$

which is the first first stage of analytic continuation. Now, to the above series apply Euler's series transformation, which, if you don't recall, is

$$\sum_{n=1}^{\infty}(-1)^{n-1}b_n=\sum_{n=0}^{\infty}\frac{1}{2^{n+1}}\sum_{m=0}^{n}(-1)^{m}\left( \begin{array}{c}n\\m\end{array}\right) b_{m+1}$$

to get the the second stage, namely

$$\boxed{\xi (s)=(1-a_22^{1-s})^{-1}\sum_{n=0}^{\infty}\frac{1}{2^{n+1}}\sum_{m=0}^{n}(-1)^{m}\left( \begin{array}{c}n\\m\end{array}\right)\frac{a_{m+1}}{(m+1)^s}}$$

when this same process of continuation is applied to the Riemann zeta it produces a series for the zeta function that converges for all s in the complex plane except s=1 (see prior thread for details.) My trouble is proving convergence in the present, more general case. Any thoughts?

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 Recognitions: Homework Help Science Advisor Not all dirichlet series have an analytic continuation. In the case of zeta, the 1-2^(1-s) is cancelling the pole at s=1. If the coefficients of $$\xi(s)$$ are real and non-negative then you have a pole at the real point on the line of convergence. I don't see anything that will gurantee this pole to be canceled in your case (could be of higher order, or at a different location, or...)- if this doesn't happen that sum cannot possibly converge.
 Recognitions: Homework Help So I should then ask, For what sequences $$a_n$$ does this favorable condition occur?

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## Senior thesis

 Quote by shmoe Not all dirichlet series have an analytic continuation. In the case of zeta, the 1-2^(1-s) is cancelling the pole at s=1. If the coefficients of $$\xi(s)$$ are real and non-negative then you have a pole at the real point on the line of convergence. I don't see anything that will gurantee this pole to be canceled in your case (could be of higher order, or at a different location, or...)- if this doesn't happen that sum cannot possibly converge.
I see the cancelation by the factor of 1-2^(1-s) at s=1, since

$$1-2^{1-s}=0 \Leftrightarrow s=1+i\frac{2k\pi}{\log 2}, k\in\mathbb{Z},$$

but we may assume principle values so that s=1 is the only point of interest. But for the factor of 1-a22^(1-s), s=1 would not be of interest unless a_2=1.

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 Quote by benorin So I should then ask, For what sequences $$a_n$$ does this favorable condition occur?
 Recognitions: Homework Help I wish to consider when exactly does Euler's series transformation provide an analytic continuation of a function defined by an alternating series. I will use a modified version of the transformation given above: For a known convergent alternating series $\sum (-1)^k b_k ,$ Euler's series transformation is given by $$\sum_{k=0}^{\infty}(-1)^{k}b_k=\sum_{k=0}^{\infty}\frac{1}{2^{k+1}}\sum_{m=0}^{k}(-1)^{m}\left( \begin{array}{c}k\\m\end{array}\right) b_{m}$$ An example: the result is trivial, yet the concept of continuation by the series transformation is rather at hand. Let z be complex. Consider the function f(z) defined by the alternating series $$f(z) = \sum_{k=0}^{\infty}(-1)^kz^k$$ which converges to $$\frac{1}{1+z}$$ on the unit disk $$|z|<1$$. Applying Euler's series transformation to f(z) we obtain $$f(z) = \sum_{k=0}^{\infty}(-1)^kz^k = \sum_{k=0}^{\infty}\frac{1}{2^{k+1}}\sum_{m=0}^{k}(-1)^{m}\left( \begin{array}{c}k\\m\end{array}\right) z^{m}$$ and since the binomial theorem gives $$\sum_{m=0}^{k}(-1)^{m}\left( \begin{array}{c}k\\m\end{array}\right) z^{m} = (1-z)^{k}$$ we may simplify this to obtain $$f(z) = \sum_{k=0}^{\infty}\frac{1}{2^{k+1}}\cdot (1-z)^{k} = \frac{1}{2}\sum_{k=0}^{\infty}\left( \frac{1-z}{2} \right) ^{k}$$ where the last series is a geometric series which converges to $$\frac{1}{1+z}$$ on the disk $$\left| \frac{1-z}{2} \right| <1 \Rightarrow |z-1|<2$$. Notice that the series thus obtained converges everywhere the given series did and on a disk twice as big! If one applies the transformation yet again, the series $$\frac{1}{4}\sum_{k=0}^{\infty}\left( \frac{3-z}{4} \right) ^{k}$$ is obtained, which is a geometric series converging to $$\frac{1}{1+z}$$ on the disk $$\left| \frac{3-z}{4} \right| <1 \Rightarrow |z-3|<4$$. I suspect that successive applications of the transformation would produce series with circles of convergence having radii that grow as powers of 2 whose left most point is z=-1. Can this process be carried out indefinely to give a series which converges in the half-plane $$\Re z >-1$$ ? But how often will it happen that an analytic continuation is obtained (necessary and sufficient conditions)? What is the maximal region of convergence thereby obtained? In "Theory and Application of Infinte Series," Knopp discusses sufficient conditions that a greater rapidity of convergence be obtained by an application of the series transformation, but I have yet to find a discussion of continuation. Any thoughts?