Is the series $sum_{n=0}^{infty}frac{B_n(z)}{2^n}$ convergent?
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$B_n(z)$ is the Bernoulli polynomial, we know that the series
$$sum_{n=0}^{infty}frac{B_n(z)}{n!}$$
is convergent for every $zinmathbb{C}$ and its closed form is equal to $$frac{e^z}{e-1}.$$ Now I deal to the series $$sum_{n=0}^{infty}frac{B_n(z)}{2^n}$$ in my research.
(a) Is the above series convergent? If yes, for which $z$?
(b) Does the series have any closed form?
(c) Does there exist any method or related books or similar reference?
sequences-and-series convergence bernoulli-numbers bernoulli-polynomials
edited Nov 21 at 16:24
Carl Schildkraut
10.6k11438
asked Nov 21 at 16:19
soodehMehboodi
52228
add a comment |
up vote
2
down vote
favorite
$B_n(z)$ is the Bernoulli polynomial, we know that the series
$$sum_{n=0}^{infty}frac{B_n(z)}{n!}$$
is convergent for every $zinmathbb{C}$ and its closed form is equal to $$frac{e^z}{e-1}.$$ Now I deal to the series $$sum_{n=0}^{infty}frac{B_n(z)}{2^n}$$ in my research.
(a) Is the above series convergent? If yes, for which $z$?
(b) Does the series have any closed form?
(c) Does there exist any method or related books or similar reference?
sequences-and-series convergence bernoulli-numbers bernoulli-polynomials
edited Nov 21 at 16:24
Carl Schildkraut
10.6k11438
asked Nov 21 at 16:19
soodehMehboodi
52228
@ Raptor thanks
– soodehMehboodi
Nov 21 at 16:23
I've changed around some of the formatting to make it more readable; please feel free to roll back if there's anything you wish to change.
– Carl Schildkraut
Nov 21 at 16:25
@Carl Schildkraut thanks for your changing.
– soodehMehboodi
Nov 21 at 16:34
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$B_n(z)$ is the Bernoulli polynomial, we know that the series
$$sum_{n=0}^{infty}frac{B_n(z)}{n!}$$
is convergent for every $zinmathbb{C}$ and its closed form is equal to $$frac{e^z}{e-1}.$$ Now I deal to the series $$sum_{n=0}^{infty}frac{B_n(z)}{2^n}$$ in my research.
(a) Is the above series convergent? If yes, for which $z$?
(b) Does the series have any closed form?
(c) Does there exist any method or related books or similar reference?
sequences-and-series convergence bernoulli-numbers bernoulli-polynomials
edited Nov 21 at 16:24
Carl Schildkraut
10.6k11438
asked Nov 21 at 16:19
soodehMehboodi
52228
$B_n(z)$ is the Bernoulli polynomial, we know that the series
$$sum_{n=0}^{infty}frac{B_n(z)}{n!}$$
is convergent for every $zinmathbb{C}$ and its closed form is equal to $$frac{e^z}{e-1}.$$ Now I deal to the series $$sum_{n=0}^{infty}frac{B_n(z)}{2^n}$$ in my research.
(a) Is the above series convergent? If yes, for which $z$?
(b) Does the series have any closed form?
(c) Does there exist any method or related books or similar reference?
sequences-and-series convergence bernoulli-numbers bernoulli-polynomials
sequences-and-series convergence bernoulli-numbers bernoulli-polynomials
edited Nov 21 at 16:24
Carl Schildkraut
10.6k11438
asked Nov 21 at 16:19
soodehMehboodi
52228
edited Nov 21 at 16:24
Carl Schildkraut
10.6k11438
asked Nov 21 at 16:19
soodehMehboodi
52228
edited Nov 21 at 16:24
Carl Schildkraut
10.6k11438
edited Nov 21 at 16:24
Carl Schildkraut
10.6k11438
edited Nov 21 at 16:24
Carl Schildkraut
10.6k11438
10.6k11438
asked Nov 21 at 16:19
soodehMehboodi
52228
asked Nov 21 at 16:19
soodehMehboodi
52228
asked Nov 21 at 16:19
soodehMehboodi
52228
52228
@ Raptor thanks
– soodehMehboodi
Nov 21 at 16:23
I've changed around some of the formatting to make it more readable; please feel free to roll back if there's anything you wish to change.
– Carl Schildkraut
Nov 21 at 16:25
@Carl Schildkraut thanks for your changing.
– soodehMehboodi
Nov 21 at 16:34
add a comment |
@ Raptor thanks
– soodehMehboodi
Nov 21 at 16:23
I've changed around some of the formatting to make it more readable; please feel free to roll back if there's anything you wish to change.
– Carl Schildkraut
Nov 21 at 16:25
@Carl Schildkraut thanks for your changing.
