Abscissa of absolute convergence for a particular Dirichlet series












0












$begingroup$



For $n=p_1^{alpha_1}p_2^{alpha_2}cdotcdotcdot p_k^{alpha_k}$ we
denote $alpha(n)=alpha_1alpha_2cdotcdotcdotalpha_k$. Show that
$F(s)=sum_{ngeq 1}frac{alpha(n)}{n^s}$ is absolutely convergent
for $sigma>1$.




My attempt:



Notice that $alpha(n)leq n$ because
begin{equation*}
alpha_ileq p_i^{alpha_i},;text{ for every $p_i$ prime}\
alpha(n)=alpha_1cdotcdotcdotalpha_kleq p_1^{alpha_1}cdotcdotcdot p_k^{alpha_k}=n
end{equation*}
Thus, $$sum_{ngeq 1}frac{alpha(n)}{n^s}leq sum_{ngeq 1}frac{n}{n^s}=sum_{ngeq 1}frac{1}{n^{s-1}}=zeta(s-1)$$
that is absolutely convergent for $sigma>2$.



But I don't find the way to ensure convergence for $sigma>1$.



Thanks for any suggestion.






number-theory analytic-number-theory dirichlet-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    $F(s) = prod_p (1+sum_{k=1}^infty k p^{-sk}) $, $frac{F(s)}{zeta(s)} = prod_p( 1+sum_{k=2}^infty p^{-sk})$ which converges for $Re(s) > 1/2$.
    $endgroup$
    – reuns
    Nov 25 '18 at 21:51












  • $begingroup$
    Doesn't that first equality hold only if the series converge absolutely?
    $endgroup$
    – Kale36
    Nov 26 '18 at 1:14










  • $begingroup$
    @reuns How did you get the expresion for $frac{F(s)}{zeta(s)}$?
    $endgroup$
    – Kale36
    Nov 26 '18 at 17:40










  • $begingroup$
    $F(s) =sum_{ngeq 1}frac{alpha(n)}{n^s}= prod_p (1+sum_{k=1}^infty k p^{-sk})$ converges for $Re(s) > 1$ and $frac{F(s)}{zeta(s)}=prod_p (1+sum_{k=1}^infty k p^{-sk})(1-p^{-s})= prod_p( 1+sum_{k=2}^infty p^{-sk})$ converges for $Re(s) > 1/2$
    $endgroup$
    – reuns
    Nov 26 '18 at 18:10


















0












$begingroup$



For $n=p_1^{alpha_1}p_2^{alpha_2}cdotcdotcdot p_k^{alpha_k}$ we
denote $alpha(n)=alpha_1alpha_2cdotcdotcdotalpha_k$. Show that
$F(s)=sum_{ngeq 1}frac{alpha(n)}{n^s}$ is absolutely convergent
for $sigma>1$.




My attempt:



Notice that $alpha(n)leq n$ because
begin{equation*}
alpha_ileq p_i^{alpha_i},;text{ for every $p_i$ prime}\
alpha(n)=alpha_1cdotcdotcdotalpha_kleq p_1^{alpha_1}cdotcdotcdot p_k^{alpha_k}=n
end{equation*}
Thus, $$sum_{ngeq 1}frac{alpha(n)}{n^s}leq sum_{ngeq 1}frac{n}{n^s}=sum_{ngeq 1}frac{1}{n^{s-1}}=zeta(s-1)$$
that is absolutely convergent for $sigma>2$.



But I don't find the way to ensure convergence for $sigma>1$.



