How to show that if $sum a_n^{3/2}$ is bounded implies $sum a_n/n$ is bounded?
0
$begingroup$
How to show that if $sum a_n^{3/2}$ is bounded implies $sum a_n/n$ is bounded?
Actually I do not able to compare both series .SO could not able to solve above problem?
What should be strategy to solve this problem?
ANy Help will be appreciated
real-analysis sequences-and-series
asked Jan 18 at 13:08
MathLoverMathLover
52510
$endgroup$
add a comment |
0
$begingroup$
How to show that if $sum a_n^{3/2}$ is bounded implies $sum a_n/n$ is bounded?
Actually I do not able to compare both series .SO could not able to solve above problem?
What should be strategy to solve this problem?
ANy Help will be appreciated
real-analysis sequences-and-series
asked Jan 18 at 13:08
MathLoverMathLover
52510
$endgroup$
add a comment |
0
0
0
$begingroup$
How to show that if $sum a_n^{3/2}$ is bounded implies $sum a_n/n$ is bounded?
Actually I do not able to compare both series .SO could not able to solve above problem?
What should be strategy to solve this problem?
ANy Help will be appreciated
real-analysis sequences-and-series
asked Jan 18 at 13:08
MathLoverMathLover
52510
$endgroup$
How to show that if $sum a_n^{3/2}$ is bounded implies $sum a_n/n$ is bounded?
Actually I do not able to compare both series .SO could not able to solve above problem?
What should be strategy to solve this problem?
ANy Help will be appreciated
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Jan 18 at 13:08
MathLoverMathLover
52510
asked Jan 18 at 13:08
MathLoverMathLover
52510
asked Jan 18 at 13:08
MathLoverMathLover
52510
asked Jan 18 at 13:08
MathLoverMathLover
52510
asked Jan 18 at 13:08
MathLoverMathLover
52510
52510
add a comment |
add a comment |
1 Answer
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2
$begingroup$
Try Hölder's inequality (assuming that $a_n$ is positive).
answered Jan 18 at 13:16
KlausKlaus
1,5349
$endgroup$
1
$begingroup$
Is it mean that $sum |a_n/n|<sum |a_n^{3/2}||sum 1/n^3|^{1/3 }$ which is bounded
$endgroup$
– MathLover
Jan 18 at 13:23
$begingroup$
You lost a power of $frac{2}{3}$ for the first sum, but otherwise yes.
$endgroup$
– Klaus
Jan 18 at 13:23
$begingroup$
yes sorry I forget to write
$endgroup$
– MathLover
Jan 18 at 13:27
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Try Hölder's inequality (assuming that $a_n$ is positive).
answered Jan 18 at 13:16
KlausKlaus
1,5349
$endgroup$
1
$begingroup$
Is it mean that $sum |a_n/n|<sum |a_n^{3/2}||sum 1/n^3|^{1/3 }$ which is bounded
$endgroup$
– MathLover
Jan 18 at 13:23
$begingroup$
You lost a power of $frac{2}{3}$ for the first sum, but otherwise yes.
$endgroup$
– Klaus
Jan 18 at 13:23
$begingroup$
yes sorry I forget to write
$endgroup$
– MathLover
Jan 18 at 13:27
add a comment |
2
$begingroup$
Try Hölder's inequality (assuming that $a_n$ is positive).
answered Jan 18 at 13:16
KlausKlaus
1,5349
$endgroup$
1
$begingroup$
Is it mean that $sum |a_n/n|<sum |a_n^{3/2}||sum 1/n^3|^{1/3 }$ which is bounded
$endgroup$
– MathLover
Jan 18 at 13:23
$begingroup$
You lost a power of $frac{2}{3}$ for the first sum, but otherwise yes.
$endgroup$
– Klaus
Jan 18 at 13:23
$begingroup$
yes sorry I forget to write
$endgroup$
– MathLover
Jan 18 at 13:27
add a comment |
2
2
2
$begingroup$
Try Hölder's inequality (assuming that $a_n$ is positive).
answered Jan 18 at 13:16
KlausKlaus
1,5349
$endgroup$
Try Hölder's inequality (assuming that $a_n$ is positive).
answered Jan 18 at 13:16
KlausKlaus
1,5349
answered Jan 18 at 13:16
KlausKlaus
1,5349
answered Jan 18 at 13:16
KlausKlaus
1,5349
answered Jan 18 at 13:16
KlausKlaus
1,5349
1,5349
1
$begingroup$
Is it mean that $sum |a_n/n|<sum |a_n^{3/2}||sum 1/n^3|^{1/3 }$ which is bounded
$endgroup$
– MathLover
Jan 18 at 13:23
$begingroup$
You lost a power of $frac{2}{3}$ for the first sum, but otherwise yes.
$endgroup$
– Klaus
Jan 18 at 13:23
$begingroup$
yes sorry I forget to write
$endgroup$
– MathLover
Jan 18 at 13:27
add a comment |
1
$begingroup$
Is it mean that $sum |a_n/n|<sum |a_n^{3/2}||sum 1/n^3|^{1/3 }$ which is bounded
$endgroup$
– MathLover
Jan 18 at 13:23
$begingroup$
You lost a power of $frac{2}{3}$ for the first sum, but otherwise yes.
$endgroup$
– Klaus
Jan 18 at 13:23
$begingroup$
yes sorry I forget to write
$endgroup$
– MathLover
Jan 18 at 13:27
1
1
$begingroup$
Is it mean that $sum |a_n/n|<sum |a_n^{3/2}||sum 1/n^3|^{1/3 }$ which is bounded
$endgroup$
– MathLover
Jan 18 at 13:23
$begingroup$
Is it mean that $sum |a_n/n|<sum |a_n^{3/2}||sum 1/n^3|^{1/3 }$ which is bounded
$endgroup$
– MathLover
Jan 18 at 13:23
$begingroup$
You lost a power of $frac{2}{3}$ for the first sum, but otherwise yes.
$endgroup$
– Klaus
Jan 18 at 13:23
$begingroup$
You lost a power of $frac{2}{3}$ for the first sum, but otherwise yes.
$endgroup$
– Klaus
Jan 18 at 13:23
$begingroup$
yes sorry I forget to write
$endgroup$
– MathLover
Jan 18 at 13:27
$begingroup$
yes sorry I forget to write
$endgroup$
– MathLover
Jan 18 at 13:27
add a comment |
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