Question about absolute convergence of series
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2
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Is this series absolutely convergent?
$sum (1/n)sin(npi /6)$.By Dirichlet test it is convergent.But how to conclude about absolutely convergent or not?
sequences-and-series
asked Nov 12 at 20:27
Believer
451214
add a comment |
up vote
2
down vote
favorite
Is this series absolutely convergent?
$sum (1/n)sin(npi /6)$.By Dirichlet test it is convergent.But how to conclude about absolutely convergent or not?
sequences-and-series
asked Nov 12 at 20:27
Believer
451214
Knowing and applying the Dirichlet test here is a little less trivial than directly calculating $sin(npi/6)$.
– Robert Wolfe
Nov 12 at 20:39
The sine values repeat periodically, so that you series can be absolutely bounded below by an Harmonic one.
– Yves Daoust
Nov 12 at 21:03
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is this series absolutely convergent?
$sum (1/n)sin(npi /6)$.By Dirichlet test it is convergent.But how to conclude about absolutely convergent or not?
sequences-and-series
asked Nov 12 at 20:27
Believer
451214
Is this series absolutely convergent?
$sum (1/n)sin(npi /6)$.By Dirichlet test it is convergent.But how to conclude about absolutely convergent or not?
sequences-and-series
sequences-and-series
asked Nov 12 at 20:27
Believer
451214
asked Nov 12 at 20:27
Believer
451214
asked Nov 12 at 20:27
Believer
451214
asked Nov 12 at 20:27
Believer
451214
asked Nov 12 at 20:27
Believer
451214
451214
Knowing and applying the Dirichlet test here is a little less trivial than directly calculating $sin(npi/6)$.
– Robert Wolfe
Nov 12 at 20:39
The sine values repeat periodically, so that you series can be absolutely bounded below by an Harmonic one.
– Yves Daoust
Nov 12 at 21:03
add a comment |
Knowing and applying the Dirichlet test here is a little less trivial than directly calculating $sin(npi/6)$.
– Robert Wolfe
Nov 12 at 20:39
The sine values repeat periodically, so that you series can be absolutely bounded below by an Harmonic one.
– Yves Daoust
Nov 12 at 21:03
Knowing and applying the Dirichlet test here is a little less trivial than directly calculating $sin(npi/6)$.
– Robert Wolfe
Nov 12 at 20:39
Knowing and applying the Dirichlet test here is a little less trivial than directly calculating $sin(npi/6)$.
– Robert Wolfe
Nov 12 at 20:39
The sine values repeat periodically, so that you series can be absolutely bounded below by an Harmonic one.
– Yves Daoust
Nov 12 at 21:03
The sine values repeat periodically, so that you series can be absolutely bounded below by an Harmonic one.
– Yves Daoust
Nov 12 at 21:03
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
$Sigma|frac{sin(frac{npi}{6})}{n}| = sin(frac{pi}{6}) + sin(frac{2pi}{6})/2 + 1/3 + ... +1/9 +...+1/15+... geq Sigma_{i = 0}^infty(frac{1}{6i + 3})$ which is not convergent.
answered Nov 12 at 20:42
araomis
3439
add a comment |
up vote
2
down vote
Note $|sin (npi/6)|= 1/2$ for $n=1,7,13,dots ...$ Thus
$$sum_{n=1}^{infty}frac{|sin (npi/6)|}{n} ge frac{1}{2}left(frac{1}{1} + frac{1}{7} + frac{1}{13} + cdotsright)=infty.$$
answered Nov 12 at 20:52
zhw.
70.3k43075
(+1) All to easy.
