Confusion if the series converges or not (alternating series test)
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I have to test for convergence and absolute convergence for the following series:
$$sum_k=1^infty (-1)^k frack1+2k^2$$
Because of the alternating series test, I have to verify if the series decreases monotonically and then show that the limit goes to zero.
I don't have any problems showing that it decreases monotonically, but I have trouble showing if the limit is zero.
$$lim_xtoinfty frack1+2k^2 = frac12k rightarrow 0$$
Therefore it converges.
But with the direct comparison test it is:
$$ mid (-1)^k frack1+2k^2mid = frack1+2k^2 =frac1frac1k+2kgeq frac1k+2k = frac13k$$
Which is similar to the harmonic series$sum_k=1^inftyfrac1k$ hence the series should diverge.
So my question is, does it diverge or converge? And how do I know if it converges absolutely?
Thank you for your time and help!
calculus convergence absolute-convergence
asked Sep 5 at 10:23
ina26
847
add a comment |Â
up vote
1
down vote
favorite
I have to test for convergence and absolute convergence for the following series:
$$sum_k=1^infty (-1)^k frack1+2k^2$$
Because of the alternating series test, I have to verify if the series decreases monotonically and then show that the limit goes to zero.
I don't have any problems showing that it decreases monotonically, but I have trouble showing if the limit is zero.
$$lim_xtoinfty frack1+2k^2 = frac12k rightarrow 0$$
Therefore it converges.
But with the direct comparison test it is:
$$ mid (-1)^k frack1+2k^2mid = frack1+2k^2 =frac1frac1k+2kgeq frac1k+2k = frac13k$$
Which is similar to the harmonic series$sum_k=1^inftyfrac1k$ hence the series should diverge.
So my question is, does it diverge or converge? And how do I know if it converges absolutely?
Thank you for your time and help!
calculus convergence absolute-convergence
asked Sep 5 at 10:23
ina26
847
Yes, the series diverges absolutely. The alternating series test tells you only if it converges. You must decide if it converges conditionally or absolutely...
– PhysicsMathsLove
Sep 5 at 10:25
1
You showed that the series converges, but it does not converge absolutely.
– mechanodroid
Sep 5 at 10:25
Indeed you have already proved that the series is convergent but not absolutely convergent.
– Rigel
Sep 5 at 10:26
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have to test for convergence and absolute convergence for the following series:
$$sum_k=1^infty (-1)^k frack1+2k^2$$
Because of the alternating series test, I have to verify if the series decreases monotonically and then show that the limit goes to zero.
I don't have any problems showing that it decreases monotonically, but I have trouble showing if the limit is zero.
$$lim_xtoinfty frack1+2k^2 = frac12k rightarrow 0$$
Therefore it converges.
But with the direct comparison test it is:
$$ mid (-1)^k frack1+2k^2mid = frack1+2k^2 =frac1frac1k+2kgeq frac1k+2k = frac13k$$
Which is similar to the harmonic series$sum_k=1^inftyfrac1k$ hence the series should diverge.
So my question is, does it diverge or converge? And how do I know if it converges absolutely?
Thank you for your time and help!
calculus convergence absolute-convergence
asked Sep 5 at 10:23
ina26
847
I have to test for convergence and absolute convergence for the following series:
$$sum_k=1^infty (-1)^k frack1+2k^2$$
Because of the alternating series test, I have to verify if the series decreases monotonically and then show that the limit goes to zero.
I don't have any problems showing that it decreases monotonically, but I have trouble showing if the limit is zero.
$$lim_xtoinfty frack1+2k^2 = frac12k rightarrow 0$$
Therefore it converges.
But with the direct comparison test it is:
$$ mid (-1)^k frack1+2k^2mid = frack1+2k^2 =frac1frac1k+2kgeq frac1k+2k = frac13k$$
Which is similar to the harmonic series$sum_k=1^inftyfrac1k$ hence the series should diverge.
So my question is, does it diverge or converge? And how do I know if it converges absolutely?
Thank you for your time and help!
calculus convergence absolute-convergence
calculus convergence absolute-convergence
asked Sep 5 at 10:23
ina26
847
asked Sep 5 at 10:23
ina26
847
asked Sep 5 at 10:23
ina26
847
asked Sep 5 at 10:23
ina26
847
asked Sep 5 at 10:23
ina26
847
847
Yes, the series diverges absolutely. The alternating series test tells you only if it converges. You must decide if it converges conditionally or absolutely...
