Why is the series function periodic if the terms tend to 0?
I am confused by something I read. Let $epsilon_n(x) = sum_{mu = -infty}^infty (x + mu)^{-n}.$ The book says that $epsilon_n$ is absolutely convergent for $n geq 2$ hence, it is obvious that $epsilon_n$ is periodic of period 1 for such $n.$ However, it then says that the same is true for $epsilon_1$ as the terms of the series tend to $0$ as $mu rightarrow pm infty.$
Why do we need absolute convergence over regular convergence? I understand that absolute convergence allows us to add the terms in whatever order we want so we can rearrange the terms of $epsilon_n(x + 1)$ to match the order of summation of $epsilon_n(x)$ for $n geq 2.$ However, is it true that regular convergence does not guarantee periodicity of period 1? This is hard to believe for some reason.
Furthermore, why does the fact that the terms converge to 0 imply that $epsilon_1$ is periodic of period 1?
sequences-and-series number-theory
asked yesterday
伽罗瓦
1,080615
add a comment |
I am confused by something I read. Let $epsilon_n(x) = sum_{mu = -infty}^infty (x + mu)^{-n}.$ The book says that $epsilon_n$ is absolutely convergent for $n geq 2$ hence, it is obvious that $epsilon_n$ is periodic of period 1 for such $n.$ However, it then says that the same is true for $epsilon_1$ as the terms of the series tend to $0$ as $mu rightarrow pm infty.$
Why do we need absolute convergence over regular convergence? I understand that absolute convergence allows us to add the terms in whatever order we want so we can rearrange the terms of $epsilon_n(x + 1)$ to match the order of summation of $epsilon_n(x)$ for $n geq 2.$ However, is it true that regular convergence does not guarantee periodicity of period 1? This is hard to believe for some reason.
Furthermore, why does the fact that the terms converge to 0 imply that $epsilon_1$ is periodic of period 1?
sequences-and-series number-theory
asked yesterday
伽罗瓦
1,080615
I think regular convergence should be enough. In regards to $epsilon_1$, just write out the definitions. $epsilon_n(x+1)-epsilon_n(x) = lim_{M to infty} sum_{mu = -M}^M (x+1+mu)^{-n} - sum_{mu=-M}^M (x+mu)^{-n} = lim_{M to infty} (x+1+M)^{-n}-(x-M)^{-n} = 0$. You should be able to work these things out yourself by unravelling definitions
– mathworker21
yesterday
add a comment |
I am confused by something I read. Let $epsilon_n(x) = sum_{mu = -infty}^infty (x + mu)^{-n}.$ The book says that $epsilon_n$ is absolutely convergent for $n geq 2$ hence, it is obvious that $epsilon_n$ is periodic of period 1 for such $n.$ However, it then says that the same is true for $epsilon_1$ as the terms of the series tend to $0$ as $mu rightarrow pm infty.$
Why do we need absolute convergence over regular convergence? I understand that absolute convergence allows us to add the terms in whatever order we want so we can rearrange the terms of $epsilon_n(x + 1)$ to match the order of summation of $epsilon_n(x)$ for $n geq 2.$ However, is it true that regular convergence does not guarantee periodicity of period 1? This is hard to believe for some reason.
Furthermore, why does the fact that the terms converge to 0 imply that $epsilon_1$ is periodic of period 1?
sequences-and-series number-theory
asked yesterday
伽罗瓦
1,080615
I am confused by something I read. Let $epsilon_n(x) = sum_{mu = -infty}^infty (x + mu)^{-n}.$ The book says that $epsilon_n$ is absolutely convergent for $n geq 2$ hence, it is obvious that $epsilon_n$ is periodic of period 1 for such $n.$ However, it then says that the same is true for $epsilon_1$ as the terms of the series tend to $0$ as $mu rightarrow pm infty.$
Why do we need absolute convergence over regular convergence? I understand that absolute convergence allows us to add the terms in whatever order we want so we can rearrange the terms of $epsilon_n(x + 1)$ to match the order of summation of $epsilon_n(x)$ for $n geq 2.$ However, is it true that regular convergence does not guarantee periodicity of period 1? This is hard to believe for some reason.
Furthermore, why does the fact that the terms converge to 0 imply that $epsilon_1$ is periodic of period 1?
sequences-and-series number-theory
sequences-and-series number-theory
asked yesterday
伽罗瓦
1,080615
asked yesterday
伽罗瓦
1,080615
asked yesterday
伽罗瓦
1,080615
asked yesterday
伽罗瓦
1,080615
asked yesterday
伽罗瓦
1,080615
1,080615
I think regular convergence should be enough. In regards to $epsilon_1$, just write out the definitions. $epsilon_n(x+1)-epsilon_n(x) = lim_{M to infty} sum_{mu = -M}^M (x+1+mu)^{-n} - sum_{mu=-M}^M (x+mu)^{-n} = lim_{M to infty} (x+1+M)^{-n}-(x-M)^{-n} = 0$. You should be able to work these things out yourself by unravelling definitions
– mathworker21
yesterday
add a comment |
I think regular convergence should be enough. In regards to $epsilon_1$, just write out the definitions. $epsilon_n(x+1)-epsilon_n(x) = lim_{M to infty} sum_{mu = -M}^M (x+1+mu)^{-n} - sum_{mu=-M}^M (x+mu)^{-n} = lim_{M to infty} (x+1+M)^{-n}-(x-M)^{-n} = 0$. You should be able to work these things out yourself by unravelling definitions
– mathworker21
yesterday
I think regular convergence should be enough. In regards to $epsilon_1$, just write out the definitions. $epsilon_n(x+1)-epsilon_n(x) = lim_{M to infty} sum_{mu = -M}^M (x+1+mu)^{-n} - sum_{mu=-M}^M (x+mu)^{-n} = lim_{M to infty} (x+1+M)^{-n}-(x-M)^{-n} = 0$. You should be able to work these things out yourself by unravelling definitions
– mathworker21
yesterday
I think regular convergence should be enough. In regards to $epsilon_1$, just write out the definitions. $epsilon_n(x+1)-epsilon_n(x) = lim_{M to infty} sum_{mu = -M}^M (x+1+mu)^{-n} - sum_{mu=-M}^M (x+mu)^{-n} = lim_{M to infty} (x+1+M)^{-n}-(x-M)^{-n} = 0$. You should be able to work these things out yourself by unravelling definitions
– mathworker21
yesterday
add a comment |
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I think regular convergence should be enough. In regards to $epsilon_1$, just write out the definitions. $epsilon_n(x+1)-epsilon_n(x) = lim_{M to infty} sum_{mu = -M}^M (x+1+mu)^{-n} - sum_{mu=-M}^M (x+mu)^{-n} = lim_{M to infty} (x+1+M)^{-n}-(x-M)^{-n} = 0$. You should be able to work these things out yourself by unravelling definitions
– mathworker21
yesterday