Theorem 3.29 in Baby Rudin
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Theorem 3.29 in Walter Rudin's Principles of Mathematical Analysis, 3rd ed., states that
If $p>1$, then the series $$sum_{n=2}^infty frac{1}{n (log n)^p} $$ converges; if $p leq 1$, the series diverges.
Now in the proof, Rudin only seems to discuss the case when $p> 0$, for it is only in this particular case that we can use the Cauchy Condensation Test.
How to deal with the case of $p<0$?
Of course, the case $p=0$ yields the divergent harmonic series.
calculus real-analysis sequences-and-series analysis convergence
asked Oct 22 '14 at 4:36
Saaqib MahmoodSaaqib Mahmood
7,95942581
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add a comment |
$begingroup$
Theorem 3.29 in Walter Rudin's Principles of Mathematical Analysis, 3rd ed., states that
If $p>1$, then the series $$sum_{n=2}^infty frac{1}{n (log n)^p} $$ converges; if $p leq 1$, the series diverges.
Now in the proof, Rudin only seems to discuss the case when $p> 0$, for it is only in this particular case that we can use the Cauchy Condensation Test.
How to deal with the case of $p<0$?
Of course, the case $p=0$ yields the divergent harmonic series.
calculus real-analysis sequences-and-series analysis convergence
asked Oct 22 '14 at 4:36
Saaqib MahmoodSaaqib Mahmood
7,95942581
$endgroup$
add a comment |
$begingroup$
Theorem 3.29 in Walter Rudin's Principles of Mathematical Analysis, 3rd ed., states that
If $p>1$, then the series $$sum_{n=2}^infty frac{1}{n (log n)^p} $$ converges; if $p leq 1$, the series diverges.
Now in the proof, Rudin only seems to discuss the case when $p> 0$, for it is only in this particular case that we can use the Cauchy Condensation Test.
How to deal with the case of $p<0$?
Of course, the case $p=0$ yields the divergent harmonic series.
calculus real-analysis sequences-and-series analysis convergence
asked Oct 22 '14 at 4:36
Saaqib MahmoodSaaqib Mahmood
7,95942581
$endgroup$
Theorem 3.29 in Walter Rudin's Principles of Mathematical Analysis, 3rd ed., states that
If $p>1$, then the series $$sum_{n=2}^infty frac{1}{n (log n)^p} $$ converges; if $p leq 1$, the series diverges.
Now in the proof, Rudin only seems to discuss the case when $p> 0$, for it is only in this particular case that we can use the Cauchy Condensation Test.
How to deal with the case of $p<0$?
Of course, the case $p=0$ yields the divergent harmonic series.
calculus real-analysis sequences-and-series analysis convergence
calculus real-analysis sequences-and-series analysis convergence
asked Oct 22 '14 at 4:36
Saaqib MahmoodSaaqib Mahmood
7,95942581
asked Oct 22 '14 at 4:36
Saaqib MahmoodSaaqib Mahmood
7,95942581
edited Dec 26 '18 at 18:22
Saaqib Mahmood
asked Oct 22 '14 at 4:36
Saaqib MahmoodSaaqib Mahmood
7,95942581
asked Oct 22 '14 at 4:36
Saaqib MahmoodSaaqib Mahmood
7,95942581
asked Oct 22 '14 at 4:36
Saaqib MahmoodSaaqib Mahmood
7,95942581
7,95942581
add a comment |
add a comment |
2 Answers
2
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oldest
votes
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The case of $p<0$ is trivial, since
$$sum_{k=2}^infty frac{ln^{-p} k}{k}gesum_{k=4}^infty frac{1}{k}.$$
answered Oct 22 '14 at 4:38
Hanul JeonHanul Jeon
17.7k42881
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add a comment |
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You can use the integral test:
$$ sum_{n geq 2 }frac{1}{n ( log n )^p} cong intlimits_2^{infty} frac{dz}{z (log z )^p} = intlimits_{2}^{infty} frac{ d(log z)}{ ( log z )^p }= ...$$
$endgroup$
2
$begingroup$
The integral test is not to be used; I'd like to do a a theorem in Rudin and so use the same armory that he's given me uptil this point.