– Mark Viola
Nov 12 at 22:47
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$Sigma|frac{sin(frac{npi}{6})}{n}| = sin(frac{pi}{6}) + sin(frac{2pi}{6})/2 + 1/3 + ... +1/9 +...+1/15+... geq Sigma_{i = 0}^infty(frac{1}{6i + 3})$ which is not convergent.
answered Nov 12 at 20:42
araomis
3439
add a comment |
up vote
0
down vote
accepted
$Sigma|frac{sin(frac{npi}{6})}{n}| = sin(frac{pi}{6}) + sin(frac{2pi}{6})/2 + 1/3 + ... +1/9 +...+1/15+... geq Sigma_{i = 0}^infty(frac{1}{6i + 3})$ which is not convergent.
answered Nov 12 at 20:42
araomis
3439
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$Sigma|frac{sin(frac{npi}{6})}{n}| = sin(frac{pi}{6}) + sin(frac{2pi}{6})/2 + 1/3 + ... +1/9 +...+1/15+... geq Sigma_{i = 0}^infty(frac{1}{6i + 3})$ which is not convergent.
answered Nov 12 at 20:42
araomis
3439
$Sigma|frac{sin(frac{npi}{6})}{n}| = sin(frac{pi}{6}) + sin(frac{2pi}{6})/2 + 1/3 + ... +1/9 +...+1/15+... geq Sigma_{i = 0}^infty(frac{1}{6i + 3})$ which is not convergent.
answered Nov 12 at 20:42
araomis
3439
answered Nov 12 at 20:42
araomis
3439
answered Nov 12 at 20:42
araomis
3439
answered Nov 12 at 20:42
araomis
3439
3439
add a comment |
add a comment |
up vote
2
down vote
Note $|sin (npi/6)|= 1/2$ for $n=1,7,13,dots ...$ Thus
$$sum_{n=1}^{infty}frac{|sin (npi/6)|}{n} ge frac{1}{2}left(frac{1}{1} + frac{1}{7} + frac{1}{13} + cdotsright)=infty.$$
answered Nov 12 at 20:52
zhw.
70.3k43075
(+1) All to easy.
– Mark Viola
Nov 12 at 22:47
add a comment |
up vote
2
down vote
Note $|sin (npi/6)|= 1/2$ for $n=1,7,13,dots ...$ Thus
$$sum_{n=1}^{infty}frac{|sin (npi/6)|}{n} ge frac{1}{2}left(frac{1}{1} + frac{1}{7} + frac{1}{13} + cdotsright)=infty.$$
answered Nov 12 at 20:52
zhw.
70.3k43075
(+1) All to easy.
– Mark Viola
Nov 12 at 22:47
add a comment |
up vote
2
down vote
up vote
2
down vote
Note $|sin (npi/6)|= 1/2$ for $n=1,7,13,dots ...$ Thus
$$sum_{n=1}^{infty}frac{|sin (npi/6)|}{n} ge frac{1}{2}left(frac{1}{1} + frac{1}{7} + frac{1}{13} + cdotsright)=infty.$$
answered Nov 12 at 20:52
zhw.
70.3k43075
Note $|sin (npi/6)|= 1/2$ for $n=1,7,13,dots ...$ Thus
$$sum_{n=1}^{infty}frac{|sin (npi/6)|}{n} ge frac{1}{2}left(frac{1}{1} + frac{1}{7} + frac{1}{13} + cdotsright)=infty.$$
answered Nov 12 at 20:52
zhw.
70.3k43075
answered Nov 12 at 20:52
zhw.
70.3k43075
answered Nov 12 at 20:52
zhw.
70.3k43075
answered Nov 12 at 20:52
zhw.
70.3k43075
70.3k43075
(+1) All to easy.
– Mark Viola
Nov 12 at 22:47
add a comment |
(+1) All to easy.
– Mark Viola
Nov 12 at 22:47
(+1) All to easy.
– Mark Viola
Nov 12 at 22:47
(+1) All to easy.
– Mark Viola
Nov 12 at 22:47
add a comment |
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Knowing and applying the Dirichlet test here is a little less trivial than directly calculating $sin(npi/6)$.
– Robert Wolfe
Nov 12 at 20:39
The sine values repeat periodically, so that you series can be absolutely bounded below by an Harmonic one.
– Yves Daoust
Nov 12 at 21:03