Behavior of $sum_{n=1}^infty frac{1}{n} z^{n!}$ on the unit circle [duplicate]
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Series with radius of convergence 1 that diverges on roots of unity, converges elsewhere on the circle.
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I'm trying to understand the behavior of $sum_{n=1}^infty frac{1}{n} z^{n!}$ on the unit circle.
Since for each $m$th root of unity $zeta_m$
$$sum_{n=1}^infty frac{1}{n} zeta_m^{n!} = C + sum_{n=m}^infty frac{1}{n} = infty$$
holds for some $C in mathbb{C}$, the series diverges for all $e^{varphi pi i}$ with $varphi in mathbb{Q}$.
But what happens for $varphi in mathbb{R} setminus mathbb{Q}$?
Does the series diverge everywhere, or are there points where it is convergent?
calculus complex-analysis
edited Nov 16 at 12:32
amWhy
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asked Nov 16 at 12:14
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marked as duplicate by kingW3, Paul Frost, Lord Shark the Unknown, Arnaud D., amWhy calculus
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Nov 16 at 20:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Series with radius of convergence 1 that diverges on roots of unity, converges elsewhere on the circle.
2 answers
I'm trying to understand the behavior of $sum_{n=1}^infty frac{1}{n} z^{n!}$ on the unit circle.
Since for each $m$th root of unity $zeta_m$
$$sum_{n=1}^infty frac{1}{n} zeta_m^{n!} = C + sum_{n=m}^infty frac{1}{n} = infty$$
holds for some $C in mathbb{C}$, the series diverges for all $e^{varphi pi i}$ with $varphi in mathbb{Q}$.
But what happens for $varphi in mathbb{R} setminus mathbb{Q}$?
Does the series diverge everywhere, or are there points where it is convergent?
calculus complex-analysis
edited Nov 16 at 12:32
amWhy
191k27223437
asked Nov 16 at 12:14
Johannes S.
314
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Nov 16 at 20:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
I have no idea. It's a cute problem, though.
– davidlowryduda♦
Nov 16 at 12:16
For the purposes of googling, such a series is said to lacunary. In that context one usually talks about gap theorems of various strengths (eg the Fabry gap theorem) and one of those may be illuminating.
– Semiclassical
Nov 16 at 14:59
add a comment |
up vote
5
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favorite
up vote
5
down vote
favorite
This question already has an answer here:
Series with radius of convergence 1 that diverges on roots of unity, converges elsewhere on the circle.
2 answers
I'm trying to understand the behavior of $sum_{n=1}^infty frac{1}{n} z^{n!}$ on the unit circle.
Since for each $m$th root of unity $zeta_m$
$$sum_{n=1}^infty frac{1}{n} zeta_m^{n!} = C + sum_{n=m}^infty frac{1}{n} = infty$$
holds for some $C in mathbb{C}$, the series diverges for all $e^{varphi pi i}$ with $varphi in mathbb{Q}$.
But what happens for $varphi in mathbb{R} setminus mathbb{Q}$?
Does the series diverge everywhere, or are there points where it is convergent?
calculus complex-analysis
edited Nov 16 at 12:32
amWhy
191k27223437
asked Nov 16 at 12:14
Johannes S.
314
New contributor
Johannes S. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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This question already has an answer here:
Series with radius of convergence 1 that diverges on roots of unity, converges elsewhere on the circle.
2 answers
I'm trying to understand the behavior of $sum_{n=1}^infty frac{1}{n} z^{n!}$ on the unit circle.
Since for each $m$th root of unity $zeta_m$
$$sum_{n=1}^infty frac{1}{n} zeta_m^{n!} = C + sum_{n=m}^infty frac{1}{n} = infty$$
holds for some $C in mathbb{C}$, the series diverges for all $e^{varphi pi i}$ with $varphi in mathbb{Q}$.
But what happens for $varphi in mathbb{R} setminus mathbb{Q}$?
Does the series diverge everywhere, or are there points where it is convergent?
This question already has an answer here:
Series with radius of convergence 1 that diverges on roots of unity, converges elsewhere on the circle.
2 answers
calculus complex-analysis
calculus complex-analysis
edited Nov 16 at 12:32
amWhy
191k27223437
asked Nov 16 at 12:14
Johannes S.
314
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edited Nov 16 at 12:32
amWhy
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asked Nov 16 at 12:14
Johannes S.
314
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edited Nov 16 at 12:32
amWhy
191k27223437
edited Nov 16 at 12:32
amWhy
191k27223437
edited Nov 16 at 12:32
amWhy
191k27223437
191k27223437
asked Nov 16 at 12:14
Johannes S.
314
New contributor
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Check out our Code of Conduct.
asked Nov 16 at 12:14
Johannes S.