$endgroup$
– Saaqib Mahmood
Oct 25 '14 at 3:38
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
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active
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votes
$begingroup$
The case of $p<0$ is trivial, since
$$sum_{k=2}^infty frac{ln^{-p} k}{k}gesum_{k=4}^infty frac{1}{k}.$$
answered Oct 22 '14 at 4:38
Hanul JeonHanul Jeon
17.7k42881
$endgroup$
add a comment |
$begingroup$
The case of $p<0$ is trivial, since
$$sum_{k=2}^infty frac{ln^{-p} k}{k}gesum_{k=4}^infty frac{1}{k}.$$
answered Oct 22 '14 at 4:38
Hanul JeonHanul Jeon
17.7k42881
$endgroup$
add a comment |
$begingroup$
The case of $p<0$ is trivial, since
$$sum_{k=2}^infty frac{ln^{-p} k}{k}gesum_{k=4}^infty frac{1}{k}.$$
answered Oct 22 '14 at 4:38
Hanul JeonHanul Jeon
17.7k42881
$endgroup$
The case of $p<0$ is trivial, since
$$sum_{k=2}^infty frac{ln^{-p} k}{k}gesum_{k=4}^infty frac{1}{k}.$$
answered Oct 22 '14 at 4:38
Hanul JeonHanul Jeon
17.7k42881
answered Oct 22 '14 at 4:38
Hanul JeonHanul Jeon
17.7k42881
answered Oct 22 '14 at 4:38
Hanul JeonHanul Jeon
17.7k42881
answered Oct 22 '14 at 4:38
Hanul JeonHanul Jeon
17.7k42881
17.7k42881
add a comment |
add a comment |
$begingroup$
You can use the integral test:
$$ sum_{n geq 2 }frac{1}{n ( log n )^p} cong intlimits_2^{infty} frac{dz}{z (log z )^p} = intlimits_{2}^{infty} frac{ d(log z)}{ ( log z )^p }= ...$$
$endgroup$
2
$begingroup$
The integral test is not to be used; I'd like to do a a theorem in Rudin and so use the same armory that he's given me uptil this point.
$endgroup$
– Saaqib Mahmood
Oct 25 '14 at 3:38
add a comment |
$begingroup$
You can use the integral test:
$$ sum_{n geq 2 }frac{1}{n ( log n )^p} cong intlimits_2^{infty} frac{dz}{z (log z )^p} = intlimits_{2}^{infty} frac{ d(log z)}{ ( log z )^p }= ...$$
$endgroup$
2
$begingroup$
The integral test is not to be used; I'd like to do a a theorem in Rudin and so use the same armory that he's given me uptil this point.
$endgroup$
– Saaqib Mahmood
Oct 25 '14 at 3:38
add a comment |
$begingroup$
You can use the integral test:
$$ sum_{n geq 2 }frac{1}{n ( log n )^p} cong intlimits_2^{infty} frac{dz}{z (log z )^p} = intlimits_{2}^{infty} frac{ d(log z)}{ ( log z )^p }= ...$$
$endgroup$
You can use the integral test:
$$ sum_{n geq 2 }frac{1}{n ( log n )^p} cong intlimits_2^{infty} frac{dz}{z (log z )^p} = intlimits_{2}^{infty} frac{ d(log z)}{ ( log z )^p }= ...$$
answered Oct 22 '14 at 4:42
user139708
2
$begingroup$
The integral test is not to be used; I'd like to do a a theorem in Rudin and so use the same armory that he's given me uptil this point.
$endgroup$
– Saaqib Mahmood
Oct 25 '14 at 3:38
add a comment |
2
$begingroup$
The integral test is not to be used; I'd like to do a a theorem in Rudin and so use the same armory that he's given me uptil this point.
$endgroup$
– Saaqib Mahmood
Oct 25 '14 at 3:38
2
2
$begingroup$
The integral test is not to be used; I'd like to do a a theorem in Rudin and so use the same armory that he's given me uptil this point.
$endgroup$
– Saaqib Mahmood
Oct 25 '14 at 3:38
$begingroup$
The integral test is not to be used; I'd like to do a a theorem in Rudin and so use the same armory that he's given me uptil this point.
$endgroup$
– Saaqib Mahmood
Oct 25 '14 at 3:38
add a comment |
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