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Math Help - Convergence of Dirichlet Series

  1. #1
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    Convergence of Dirichlet Series

    Show that L(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)^s} converges for Re(s) > 0.
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  2. #2
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    Use the alternating series test.

    \sum_{n=0}^{\infty} (-1)^n a_n converges if \lim_{n \to \infty} a_n \to 0 and all the a_n are non-negative.

    For your example, a_n = \frac{1}{(2n-1)^s}.

    Can you finish..? (note that your example starts with n=1 not n=0 as in the definition but it won't make a difference).
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  3. #3
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    What do you mean by non-negative, I mean the a_n are complex numbers?
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  4. #4
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    Quote Originally Posted by EinStone View Post
    Show that L(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)^s} converges for Re(s) > 0.
    Group the terms by pairs : \sum_{k=1}^\infty \left(\frac{1}{(2(2k-1)+1)^s}-\frac{1}{(2(2k)+1)^s}\right), and prove that this series converges absolutely.
    (you'll need an asymptotic expansion as n\to\infty to see that : \frac{1}{x^s}-\frac{1}{(x+2)^s}=\frac{1}{x^s}\left(1-\left(1+\frac{2}{x}\right)^{-s}\right)\sim \frac{2s}{x^{s+1}} as x\to\infty

    , and |a^s|=a^{{\rm Re}(s)} in the end)
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  5. #5
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    But the series converges absolutely only for Re(s) > 1 (this is given), but I need to show convergence for Re(s) > 0.
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  6. #6
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    Quote Originally Posted by EinStone View Post
    But the series converges absolutely only for Re(s) > 1 (this is given), but I need to show convergence for Re(s) > 0.
    Read my post more carefully, it does answer the question (notice the \frac{1}{x^{s+1}}...). The initial series doesn't converge absolutely, but the one I wrote does (and its convergence is equivalent to that of the initial series, I let you find why).
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  7. #7
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    Sorry, but what you are doing is super confusing, I dont get anything -_-.
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  8. #8
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    Quote Originally Posted by EinStone View Post
    Sorry, but what you are doing is super confusing, I dont get anything -_-.
    Fine. Your series is \frac{1}{1^s}-\frac{1}{3^s}+\frac{1}{5^s}-\frac{1}{7^s}+\cdots. I say: this series converges if and only if the series \left(\frac{1}{1^s}-\frac{1}{3^s}\right)+\left(\frac{1}{5^s}-\frac{1}{7^s}\right)+\cdots (the terms are inside parentheses) converges.

    Indeed, if S_n=\sum_{k=1}^n \frac{(-1)^{k+1}}{(2k-1)^s} denotes the sequence of partial sums of the initial series, the second series has partial sums T_n=S_{2n}. If S_n\to\ell (i.e. if the first series converges) then T_n as well (as a subsequence). And if T_n\to\ell, then S_{2n}\to\ell and S_{2n+1}=S_{2n}+\frac{1}{(2(2n+1)-1)^s}\to \ell as well, hence classically S_n\to\ell.

    Thus, it suffices to prove the convergence of the series where the terms are grouped two by two. And, as a matter of fact, this new series converges absolutely, as we will prove.

    A short computation shows \frac{1}{x^s}-\frac{1}{(x+2)^s}=\frac{1}{x^s}\left(1-\left(1+\frac{2}{x}\right)^{-s}\right)\sim\frac{2s}{x^{s+1}} as x\to\infty, hence \left|\frac{1}{x^s}-\frac{1}{(x+2)^s}\right|\sim\frac{2|s|}{x^{{\rm Re}(s)+1}}. Substituting x=2(2n+1)-1 shows that the absolute value of the general term of the new series is asymptotically equivalent to \frac{2|s|}{(4n)^{{\rm Re}(s)+1}}. Since {\rm Re}(s)+1>1, this series converges (usual Riemann series). Thus the new series converges absolutely, hence it converges, and the inital series as well.
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  9. #9
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    Ah now I got the connection, nice thanks!!
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