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(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\xi (s)=\sum_{n=1}^{\infty}\frac{a_n}{n^s}[/tex]

where [tex]a_n,s\in\mathbb{C}[/tex] and [tex]a_n[/tex] is multiplicative, hence

[tex]\forall n,m\in\mathbb{N}, \, a_{nm}=a_{n}a_{m}[/tex]

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

[tex]\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}^{\infty}(-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}[/tex]

so that

[tex]\xi (s)=(1-a_22^{1-s})^{-1}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{a_n}{n^s}[/tex]

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

[tex]\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}[/tex]

to get the the second stage, namely

[tex]\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}}[/tex]

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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# Senior thesis

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