Thanks for any suggestion.






number-theory analytic-number-theory dirichlet-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    $F(s) = prod_p (1+sum_{k=1}^infty k p^{-sk}) $, $frac{F(s)}{zeta(s)} = prod_p( 1+sum_{k=2}^infty p^{-sk})$ which converges for $Re(s) > 1/2$.
    $endgroup$
    – reuns
    Nov 25 '18 at 21:51












  • $begingroup$
    Doesn't that first equality hold only if the series converge absolutely?
    $endgroup$
    – Kale36
    Nov 26 '18 at 1:14










  • $begingroup$
    @reuns How did you get the expresion for $frac{F(s)}{zeta(s)}$?
    $endgroup$
    – Kale36
    Nov 26 '18 at 17:40










  • $begingroup$
    $F(s) =sum_{ngeq 1}frac{alpha(n)}{n^s}= prod_p (1+sum_{k=1}^infty k p^{-sk})$ converges for $Re(s) > 1$ and $frac{F(s)}{zeta(s)}=prod_p (1+sum_{k=1}^infty k p^{-sk})(1-p^{-s})= prod_p( 1+sum_{k=2}^infty p^{-sk})$ converges for $Re(s) > 1/2$
    $endgroup$
    – reuns
    Nov 26 '18 at 18:10
















0












0








0





$begingroup$



For $n=p_1^{alpha_1}p_2^{alpha_2}cdotcdotcdot p_k^{alpha_k}$ we
denote $alpha(n)=alpha_1alpha_2cdotcdotcdotalpha_k$. Show that
$F(s)=sum_{ngeq 1}frac{alpha(n)}{n^s}$ is absolutely convergent
for $sigma>1$.




My attempt:



Notice that $alpha(n)leq n$ because
begin{equation*}
alpha_ileq p_i^{alpha_i},;text{ for every $p_i$ prime}\
alpha(n)=alpha_1cdotcdotcdotalpha_kleq p_1^{alpha_1}cdotcdotcdot p_k^{alpha_k}=n
end{equation*}
Thus, $$sum_{ngeq 1}frac{alpha(n)}{n^s}leq sum_{ngeq 1}frac{n}{n^s}=sum_{ngeq 1}frac{1}{n^{s-1}}=zeta(s-1)$$
that is absolutely convergent for $sigma>2$.



But I don't find the way to ensure convergence for $sigma>1$.



Thanks for any suggestion.






number-theory analytic-number-theory dirichlet-series







share|cite|improve this question











$endgroup$





For $n=p_1^{alpha_1}p_2^{alpha_2}cdotcdotcdot p_k^{alpha_k}$ we
denote $alpha(n)=alpha_1alpha_2cdotcdotcdotalpha_k$. Show that
$F(s)=sum_{ngeq 1}frac{alpha(n)}{n^s}$ is absolutely convergent
for $sigma>1$.




My attempt:



Notice that $alpha(n)leq n$ because
begin{equation*}
alpha_ileq p_i^{alpha_i},;text{ for every $p_i$ prime}\
alpha(n)=alpha_1cdotcdotcdotalpha_kleq p_1^{alpha_1}cdotcdotcdot p_k^{alpha_k}=n
end{equation*}
Thus, $$sum_{ngeq 1}frac{alpha(n)}{n^s}leq sum_{ngeq 1}frac{n}{n^s}=sum_{ngeq 1}frac{1}{n^{s-1}}=zeta(s-1)$$
that is absolutely convergent for $sigma>2$.



But I don't find the way to ensure convergence for $sigma>1$.



Thanks for any suggestion.






number-theory analytic-number-theory dirichlet-series




number-theory analytic-number-theory dirichlet-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 18:20









Bernard

119k740113




119k740113










asked Nov 25 '18 at 18:06









Kale36Kale36

454




454












  • $begingroup$
    $F(s) = prod_p (1+sum_{k=1}^infty k p^{-sk}) $, $frac{F(s)}{zeta(s)} = prod_p( 1+sum_{k=2}^infty p^{-sk})$ which converges for $Re(s) > 1/2$.
    $endgroup$
    – reuns
    Nov 25 '18 at 21:51












  • $begingroup$
    Doesn't that first equality hold only if the series converge absolutely?
    $endgroup$
    – Kale36
    Nov 26 '18 at 1:14










  • $begingroup$
    @reuns How did you get the expresion for $frac{F(s)}{zeta(s)}$?
    $endgroup$
    – Kale36
    Nov 26 '18 at 17:40