– soodehMehboodi
Nov 21 at 16:34
@ Raptor thanks
– soodehMehboodi
Nov 21 at 16:23
@ Raptor thanks
– soodehMehboodi
Nov 21 at 16:23
I've changed around some of the formatting to make it more readable; please feel free to roll back if there's anything you wish to change.
– Carl Schildkraut
Nov 21 at 16:25
I've changed around some of the formatting to make it more readable; please feel free to roll back if there's anything you wish to change.
– Carl Schildkraut
Nov 21 at 16:25
@Carl Schildkraut thanks for your changing.
– soodehMehboodi
Nov 21 at 16:34
@Carl Schildkraut thanks for your changing.
– soodehMehboodi
Nov 21 at 16:34
add a comment |
1 Answer
1
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oldest
votes
up vote
2
down vote
Formally, the o.g.f. comes from the Laplace transform of the e.g.f. That is,
given $$ frac{t e^{zt}}{e^t-1} = sum_{n=0}^infty B_n(z) frac{t^n}{n!}$$
$$ sum_{n=0}^infty B_n(z) s^{-1-n} = int_0^infty frac{t e^{zt}}{e^t-1} e^{-st}; dt = Psi^{(1)}(1-z+s) $$
The right side is singular when $1-z+s$ is a nonpositive integer.
If the series had a positive radius of convergence, the set of singularities would have to be bounded. Since it is not, we conclude that the radius of convergence is $0$. In particular,
your series $sum_{n=0}^infty B_n(z) 2^{-n}$ diverges for all $z$.
EDIT: There' a simpler argument. The fact that the e.g.f.
$frac{t e^{zt}}{e^t-1}$ has finite radius of convergence (as it has singularities $t = 2 n pi i$ for nonzero integers $n$) means
that $limsup_n B_n r^n/n! > 0$ for some $r$. But that says
for some $r$ and $epsilon > 0$, $|B_n| ge epsilon n! r^{-n}$ infinitely often. For any $z ne 0$, $epsilon n! |z/r|^n to infty$, so your series can't converge.
More generally, whenever the e.g.f. of a sequence has a finite radius of convergence, the o.g.f. has radius of convergence $0$.
answered Nov 21 at 16:33
Robert Israel
314k23206453
would you tell me, How can I get more information about the mentioned series and your solution, Also, particularly your edit . Thanks a lot.
– soodehMehboodi
Nov 22 at 17:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Formally, the o.g.f. comes from the Laplace transform of the e.g.f. That is,
given $$ frac{t e^{zt}}{e^t-1} = sum_{n=0}^infty B_n(z) frac{t^n}{n!}$$
$$ sum_{n=0}^infty B_n(z) s^{-1-n} = int_0^infty frac{t e^{zt}}{e^t-1} e^{-st}; dt = Psi^{(1)}(1-z+s) $$
The right side is singular when $1-z+s$ is a nonpositive integer.
If the series had a positive radius of convergence, the set of singularities would have to be bounded. Since it is not, we conclude that the radius of convergence is $0$. In particular,
your series $sum_{n=0}^infty B_n(z) 2^{-n}$ diverges for all $z$.
EDIT: There' a simpler argument. The fact that the e.g.f.
$frac{t e^{zt}}{e^t-1}$ has finite radius of convergence (as it has singularities $t = 2 n pi i$ for nonzero integers $n$) means
that $limsup_n B_n r^n/n! > 0$ for some $r$. But that says
for some $r$ and $epsilon > 0$, $|B_n| ge epsilon n! r^{-n}$ infinitely often. For any $z ne 0$, $epsilon n! |z/r|^n to infty$, so your series can't converge.
More generally, whenever the e.g.f. of a sequence has a finite radius of convergence, the o.g.f. has radius of convergence $0$.
answered Nov 21 at 16:33
Robert Israel
314k23206453
would you tell me, How can I get more information about the mentioned series and your solution, Also, particularly your edit . Thanks a lot.
– soodehMehboodi
Nov 22 at 17:48
add a comment |
up vote
2
down vote
Formally, the o.g.f. comes from the Laplace transform of the e.g.f. That is,
given $$ frac{t e^{zt}}{e^t-1} = sum_{n=0}^infty B_n(z) frac{t^n}{n!}$$
$$ sum_{n=0}^infty B_n(z) s^{-1-n} = int_0^infty frac{t e^{zt}}{e^t-1} e^{-st}; dt = Psi^{(1)}(1-z+s) $$
The right side is singular when $1-z+s$ is a nonpositive integer.
If the series had a positive radius of convergence, the set of singularities would have to be bounded. Since it is not, we conclude that the radius of convergence is $0$. In particular,
your series $sum_{n=0}^infty B_n(z) 2^{-n}$ diverges for all $z$.
EDIT: There' a simpler argument. The fact that the e.g.f.