– PhysicsMathsLove
Sep 5 at 10:25
1
You showed that the series converges, but it does not converge absolutely.
– mechanodroid
Sep 5 at 10:25
Indeed you have already proved that the series is convergent but not absolutely convergent.
– Rigel
Sep 5 at 10:26
add a comment |Â
Yes, the series diverges absolutely. The alternating series test tells you only if it converges. You must decide if it converges conditionally or absolutely...
– PhysicsMathsLove
Sep 5 at 10:25
1
You showed that the series converges, but it does not converge absolutely.
– mechanodroid
Sep 5 at 10:25
Indeed you have already proved that the series is convergent but not absolutely convergent.
– Rigel
Sep 5 at 10:26
Yes, the series diverges absolutely. The alternating series test tells you only if it converges. You must decide if it converges conditionally or absolutely...
– PhysicsMathsLove
Sep 5 at 10:25
Yes, the series diverges absolutely. The alternating series test tells you only if it converges. You must decide if it converges conditionally or absolutely...
– PhysicsMathsLove
Sep 5 at 10:25
1
1
You showed that the series converges, but it does not converge absolutely.
– mechanodroid
Sep 5 at 10:25
You showed that the series converges, but it does not converge absolutely.
– mechanodroid
Sep 5 at 10:25
Indeed you have already proved that the series is convergent but not absolutely convergent.
– Rigel
Sep 5 at 10:26
Indeed you have already proved that the series is convergent but not absolutely convergent.
– Rigel
Sep 5 at 10:26
add a comment |Â
1 Answer
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You're doing nothing wrong.
Since $|a_k|$ decreases monotonically, and $a_k to 0$, you can conclude that $sum a_k$ converges, by the alternating series test.
Now, we say that a series $sum a_k$ coverges absolutely if $sum |a_k|$ converges. But as you show, $sum |a_k|$ doesn't converge, so we have a conditionally convergent series, one that converges, but not absolutely.
answered Sep 5 at 10:36
JuliusL33t
1,260817
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You're doing nothing wrong.
Since $|a_k|$ decreases monotonically, and $a_k to 0$, you can conclude that $sum a_k$ converges, by the alternating series test.
Now, we say that a series $sum a_k$ coverges absolutely if $sum |a_k|$ converges. But as you show, $sum |a_k|$ doesn't converge, so we have a conditionally convergent series, one that converges, but not absolutely.
answered Sep 5 at 10:36
JuliusL33t
1,260817
add a comment |Â
up vote
2
down vote
accepted
You're doing nothing wrong.
Since $|a_k|$ decreases monotonically, and $a_k to 0$, you can conclude that $sum a_k$ converges, by the alternating series test.
Now, we say that a series $sum a_k$ coverges absolutely if $sum |a_k|$ converges. But as you show, $sum |a_k|$ doesn't converge, so we have a conditionally convergent series, one that converges, but not absolutely.
answered Sep 5 at 10:36
JuliusL33t
1,260817
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You're doing nothing wrong.
Since $|a_k|$ decreases monotonically, and $a_k to 0$, you can conclude that $sum a_k$ converges, by the alternating series test.
Now, we say that a series $sum a_k$ coverges absolutely if $sum |a_k|$ converges. But as you show, $sum |a_k|$ doesn't converge, so we have a conditionally convergent series, one that converges, but not absolutely.
answered Sep 5 at 10:36
JuliusL33t
1,260817
You're doing nothing wrong.
Since $|a_k|$ decreases monotonically, and $a_k to 0$, you can conclude that $sum a_k$ converges, by the alternating series test.
Now, we say that a series $sum a_k$ coverges absolutely if $sum |a_k|$ converges. But as you show, $sum |a_k|$ doesn't converge, so we have a conditionally convergent series, one that converges, but not absolutely.
answered Sep 5 at 10:36
JuliusL33t
1,260817
answered Sep 5 at 10:36
JuliusL33t
1,260817
answered Sep 5 at 10:36
JuliusL33t
1,260817
answered Sep 5 at 10:36
JuliusL33t
1,260817
1,260817
add a comment |Â
add a comment |Â
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Yes, the series diverges absolutely. The alternating series test tells you only if it converges. You must decide if it converges conditionally or absolutely...
– PhysicsMathsLove
Sep 5 at 10:25
1
You showed that the series converges, but it does not converge absolutely.
– mechanodroid
Sep 5 at 10:25
Indeed you have already proved that the series is convergent but not absolutely convergent.
– Rigel
Sep 5 at 10:26