  • $begingroup$
    $F(s) =sum_{ngeq 1}frac{alpha(n)}{n^s}= prod_p (1+sum_{k=1}^infty k p^{-sk})$ converges for $Re(s) > 1$ and $frac{F(s)}{zeta(s)}=prod_p (1+sum_{k=1}^infty k p^{-sk})(1-p^{-s})= prod_p( 1+sum_{k=2}^infty p^{-sk})$ converges for $Re(s) > 1/2$
    $endgroup$
    – reuns
    Nov 26 '18 at 18:10




















  • $begingroup$
    $F(s) = prod_p (1+sum_{k=1}^infty k p^{-sk}) $, $frac{F(s)}{zeta(s)} = prod_p( 1+sum_{k=2}^infty p^{-sk})$ which converges for $Re(s) > 1/2$.
    $endgroup$
    – reuns
    Nov 25 '18 at 21:51












  • $begingroup$
    Doesn't that first equality hold only if the series converge absolutely?
    $endgroup$
    – Kale36
    Nov 26 '18 at 1:14










  • $begingroup$
    @reuns How did you get the expresion for $frac{F(s)}{zeta(s)}$?
    $endgroup$
    – Kale36
    Nov 26 '18 at 17:40










  • $begingroup$
    $F(s) =sum_{ngeq 1}frac{alpha(n)}{n^s}= prod_p (1+sum_{k=1}^infty k p^{-sk})$ converges for $Re(s) > 1$ and $frac{F(s)}{zeta(s)}=prod_p (1+sum_{k=1}^infty k p^{-sk})(1-p^{-s})= prod_p( 1+sum_{k=2}^infty p^{-sk})$ converges for $Re(s) > 1/2$
    $endgroup$
    – reuns
    Nov 26 '18 at 18:10


















$begingroup$
$F(s) = prod_p (1+sum_{k=1}^infty k p^{-sk}) $, $frac{F(s)}{zeta(s)} = prod_p( 1+sum_{k=2}^infty p^{-sk})$ which converges for $Re(s) > 1/2$.
$endgroup$
– reuns
Nov 25 '18 at 21:51






$begingroup$
$F(s) = prod_p (1+sum_{k=1}^infty k p^{-sk}) $, $frac{F(s)}{zeta(s)} = prod_p( 1+sum_{k=2}^infty p^{-sk})$ which converges for $Re(s) > 1/2$.
$endgroup$
– reuns
Nov 25 '18 at 21:51














$begingroup$
Doesn't that first equality hold only if the series converge absolutely?
$endgroup$
– Kale36
Nov 26 '18 at 1:14




$begingroup$
Doesn't that first equality hold only if the series converge absolutely?
$endgroup$
– Kale36
Nov 26 '18 at 1:14












$begingroup$
@reuns How did you get the expresion for $frac{F(s)}{zeta(s)}$?
$endgroup$
– Kale36
Nov 26 '18 at 17:40




$begingroup$
@reuns How did you get the expresion for $frac{F(s)}{zeta(s)}$?
$endgroup$
– Kale36
Nov 26 '18 at 17:40












$begingroup$
$F(s) =sum_{ngeq 1}frac{alpha(n)}{n^s}= prod_p (1+sum_{k=1}^infty k p^{-sk})$ converges for $Re(s) > 1$ and $frac{F(s)}{zeta(s)}=prod_p (1+sum_{k=1}^infty k p^{-sk})(1-p^{-s})= prod_p( 1+sum_{k=2}^infty p^{-sk})$ converges for $Re(s) > 1/2$
$endgroup$
– reuns
Nov 26 '18 at 18:10






$begingroup$
$F(s) =sum_{ngeq 1}frac{alpha(n)}{n^s}= prod_p (1+sum_{k=1}^infty k p^{-sk})$ converges for $Re(s) > 1$ and $frac{F(s)}{zeta(s)}=prod_p (1+sum_{k=1}^infty k p^{-sk})(1-p^{-s})= prod_p( 1+sum_{k=2}^infty p^{-sk})$ converges for $Re(s) > 1/2$
$endgroup$
– reuns
Nov 26 '18 at 18:10












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