Series of reciprocal of integers
$begingroup$
This is a question I asked myself today...
$ $
Do you know if it is possible to build a strictly-increasing sequence $(u_n)_{ninmathbb{N}^star}$ of positive integers such that $displaystylesum_{n=1}^{+infty}frac{1}{u_n}<+infty$ and such that for any given $0<varepsilon<1$, one has $displaystylesum_{n=1}^{+infty}frac{1}{u_n^varepsilon}=+infty$ ?
$ $
This is looking for a Dirichlet series (with reciprocal of $u_n$ coefficients) with abscissa of convergence $sigma=1$ and that has a finite limit at $s=1$, does that exist ? It would mean that the corresponding L-function (if the series was extendable around $s=1$) would have no pole at $s=1$. This is tough because the L-function would not be in the Selberg class, and those are hard to study...
sequences-and-series l-functions
asked Dec 7 '18 at 1:26
AnthonyAnthony
628
$endgroup$
add a comment |
$begingroup$
This is a question I asked myself today...
$ $
Do you know if it is possible to build a strictly-increasing sequence $(u_n)_{ninmathbb{N}^star}$ of positive integers such that $displaystylesum_{n=1}^{+infty}frac{1}{u_n}<+infty$ and such that for any given $0<varepsilon<1$, one has $displaystylesum_{n=1}^{+infty}frac{1}{u_n^varepsilon}=+infty$ ?
$ $
This is looking for a Dirichlet series (with reciprocal of $u_n$ coefficients) with abscissa of convergence $sigma=1$ and that has a finite limit at $s=1$, does that exist ? It would mean that the corresponding L-function (if the series was extendable around $s=1$) would have no pole at $s=1$. This is tough because the L-function would not be in the Selberg class, and those are hard to study...
sequences-and-series l-functions
asked Dec 7 '18 at 1:26
AnthonyAnthony
628
$endgroup$
add a comment |
$begingroup$
This is a question I asked myself today...
$ $
Do you know if it is possible to build a strictly-increasing sequence $(u_n)_{ninmathbb{N}^star}$ of positive integers such that $displaystylesum_{n=1}^{+infty}frac{1}{u_n}<+infty$ and such that for any given $0<varepsilon<1$, one has $displaystylesum_{n=1}^{+infty}frac{1}{u_n^varepsilon}=+infty$ ?
$ $
This is looking for a Dirichlet series (with reciprocal of $u_n$ coefficients) with abscissa of convergence $sigma=1$ and that has a finite limit at $s=1$, does that exist ? It would mean that the corresponding L-function (if the series was extendable around $s=1$) would have no pole at $s=1$. This is tough because the L-function would not be in the Selberg class, and those are hard to study...
sequences-and-series l-functions
asked Dec 7 '18 at 1:26
AnthonyAnthony
628
$endgroup$
This is a question I asked myself today...
$ $
Do you know if it is possible to build a strictly-increasing sequence $(u_n)_{ninmathbb{N}^star}$ of positive integers such that $displaystylesum_{n=1}^{+infty}frac{1}{u_n}<+infty$ and such that for any given $0<varepsilon<1$, one has $displaystylesum_{n=1}^{+infty}frac{1}{u_n^varepsilon}=+infty$ ?
$ $
This is looking for a Dirichlet series (with reciprocal of $u_n$ coefficients) with abscissa of convergence $sigma=1$ and that has a finite limit at $s=1$, does that exist ? It would mean that the corresponding L-function (if the series was extendable around $s=1$) would have no pole at $s=1$. This is tough because the L-function would not be in the Selberg class, and those are hard to study...
sequences-and-series l-functions
sequences-and-series l-functions
asked Dec 7 '18 at 1:26
AnthonyAnthony
628
asked Dec 7 '18 at 1:26
AnthonyAnthony
628
edited Dec 7 '18 at 1:46
Anthony
asked Dec 7 '18 at 1:26
AnthonyAnthony
628
asked Dec 7 '18 at 1:26
AnthonyAnthony
628
asked Dec 7 '18 at 1:26
AnthonyAnthony
628
628
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can take $u_n = lceil n log^2 nrceil$:
$$
sum_{n=1}^infty frac{1}{n log^2 n} < infty
$$
but
$$
sum_{n=1}^infty frac{1}{n^varepsilon log^{2varepsilon} n} = infty
$$
for every $varepsilon < 1$.
answered Dec 7 '18 at 2:00
Clement C.Clement C.