$frac{t e^{zt}}{e^t-1}$ has finite radius of convergence (as it has singularities $t = 2 n pi i$ for nonzero integers $n$) means
that $limsup_n B_n r^n/n! > 0$ for some $r$. But that says
for some $r$ and $epsilon > 0$, $|B_n| ge epsilon n! r^{-n}$ infinitely often. For any $z ne 0$, $epsilon n! |z/r|^n to infty$, so your series can't converge.
More generally, whenever the e.g.f. of a sequence has a finite radius of convergence, the o.g.f. has radius of convergence $0$.
answered Nov 21 at 16:33
Robert Israel
314k23206453
would you tell me, How can I get more information about the mentioned series and your solution, Also, particularly your edit . Thanks a lot.
– soodehMehboodi
Nov 22 at 17:48
add a comment |
up vote
2
down vote
up vote
2
down vote
Formally, the o.g.f. comes from the Laplace transform of the e.g.f. That is,
given $$ frac{t e^{zt}}{e^t-1} = sum_{n=0}^infty B_n(z) frac{t^n}{n!}$$
$$ sum_{n=0}^infty B_n(z) s^{-1-n} = int_0^infty frac{t e^{zt}}{e^t-1} e^{-st}; dt = Psi^{(1)}(1-z+s) $$
The right side is singular when $1-z+s$ is a nonpositive integer.
If the series had a positive radius of convergence, the set of singularities would have to be bounded. Since it is not, we conclude that the radius of convergence is $0$. In particular,
your series $sum_{n=0}^infty B_n(z) 2^{-n}$ diverges for all $z$.
EDIT: There' a simpler argument. The fact that the e.g.f.
$frac{t e^{zt}}{e^t-1}$ has finite radius of convergence (as it has singularities $t = 2 n pi i$ for nonzero integers $n$) means
that $limsup_n B_n r^n/n! > 0$ for some $r$. But that says
for some $r$ and $epsilon > 0$, $|B_n| ge epsilon n! r^{-n}$ infinitely often. For any $z ne 0$, $epsilon n! |z/r|^n to infty$, so your series can't converge.
More generally, whenever the e.g.f. of a sequence has a finite radius of convergence, the o.g.f. has radius of convergence $0$.
answered Nov 21 at 16:33
Robert Israel
314k23206453
Formally, the o.g.f. comes from the Laplace transform of the e.g.f. That is,
given $$ frac{t e^{zt}}{e^t-1} = sum_{n=0}^infty B_n(z) frac{t^n}{n!}$$
$$ sum_{n=0}^infty B_n(z) s^{-1-n} = int_0^infty frac{t e^{zt}}{e^t-1} e^{-st}; dt = Psi^{(1)}(1-z+s) $$
The right side is singular when $1-z+s$ is a nonpositive integer.
If the series had a positive radius of convergence, the set of singularities would have to be bounded. Since it is not, we conclude that the radius of convergence is $0$. In particular,
your series $sum_{n=0}^infty B_n(z) 2^{-n}$ diverges for all $z$.
EDIT: There' a simpler argument. The fact that the e.g.f.
$frac{t e^{zt}}{e^t-1}$ has finite radius of convergence (as it has singularities $t = 2 n pi i$ for nonzero integers $n$) means
that $limsup_n B_n r^n/n! > 0$ for some $r$. But that says
for some $r$ and $epsilon > 0$, $|B_n| ge epsilon n! r^{-n}$ infinitely often. For any $z ne 0$, $epsilon n! |z/r|^n to infty$, so your series can't converge.
More generally, whenever the e.g.f. of a sequence has a finite radius of convergence, the o.g.f. has radius of convergence $0$.
answered Nov 21 at 16:33
Robert Israel
314k23206453
edited Nov 22 at 14:26
answered Nov 21 at 16:33
Robert Israel
314k23206453
answered Nov 21 at 16:33
Robert Israel
314k23206453
answered Nov 21 at 16:33
Robert Israel
314k23206453
314k23206453
would you tell me, How can I get more information about the mentioned series and your solution, Also, particularly your edit . Thanks a lot.
– soodehMehboodi
Nov 22 at 17:48
add a comment |
would you tell me, How can I get more information about the mentioned series and your solution, Also, particularly your edit . Thanks a lot.
– soodehMehboodi
Nov 22 at 17:48
would you tell me, How can I get more information about the mentioned series and your solution, Also, particularly your edit . Thanks a lot.
– soodehMehboodi
Nov 22 at 17:48
would you tell me, How can I get more information about the mentioned series and your solution, Also, particularly your edit . Thanks a lot.
– soodehMehboodi
Nov 22 at 17:48
add a comment |
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@ Raptor thanks
– soodehMehboodi
Nov 21 at 16:23
I've changed around some of the formatting to make it more readable; please feel free to roll back if there's anything you wish to change.
– Carl Schildkraut
Nov 21 at 16:25
@Carl Schildkraut thanks for your changing.
– soodehMehboodi
Nov 21 at 16:34