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Hampus Nyberg
Hampus Nyberg
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Consider the series $$ \sum_{n=1}^{\infty} ( \{ n \xi \} - \frac{1}{2}) $$ where $\{ \}$ denotes the fractional part and $\xi$ is some irrational number. It clearly does not converge but can we show that it does not tend to $\infty$ or $-\infty$? Would it also be true that it bounded?

The problem seems to be related to equidistributed sequences. Since ${(\{ n \xi \} )}_{n \geq 1}$ is equidistributed we have that

$$ \frac{1}{N}\sum_{n=1}^{N}(\{n\xi\}- \frac{1}{2}) \to 0 $$ as $N \to \infty$. However, just the equidistribution property cannot be enough. Consider for example a sequence $(a_n)$ given by $a_n = 1$ if $n$ is odd or a power of $2$ and $a_n= -1$ otherwise. Then $(a_n)$ is equidistributed on $ \\{ -1,1 \\}$$ \{ -1,1 \}$ but

$$ \sum_{n=0}^{\infty} a_n = \infty. $$

Consider the series $$ \sum_{n=1}^{\infty} ( \{ n \xi \} - \frac{1}{2}) $$ where $\{ \}$ denotes the fractional part and $\xi$ is some irrational number. It clearly does not converge but can we show that it does not tend to $\infty$ or $-\infty$? Would it also be true that it bounded?

The problem seems to be related to equidistributed sequences. Since ${(\{ n \xi \} )}_{n \geq 1}$ is equidistributed we have that

$$ \frac{1}{N}\sum_{n=1}^{N}(\{n\xi\}- \frac{1}{2}) \to 0 $$ as $N \to \infty$. However, just the equidistribution property cannot be enough. Consider for example a sequence $(a_n)$ given by $a_n = 1$ if $n$ is odd or a power of $2$ and $a_n= -1$ otherwise. Then $(a_n)$ is equidistributed on $ \\{ -1,1 \\}$ but

$$ \sum_{n=0}^{\infty} a_n = \infty. $$

Consider the series $$ \sum_{n=1}^{\infty} ( \{ n \xi \} - \frac{1}{2}) $$ where $\{ \}$ denotes the fractional part and $\xi$ is some irrational number. It clearly does not converge but can we show that it does not tend to $\infty$ or $-\infty$? Would it also be true that it bounded?

The problem seems to be related to equidistributed sequences. Since ${(\{ n \xi \} )}_{n \geq 1}$ is equidistributed we have that

$$ \frac{1}{N}\sum_{n=1}^{N}(\{n\xi\}- \frac{1}{2}) \to 0 $$ as $N \to \infty$. However, just the equidistribution property cannot be enough. Consider for example a sequence $(a_n)$ given by $a_n = 1$ if $n$ is odd or a power of $2$ and $a_n= -1$ otherwise. Then $(a_n)$ is equidistributed on $ \{ -1,1 \}$ but

$$ \sum_{n=0}^{\infty} a_n = \infty. $$

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Hampus Nyberg
Hampus Nyberg
  • 177
  • 5

Consider the series $$ \sum_{n=1}^{\infty} ( \\{ n \xi \\} - \frac{1}{2}) $$$$ \sum_{n=1}^{\infty} ( \{ n \xi \} - \frac{1}{2}) $$ where $\\{ \\}$$\{ \}$ denotes the fractional part and $\xi$ is some irrational number. It clearly does not converge but can we show that it does not tend to $\infty$ or $-\infty$? Would it also be true that it bounded?

The problem seems to be related to equidistributed sequences. Since ${(\\{ n \xi \\} )}_{n \geq 1}$${(\{ n \xi \} )}_{n \geq 1}$ is equidistributed we have that

$$ \frac{1}{N}\sum_{n=1}^{N}(\\{n\xi\\}- \frac{1}{2}) \to 0 $$$$ \frac{1}{N}\sum_{n=1}^{N}(\{n\xi\}- \frac{1}{2}) \to 0 $$ as $N \to \infty$. However, just the equidistribution property cannot be enough. Consider for example a sequence $(a_n)$ given by $a_n = 1$ if $n$ is odd or a power of $2$ and $a_n= -1$ otherwise. Then $(a_n)$ is equidistributed on $ \\{ -1,1 \\}$ but

$$ \sum_{n=0}^{\infty} a_n = \infty. $$

Consider the series $$ \sum_{n=1}^{\infty} ( \\{ n \xi \\} - \frac{1}{2}) $$ where $\\{ \\}$ denotes the fractional part and $\xi$ is some irrational number. It clearly does not converge but can we show that it does not tend to $\infty$ or $-\infty$? Would it also be true that it bounded?