49.9k33887
$endgroup$
$begingroup$
Oh that's totally true ! I tried $u_n=nlog(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks !
$endgroup$
– Anthony
Dec 7 '18 at 2:01
$begingroup$
@Anthony Glad this helped!
$endgroup$
– Clement C.
Dec 7 '18 at 2:02
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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votes
$begingroup$
You can take $u_n = lceil n log^2 nrceil$:
$$
sum_{n=1}^infty frac{1}{n log^2 n} < infty
$$
but
$$
sum_{n=1}^infty frac{1}{n^varepsilon log^{2varepsilon} n} = infty
$$
for every $varepsilon < 1$.
answered Dec 7 '18 at 2:00
Clement C.Clement C.
49.9k33887
$endgroup$
$begingroup$
Oh that's totally true ! I tried $u_n=nlog(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks !
$endgroup$
– Anthony
Dec 7 '18 at 2:01
$begingroup$
@Anthony Glad this helped!
$endgroup$
– Clement C.
Dec 7 '18 at 2:02
add a comment |
$begingroup$
You can take $u_n = lceil n log^2 nrceil$:
$$
sum_{n=1}^infty frac{1}{n log^2 n} < infty
$$
but
$$
sum_{n=1}^infty frac{1}{n^varepsilon log^{2varepsilon} n} = infty
$$
for every $varepsilon < 1$.
answered Dec 7 '18 at 2:00
Clement C.Clement C.
49.9k33887
$endgroup$
$begingroup$
Oh that's totally true ! I tried $u_n=nlog(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks !
$endgroup$
– Anthony
Dec 7 '18 at 2:01
$begingroup$
@Anthony Glad this helped!
$endgroup$
– Clement C.
Dec 7 '18 at 2:02
add a comment |
$begingroup$
You can take $u_n = lceil n log^2 nrceil$:
$$
sum_{n=1}^infty frac{1}{n log^2 n} < infty
$$
but
$$
sum_{n=1}^infty frac{1}{n^varepsilon log^{2varepsilon} n} = infty
$$
for every $varepsilon < 1$.
answered Dec 7 '18 at 2:00
Clement C.Clement C.
49.9k33887
$endgroup$
You can take $u_n = lceil n log^2 nrceil$:
$$
sum_{n=1}^infty frac{1}{n log^2 n} < infty
$$
but
$$
sum_{n=1}^infty frac{1}{n^varepsilon log^{2varepsilon} n} = infty
$$
for every $varepsilon < 1$.
answered Dec 7 '18 at 2:00
Clement C.Clement C.
49.9k33887
answered Dec 7 '18 at 2:00
Clement C.Clement C.
49.9k33887
answered Dec 7 '18 at 2:00
Clement C.Clement C.
49.9k33887
answered Dec 7 '18 at 2:00
Clement C.Clement C.
49.9k33887
49.9k33887
$begingroup$
Oh that's totally true ! I tried $u_n=nlog(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks !
$endgroup$
– Anthony
Dec 7 '18 at 2:01
$begingroup$
@Anthony Glad this helped!
$endgroup$
– Clement C.
Dec 7 '18 at 2:02
add a comment |
$begingroup$
Oh that's totally true ! I tried $u_n=nlog(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks !
$endgroup$
– Anthony
Dec 7 '18 at 2:01
$begingroup$
@Anthony Glad this helped!
$endgroup$
– Clement C.
Dec 7 '18 at 2:02
$begingroup$
Oh that's totally true ! I tried $u_n=nlog(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks !
$endgroup$
– Anthony
Dec 7 '18 at 2:01
$begingroup$
Oh that's totally true ! I tried $u_n=nlog(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks !
$endgroup$
– Anthony
Dec 7 '18 at 2:01
$begingroup$
@Anthony Glad this helped!
$endgroup$
– Clement C.
Dec 7 '18 at 2:02
$begingroup$
@Anthony Glad this helped!
$endgroup$
– Clement C.
Dec 7 '18 at 2:02
add a comment |
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