Dirichlet series of convolution of conditionally convergent Dirichlet series is not necessarily convergent
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In Montgomery and Vaughan's Multiplicative Number Theory I. Classical Theory, they claim the statement in this question's title, using the example $$ alpha(s) = sum_{n=1}^infty (-1)^{n-1}n^{-s} $$ as the conditionally convergent series, for $0< sigma < 1$, and state that the Dirichlet series of $alpha(s)^2$ has abscissa of convergence $1/4$. However, I cannot see why the abscissa of convergence is $1/4$.
sequences-and-series convergence
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In Montgomery and Vaughan's Multiplicative Number Theory I. Classical Theory, they claim the statement in this question's title, using the example $$ alpha(s) = sum_{n=1}^infty (-1)^{n-1}n^{-s} $$ as the conditionally convergent series, for $0< sigma < 1$, and state that the Dirichlet series of $alpha(s)^2$ has abscissa of convergence $1/4$. However, I cannot see why the abscissa of convergence is $1/4$.
sequences-and-series convergence
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Do you mean $sum_{s=1}^infty{alpha^2(s)}$?
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– Mostafa Ayaz
Jan 6 at 9:59
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No, I mean the Dirichlet series $$sum_{n=1}^infty c(n)n^{-s},$$ where $$c(n)=sum_{d|n} (-1)^{d+n/d}$$ is the Dirichlet convolution of $(-1)^{n-1}$ with itself.
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– 1000teslas
Jan 7 at 4:45
add a comment |
$begingroup$
In Montgomery and Vaughan's Multiplicative Number Theory I. Classical Theory, they claim the statement in this question's title, using the example $$ alpha(s) = sum_{n=1}^infty (-1)^{n-1}n^{-s} $$ as the conditionally convergent series, for $0< sigma < 1$, and state that the Dirichlet series of $alpha(s)^2$ has abscissa of convergence $1/4$. However, I cannot see why the abscissa of convergence is $1/4$.
sequences-and-series convergence
$endgroup$
In Montgomery and Vaughan's Multiplicative Number Theory I. Classical Theory, they claim the statement in this question's title, using the example $$ alpha(s) = sum_{n=1}^infty (-1)^{n-1}n^{-s} $$ as the conditionally convergent series, for $0< sigma < 1$, and state that the Dirichlet series of $alpha(s)^2$ has abscissa of convergence $1/4$. However, I cannot see why the abscissa of convergence is $1/4$.
sequences-and-series convergence
sequences-and-series convergence
asked Jan 6 at 8:10
1000teslas1000teslas
1
1
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Do you mean $sum_{s=1}^infty{alpha^2(s)}$?
$endgroup$
– Mostafa Ayaz
Jan 6 at 9:59
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No, I mean the Dirichlet series $$sum_{n=1}^infty c(n)n^{-s},$$ where $$c(n)=sum_{d|n} (-1)^{d+n/d}$$ is the Dirichlet convolution of $(-1)^{n-1}$ with itself.
$endgroup$
– 1000teslas
Jan 7 at 4:45
add a comment |
$begingroup$
Do you mean $sum_{s=1}^infty{alpha^2(s)}$?
$endgroup$
– Mostafa Ayaz
Jan 6 at 9:59
$begingroup$
No, I mean the Dirichlet series $$sum_{n=1}^infty c(n)n^{-s},$$ where $$c(n)=sum_{d|n} (-1)^{d+n/d}$$ is the Dirichlet convolution of $(-1)^{n-1}$ with itself.
$endgroup$
– 1000teslas
Jan 7 at 4:45
$begingroup$
Do you mean $sum_{s=1}^infty{alpha^2(s)}$?
$endgroup$
– Mostafa Ayaz
Jan 6 at 9:59
$begingroup$
Do you mean $sum_{s=1}^infty{alpha^2(s)}$?
$endgroup$
– Mostafa Ayaz
Jan 6 at 9:59
$begingroup$
No, I mean the Dirichlet series $$sum_{n=1}^infty c(n)n^{-s},$$ where $$c(n)=sum_{d|n} (-1)^{d+n/d}$$ is the Dirichlet convolution of $(-1)^{n-1}$ with itself.
$endgroup$
– 1000teslas
Jan 7 at 4:45
$begingroup$
No, I mean the Dirichlet series $$sum_{n=1}^infty c(n)n^{-s},$$ where $$c(n)=sum_{d|n} (-1)^{d+n/d}$$ is the Dirichlet convolution of $(-1)^{n-1}$ with itself.
$endgroup$
– 1000teslas
Jan 7 at 4:45
add a comment |
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$begingroup$
Do you mean $sum_{s=1}^infty{alpha^2(s)}$?
$endgroup$
– Mostafa Ayaz
Jan 6 at 9:59
$begingroup$
No, I mean the Dirichlet series $$sum_{n=1}^infty c(n)n^{-s},$$ where $$c(n)=sum_{d|n} (-1)^{d+n/d}$$ is the Dirichlet convolution of $(-1)^{n-1}$ with itself.
$endgroup$
– 1000teslas
Jan 7 at 4:45