Show $sum_{i=1}^infty x_i y_i$ is absolutely convergent
Let $sum_{i=1}^infty x_i$ be absolutely convergent and let $(y_n)$ be a sequence satisying $forall i in mathbb N, exists M in mathbb R$ such that $|y_i| leq M$. Then $sum_{i=1}^infty x_i y_i$ is absolutely convergent.
So $sum_{i=1}^infty x_i$ is absolutely convergent means both $sum_{i=1}^infty x_i$ and $sum_{i=1}^infty |x_i|$ is convergent. And the condition on y indicates y is bounded above.
For the proof, I have this idea: since if some series is convergent/divergent multiplication by a constant such as M does not change the convergence/divergence, I might be able to say that at worst case where $y_i$ is M, $sum_{i=1}^infty x_i y_i$ should have same convergent/divergent property with $x_n$. But I do not know how to show this proof formally.
real-analysis sequences-and-series convergence
asked Nov 23 '18 at 7:54
PumpkinPumpkin
5021417
add a comment |
Let $sum_{i=1}^infty x_i$ be absolutely convergent and let $(y_n)$ be a sequence satisying $forall i in mathbb N, exists M in mathbb R$ such that $|y_i| leq M$. Then $sum_{i=1}^infty x_i y_i$ is absolutely convergent.
So $sum_{i=1}^infty x_i$ is absolutely convergent means both $sum_{i=1}^infty x_i$ and $sum_{i=1}^infty |x_i|$ is convergent. And the condition on y indicates y is bounded above.
For the proof, I have this idea: since if some series is convergent/divergent multiplication by a constant such as M does not change the convergence/divergence, I might be able to say that at worst case where $y_i$ is M, $sum_{i=1}^infty x_i y_i$ should have same convergent/divergent property with $x_n$. But I do not know how to show this proof formally.
real-analysis sequences-and-series convergence
asked Nov 23 '18 at 7:54
PumpkinPumpkin
5021417
add a comment |
Let $sum_{i=1}^infty x_i$ be absolutely convergent and let $(y_n)$ be a sequence satisying $forall i in mathbb N, exists M in mathbb R$ such that $|y_i| leq M$. Then $sum_{i=1}^infty x_i y_i$ is absolutely convergent.
So $sum_{i=1}^infty x_i$ is absolutely convergent means both $sum_{i=1}^infty x_i$ and $sum_{i=1}^infty |x_i|$ is convergent. And the condition on y indicates y is bounded above.
For the proof, I have this idea: since if some series is convergent/divergent multiplication by a constant such as M does not change the convergence/divergence, I might be able to say that at worst case where $y_i$ is M, $sum_{i=1}^infty x_i y_i$ should have same convergent/divergent property with $x_n$. But I do not know how to show this proof formally.
real-analysis sequences-and-series convergence
asked Nov 23 '18 at 7:54
PumpkinPumpkin
5021417
Let $sum_{i=1}^infty x_i$ be absolutely convergent and let $(y_n)$ be a sequence satisying $forall i in mathbb N, exists M in mathbb R$ such that $|y_i| leq M$. Then $sum_{i=1}^infty x_i y_i$ is absolutely convergent.
So $sum_{i=1}^infty x_i$ is absolutely convergent means both $sum_{i=1}^infty x_i$ and $sum_{i=1}^infty |x_i|$ is convergent. And the condition on y indicates y is bounded above.
For the proof, I have this idea: since if some series is convergent/divergent multiplication by a constant such as M does not change the convergence/divergence, I might be able to say that at worst case where $y_i$ is M, $sum_{i=1}^infty x_i y_i$ should have same convergent/divergent property with $x_n$. But I do not know how to show this proof formally.
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
asked Nov 23 '18 at 7:54
PumpkinPumpkin
5021417
asked Nov 23 '18 at 7:54
PumpkinPumpkin
5021417
asked Nov 23 '18 at 7:54
PumpkinPumpkin
5021417
asked Nov 23 '18 at 7:54
PumpkinPumpkin
5021417
asked Nov 23 '18 at 7:54
PumpkinPumpkin
5021417
5021417
add a comment |
add a comment |
1 Answer
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A series $sum a_i$ is absolutely convergent if $sum |a_i| <infty$. $sum |x_iy_i| leq Msum |x_i| <infty$. So $sum x_iy_i$ is absolutely convergent. This is a complete proof.
answered Nov 23 '18 at 7:55
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
add a comment |
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1 Answer
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1 Answer
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active
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votes
A series $sum a_i$ is absolutely convergent if $sum |a_i| <infty$. $sum |x_iy_i| leq Msum |x_i| <infty$. So $sum x_iy_i$ is absolutely convergent. This is a complete proof.
answered Nov 23 '18 at 7:55
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
add a comment |
A series $sum a_i$ is absolutely convergent if $sum |a_i| <infty$. $sum |x_iy_i| leq Msum |x_i| <infty$. So $sum x_iy_i$ is absolutely convergent. This is a complete proof.
answered Nov 23 '18 at 7:55
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
add a comment |
A series $sum a_i$ is absolutely convergent if $sum |a_i| <infty$. $sum |x_iy_i| leq Msum |x_i| <infty$. So $sum x_iy_i$ is absolutely convergent. This is a complete proof.
answered Nov 23 '18 at 7:55
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
A series $sum a_i$ is absolutely convergent if $sum |a_i| <infty$. $sum |x_iy_i| leq Msum |x_i| <infty$. So $sum x_iy_i$ is absolutely convergent. This is a complete proof.
answered Nov 23 '18 at 7:55
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
answered Nov 23 '18 at 7:55
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
answered Nov 23 '18 at 7:55
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
answered Nov 23 '18 at 7:55
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
52.1k32055
add a comment |
add a comment |
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