314
asked Nov 16 at 12:14
Johannes S.
314
314
New contributor
Johannes S. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Johannes S. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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marked as duplicate by kingW3, Paul Frost, Lord Shark the Unknown, Arnaud D., amWhy calculus
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Nov 16 at 20:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Nov 16 at 20:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
I have no idea. It's a cute problem, though.
– davidlowryduda♦
Nov 16 at 12:16
For the purposes of googling, such a series is said to lacunary. In that context one usually talks about gap theorems of various strengths (eg the Fabry gap theorem) and one of those may be illuminating.
– Semiclassical
Nov 16 at 14:59
add a comment |
I have no idea. It's a cute problem, though.
– davidlowryduda♦
Nov 16 at 12:16
For the purposes of googling, such a series is said to lacunary. In that context one usually talks about gap theorems of various strengths (eg the Fabry gap theorem) and one of those may be illuminating.
– Semiclassical
Nov 16 at 14:59
I have no idea. It's a cute problem, though.
– davidlowryduda♦
Nov 16 at 12:16
I have no idea. It's a cute problem, though.
– davidlowryduda♦
Nov 16 at 12:16
For the purposes of googling, such a series is said to lacunary. In that context one usually talks about gap theorems of various strengths (eg the Fabry gap theorem) and one of those may be illuminating.
– Semiclassical
Nov 16 at 14:59
For the purposes of googling, such a series is said to lacunary. In that context one usually talks about gap theorems of various strengths (eg the Fabry gap theorem) and one of those may be illuminating.
– Semiclassical
Nov 16 at 14:59
add a comment |
1 Answer
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The sum converges for $z = e^{2pi ivarphi}$ for $varphi = frac{1}{2}frac{1}{1!}+frac{1}{2}frac{1}{3!}+frac{1}{2}frac{1}{5!}+frac{1}{2}frac{1}{7!}+...$. The reason is that for $n$ odd, $n!varphi pmod{1}$ is basically $frac{1}{2}$, while for $n$ even, $n!varphi pmod{1}$ is basically $0$.
answered Nov 16 at 12:32
mathworker21
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
up vote
5
down vote
accepted
The sum converges for $z = e^{2pi ivarphi}$ for $varphi = frac{1}{2}frac{1}{1!}+frac{1}{2}frac{1}{3!}+frac{1}{2}frac{1}{5!}+frac{1}{2}frac{1}{7!}+...$. The reason is that for $n$ odd, $n!varphi pmod{1}$ is basically $frac{1}{2}$, while for $n$ even, $n!varphi pmod{1}$ is basically $0$.
answered Nov 16 at 12:32
mathworker21
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up vote
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The sum converges for $z = e^{2pi ivarphi}$ for $varphi = frac{1}{2}frac{1}{1!}+frac{1}{2}frac{1}{3!}+frac{1}{2}frac{1}{5!}+frac{1}{2}frac{1}{7!}+...$. The reason is that for $n$ odd, $n!varphi pmod{1}$ is basically $frac{1}{2}$, while for $n$ even, $n!varphi pmod{1}$ is basically $0$.
answered Nov 16 at 12:32
mathworker21
7,9481827
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
The sum converges for $z = e^{2pi ivarphi}$ for $varphi = frac{1}{2}frac{1}{1!}+frac{1}{2}frac{1}{3!}+frac{1}{2}frac{1}{5!}+frac{1}{2}frac{1}{7!}+...$. The reason is that for $n$ odd, $n!varphi pmod{1}$ is basically $frac{1}{2}$, while for $n$ even, $n!varphi pmod{1}$ is basically $0$.
answered Nov 16 at 12:32
mathworker21
7,9481827
The sum converges for $z = e^{2pi ivarphi}$ for $varphi = frac{1}{2}frac{1}{1!}+frac{1}{2}frac{1}{3!}+frac{1}{2}frac{1}{5!}+frac{1}{2}frac{1}{7!}+...$. The reason is that for $n$ odd, $n!varphi pmod{1}$ is basically $frac{1}{2}$, while for $n$ even, $n!varphi pmod{1}$ is basically $0$.
answered Nov 16 at 12:32
mathworker21
7,9481827
edited yesterday
answered Nov 16 at 12:32
mathworker21
7,9481827
answered Nov 16 at 12:32
mathworker21
7,9481827
answered Nov 16 at 12:32
mathworker21
7,9481827
7,9481827
add a comment |
add a comment |
I have no idea. It's a cute problem, though.
– davidlowryduda♦
Nov 16 at 12:16
For the purposes of googling, such a series is said to lacunary. In that context one usually talks about gap theorems of various strengths (eg the Fabry gap theorem) and one of those may be illuminating.
– Semiclassical
Nov 16 at 14:59