How to prove that for any $nin mathbb{N}$ and any $p>0$, $sum_{k=1}^{n}k^{-p} > int_{1}^{n+1}x^{-p}dx...
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Here is a question for my Math homework,
Prove that for any $nin mathbb{N}$ and any $p>0$, $sum_{k=1}^{n}k^{-p} > int_{1}^{n+1}x^{-p}dx > sum_{k=2}^{n+1}k^{-p}$
I want to use methods learned from Sequence and Series and limits to prove this.
calculus sequences-and-series
asked Jan 18 at 8:37
tjysdsgtjysdsg
33
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add a comment |
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Here is a question for my Math homework,
Prove that for any $nin mathbb{N}$ and any $p>0$, $sum_{k=1}^{n}k^{-p} > int_{1}^{n+1}x^{-p}dx > sum_{k=2}^{n+1}k^{-p}$
I want to use methods learned from Sequence and Series and limits to prove this.
calculus sequences-and-series
asked Jan 18 at 8:37
tjysdsgtjysdsg
33
$endgroup$
2
$begingroup$
elcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
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– José Carlos Santos
Jan 18 at 8:41
add a comment |
$begingroup$
Here is a question for my Math homework,
Prove that for any $nin mathbb{N}$ and any $p>0$, $sum_{k=1}^{n}k^{-p} > int_{1}^{n+1}x^{-p}dx > sum_{k=2}^{n+1}k^{-p}$
I want to use methods learned from Sequence and Series and limits to prove this.
calculus sequences-and-series
asked Jan 18 at 8:37
tjysdsgtjysdsg
33
$endgroup$
Here is a question for my Math homework,
Prove that for any $nin mathbb{N}$ and any $p>0$, $sum_{k=1}^{n}k^{-p} > int_{1}^{n+1}x^{-p}dx > sum_{k=2}^{n+1}k^{-p}$
I want to use methods learned from Sequence and Series and limits to prove this.
calculus sequences-and-series
calculus sequences-and-series
asked Jan 18 at 8:37
tjysdsgtjysdsg
33
asked Jan 18 at 8:37
tjysdsgtjysdsg
33
edited Jan 18 at 8:43
tjysdsg
asked Jan 18 at 8:37
tjysdsgtjysdsg
33
asked Jan 18 at 8:37
tjysdsgtjysdsg
33
asked Jan 18 at 8:37
tjysdsgtjysdsg
33
33
2
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elcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
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– José Carlos Santos
Jan 18 at 8:41
add a comment |
2
$begingroup$
elcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Jan 18 at 8:41
2
2
$begingroup$
elcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
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– José Carlos Santos
Jan 18 at 8:41
$begingroup$
elcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
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– José Carlos Santos
Jan 18 at 8:41
add a comment |
2 Answers
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$$sum_{k=1}^{n}int_{k}^{k+1}(k+1)^{-p}dx<int_{1}^{n+1}x^{-p}dx=sum_{k=1}^{n}int_{k}^{k+1}x^{-p}dx<sum_{k=1}^{n}int_{k}^{k+1}k^{-p}dx,$$
$$ sum_{k=2}^{n+1}k^{-p}< int_{1}^{n+1}x^{-p}dx <sum_{k=1}^{n}k^{-p}. $$
answered Jan 18 at 8:43
RiemannRiemann
3,5671422
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Just split the integral into integrals over $(j,j+1)$ and note that $(j+1)^{-p}<x^{-p} <j^{-p}$ on $(j,j+1)$.
answered Jan 18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
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2 Answers
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2 Answers
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$begingroup$
$$sum_{k=1}^{n}int_{k}^{k+1}(k+1)^{-p}dx<int_{1}^{n+1}x^{-p}dx=sum_{k=1}^{n}int_{k}^{k+1}x^{-p}dx<sum_{k=1}^{n}int_{k}^{k+1}k^{-p}dx,$$
$$ sum_{k=2}^{n+1}k^{-p}< int_{1}^{n+1}x^{-p}dx <sum_{k=1}^{n}k^{-p}. $$
answered Jan 18 at 8:43
RiemannRiemann
3,5671422
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$begingroup$
$$sum_{k=1}^{n}int_{k}^{k+1}(k+1)^{-p}dx<int_{1}^{n+1}x^{-p}dx=sum_{k=1}^{n}int_{k}^{k+1}x^{-p}dx<sum_{k=1}^{n}int_{k}^{k+1}k^{-p}dx,$$
$$ sum_{k=2}^{n+1}k^{-p}< int_{1}^{n+1}x^{-p}dx <sum_{k=1}^{n}k^{-p}. $$
answered Jan 18 at 8:43
RiemannRiemann
3,5671422
$endgroup$
add a comment |
$begingroup$
$$sum_{k=1}^{n}int_{k}^{k+1}(k+1)^{-p}dx<int_{1}^{n+1}x^{-p}dx=sum_{k=1}^{n}int_{k}^{k+1}x^{-p}dx<sum_{k=1}^{n}int_{k}^{k+1}k^{-p}dx,$$
$$ sum_{k=2}^{n+1}k^{-p}< int_{1}^{n+1}x^{-p}dx <sum_{k=1}^{n}k^{-p}. $$
answered Jan 18 at 8:43
RiemannRiemann
3,5671422
$endgroup$
$$sum_{k=1}^{n}int_{k}^{k+1}(k+1)^{-p}dx<int_{1}^{n+1}x^{-p}dx=sum_{k=1}^{n}int_{k}^{k+1}x^{-p}dx<sum_{k=1}^{n}int_{k}^{k+1}k^{-p}dx,$$
$$ sum_{k=2}^{n+1}k^{-p}< int_{1}^{n+1}x^{-p}dx <sum_{k=1}^{n}k^{-p}. $$
answered Jan 18 at 8:43
RiemannRiemann
3,5671422
answered Jan 18 at 8:43
RiemannRiemann
3,5671422
answered Jan 18 at 8:43
RiemannRiemann
3,5671422
answered Jan 18 at 8:43
RiemannRiemann
3,5671422
3,5671422
add a comment |
add a comment |
$begingroup$
Just split the integral into integrals over $(j,j+1)$ and note that $(j+1)^{-p}<x^{-p} <j^{-p}$ on $(j,j+1)$.
answered Jan 18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
$endgroup$
add a comment |
$begingroup$
Just split the integral into integrals over $(j,j+1)$ and note that $(j+1)^{-p}<x^{-p} <j^{-p}$ on $(j,j+1)$.
answered Jan 18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
$endgroup$
add a comment |
$begingroup$
Just split the integral into integrals over $(j,j+1)$ and note that $(j+1)^{-p}<x^{-p} <j^{-p}$ on $(j,j+1)$.
answered Jan 18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
$endgroup$
Just split the integral into integrals over $(j,j+1)$ and note that $(j+1)^{-p}<x^{-p} <j^{-p}$ on $(j,j+1)$.
answered Jan 18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
answered Jan 18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
answered Jan 18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
answered Jan 18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
75.5k53270
add a comment |
add a comment |
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elcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
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– José Carlos Santos
Jan 18 at 8:41