The problem seems to be related to equidistributed sequences. Since ${(\\{ n \xi \\} )}_{n \geq 1}$ is equidistributed we have that

$$ \frac{1}{N}\sum_{n=1}^{N}(\\{n\xi\\}- \frac{1}{2}) \to 0 $$ as $N \to \infty$. However, just the equidistribution property cannot be enough. Consider for example a sequence $(a_n)$ given by $a_n = 1$ if $n$ is odd or a power of $2$ and $a_n= -1$ otherwise. Then $(a_n)$ is equidistributed on $ \\{ -1,1 \\}$ but

$$ \sum_{n=0}^{\infty} a_n = \infty. $$

Consider the series $$ \sum_{n=1}^{\infty} ( \{ n \xi \} - \frac{1}{2}) $$ where $\{ \}$ denotes the fractional part and $\xi$ is some irrational number. It clearly does not converge but can we show that it does not tend to $\infty$ or $-\infty$? Would it also be true that it bounded?

The problem seems to be related to equidistributed sequences. Since ${(\{ n \xi \} )}_{n \geq 1}$ is equidistributed we have that

$$ \frac{1}{N}\sum_{n=1}^{N}(\{n\xi\}- \frac{1}{2}) \to 0 $$ as $N \to \infty$. However, just the equidistribution property cannot be enough. Consider for example a sequence $(a_n)$ given by $a_n = 1$ if $n$ is odd or a power of $2$ and $a_n= -1$ otherwise. Then $(a_n)$ is equidistributed on $ \\{ -1,1 \\}$ but

$$ \sum_{n=0}^{\infty} a_n = \infty. $$

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Hampus Nyberg
Hampus Nyberg
  • 177
  • 5

Consider the series $$ \sum_{n=1}^{\infty} ( \\{ n \xi \\} - \frac{1}{2}) $$ where $\\{ \\}$ denotes the fractional part and $\xi$ is some irrational number. It clearly does not converge but can we show that it does not tend to $\infty$ or $-\infty$? Would it also be true that it bounded?

The problem seems to be related to equidistributed sequences. Since $(\\{n \xi \\})_{n \geq 1}$${(\\{ n \xi \\} )}_{n \geq 1}$ is equidistributed we have that

$$ \frac{1}{N}\sum_{n=1}^{N}(\\{n\xi\\}- \frac{1}{2}) \to 0 $$ as $N \to \infty$. However, just the equidistribution property cannot be enough. Consider for example a sequence $(a_n)$ given by $a_n = 1$ if $n$ is odd or a power of $2$ and $a_n= -1$ otherwise. Then $(a_n)$ is equidistributed on $ \\{ -1,1 \\}$ but

$$ \sum_{n=0}^{\infty} a_n = \infty. $$

Consider the series $$ \sum_{n=1}^{\infty} ( \\{ n \xi \\} - \frac{1}{2}) $$ where $\\{ \\}$ denotes the fractional part and $\xi$ is some irrational number. It clearly does not converge but can we show that it does not tend to $\infty$ or $-\infty$? Would it also be true that it bounded?

The problem seems to be related to equidistributed sequences. Since $(\\{n \xi \\})_{n \geq 1}$ is equidistributed we have that

$$ \frac{1}{N}\sum_{n=1}^{N}(\\{n\xi\\}- \frac{1}{2}) \to 0 $$ as $N \to \infty$. However, just the equidistribution property cannot be enough. Consider for example a sequence $(a_n)$ given by $a_n = 1$ if $n$ is odd or a power of $2$ and $a_n= -1$ otherwise. Then $(a_n)$ is equidistributed on $ \\{ -1,1 \\}$ but

$$ \sum_{n=0}^{\infty} a_n = \infty. $$

Consider the series $$ \sum_{n=1}^{\infty} ( \\{ n \xi \\} - \frac{1}{2}) $$ where $\\{ \\}$ denotes the fractional part and $\xi$ is some irrational number. It clearly does not converge but can we show that it does not tend to $\infty$ or $-\infty$? Would it also be true that it bounded?

The problem seems to be related to equidistributed sequences. Since ${(\\{ n \xi \\} )}_{n \geq 1}$ is equidistributed we have that

$$ \frac{1}{N}\sum_{n=1}^{N}(\\{n\xi\\}- \frac{1}{2}) \to 0 $$ as $N \to \infty$. However, just the equidistribution property cannot be enough. Consider for example a sequence $(a_n)$ given by $a_n = 1$ if $n$ is odd or a power of $2$ and $a_n= -1$ otherwise. Then $(a_n)$ is equidistributed on $ \\{ -1,1 \\}$ but

$$ \sum_{n=0}^{\infty} a_n = \infty. $$

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Hampus Nyberg
Hampus Nyberg
  • 177
  • 5