Infinite Series Diverges By Divergence Test But Converges By Limit Comparison Test
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I understand why this infinite series diverges by the divergence test but I can't find fault in my limit comparison test which says it diverges. Please help. Thanks
P.S. if my handwriting threw you off the original equation is
$$sum_{n=0}^{infty} (-1)^n cdot frac{n^4}{n^3 + 1}.$$
calculus sequences-and-series convergence divergent-series absolute-convergence
edited Nov 22 at 6:17
Robert Z
92.1k1058129
asked Nov 22 at 6:11
SeanCorc
1
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up vote
0
down vote
favorite
Image of My Work
I understand why this infinite series diverges by the divergence test but I can't find fault in my limit comparison test which says it diverges. Please help. Thanks
P.S. if my handwriting threw you off the original equation is
$$sum_{n=0}^{infty} (-1)^n cdot frac{n^4}{n^3 + 1}.$$
calculus sequences-and-series convergence divergent-series absolute-convergence
edited Nov 22 at 6:17
Robert Z
92.1k1058129
asked Nov 22 at 6:11
SeanCorc
1
I think you have a fundamental misunderstanding. First, the limit comparison is only going to work if all of the terms are positive (you might take absolute values, but this changes the problem. In some cases, this change means the difference between convergence and divergence.) Second, the limit of the sequence converges, but the limit of the sum doesn’t converge... At any rate, you compared with $b_n=n$. $sum b_n$ diverges (you might need to think about the statement of the tests to wrap your head around what I mean.)
– Clayton
Nov 22 at 6:21
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Image of My Work
I understand why this infinite series diverges by the divergence test but I can't find fault in my limit comparison test which says it diverges. Please help. Thanks
P.S. if my handwriting threw you off the original equation is
$$sum_{n=0}^{infty} (-1)^n cdot frac{n^4}{n^3 + 1}.$$
calculus sequences-and-series convergence divergent-series absolute-convergence
edited Nov 22 at 6:17
Robert Z
92.1k1058129
asked Nov 22 at 6:11
SeanCorc
1
Image of My Work
I understand why this infinite series diverges by the divergence test but I can't find fault in my limit comparison test which says it diverges. Please help. Thanks
P.S. if my handwriting threw you off the original equation is
$$sum_{n=0}^{infty} (-1)^n cdot frac{n^4}{n^3 + 1}.$$
calculus sequences-and-series convergence divergent-series absolute-convergence
calculus sequences-and-series convergence divergent-series absolute-convergence
edited Nov 22 at 6:17
Robert Z
92.1k1058129
asked Nov 22 at 6:11
SeanCorc
1
edited Nov 22 at 6:17
Robert Z
92.1k1058129
asked Nov 22 at 6:11
SeanCorc
1
edited Nov 22 at 6:17
Robert Z
92.1k1058129
edited Nov 22 at 6:17
Robert Z
92.1k1058129
edited Nov 22 at 6:17
Robert Z
92.1k1058129
92.1k1058129
asked Nov 22 at 6:11
SeanCorc
1
asked Nov 22 at 6:11
SeanCorc
1
asked Nov 22 at 6:11
SeanCorc
1
1
I think you have a fundamental misunderstanding. First, the limit comparison is only going to work if all of the terms are positive (you might take absolute values, but this changes the problem. In some cases, this change means the difference between convergence and divergence.) Second, the limit of the sequence converges, but the limit of the sum doesn’t converge... At any rate, you compared with $b_n=n$. $sum b_n$ diverges (you might need to think about the statement of the tests to wrap your head around what I mean.)
– Clayton
Nov 22 at 6:21
add a comment |
I think you have a fundamental misunderstanding. First, the limit comparison is only going to work if all of the terms are positive (you might take absolute values, but this changes the problem. In some cases, this change means the difference between convergence and divergence.) Second, the limit of the sequence converges, but the limit of the sum doesn’t converge... At any rate, you compared with $b_n=n$. $sum b_n$ diverges (you might need to think about the statement of the tests to wrap your head around what I mean.)
– Clayton
Nov 22 at 6:21
I think you have a fundamental misunderstanding. First, the limit comparison is only going to work if all of the terms are positive (you might take absolute values, but this changes the problem. In some cases, this change means the difference between convergence and divergence.) Second, the limit of the sequence converges, but the limit of the sum doesn’t converge... At any rate, you compared with $b_n=n$. $sum b_n$ diverges (you might need to think about the statement of the tests to wrap your head around what I mean.)
– Clayton
Nov 22 at 6:21
I think you have a fundamental misunderstanding. First, the limit comparison is only going to work if all of the terms are positive (you might take absolute values, but this changes the problem. In some cases, this change means the difference between convergence and divergence.) Second, the limit of the sequence converges, but the limit of the sum doesn’t converge... At any rate, you compared with $b_n=n$. $sum b_n$ diverges (you might need to think about the statement of the tests to wrap your head around what I mean.)
– Clayton
Nov 22 at 6:21
add a comment |
2 Answers
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up vote
1
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There's nothing wrong with your calculations. You did well to find out the value of $b_n=x$, but it actually diverges, so when you compute the limit of your series over $b_n$, finding it to be 1 actually means that it diverges rather than converges.
answered Nov 22 at 6:23
Muchang Bahng
562
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up vote
0
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Is one of your "diverges" supposed to be "converges"?
Your sum is essentially
$sum_{n=1}^{infty} (-1)^nn
$
or
$-1+2-3+4-5+6...$
with partial sums
$-1,1,-2,2,-3,3,...$.
This diverges.
In what sense can the series converge?
If you are looking at Cesaro sums,
the even ones are zero
and the odd ones
go to -1/2,
so these do not converge.
answered Nov 22 at 6:34
marty cohen
71.9k547126
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
There's nothing wrong with your calculations. You did well to find out the value of $b_n=x$, but it actually diverges, so when you compute the limit of your series over $b_n$, finding it to be 1 actually means that it diverges rather than converges.
answered Nov 22 at 6:23
Muchang Bahng
562
add a comment |
up vote
1
down vote
There's nothing wrong with your calculations. You did well to find out the value of $b_n=x$, but it actually diverges, so when you compute the limit of your series over $b_n$, finding it to be 1 actually means that it diverges rather than converges.
answered Nov 22 at 6:23
Muchang Bahng
562
add a comment |
up vote
1
down vote
up vote
1
down vote
There's nothing wrong with your calculations. You did well to find out the value of $b_n=x$, but it actually diverges, so when you compute the limit of your series over $b_n$, finding it to be 1 actually means that it diverges rather than converges.
answered Nov 22 at 6:23
Muchang Bahng
562
There's nothing wrong with your calculations. You did well to find out the value of $b_n=x$, but it actually diverges, so when you compute the limit of your series over $b_n$, finding it to be 1 actually means that it diverges rather than converges.
answered Nov 22 at 6:23
Muchang Bahng
562
answered Nov 22 at 6:23
Muchang Bahng
562
answered Nov 22 at 6:23
Muchang Bahng
562
answered Nov 22 at 6:23
Muchang Bahng
562
562
add a comment |
add a comment |
up vote
0
down vote
Is one of your "diverges" supposed to be "converges"?
Your sum is essentially
$sum_{n=1}^{infty} (-1)^nn
$
or
$-1+2-3+4-5+6...$
with partial sums
$-1,1,-2,2,-3,3,...$.
This diverges.
In what sense can the series converge?
If you are looking at Cesaro sums,
the even ones are zero
and the odd ones
go to -1/2,
so these do not converge.
answered Nov 22 at 6:34
marty cohen
71.9k547126
add a comment |
up vote
0
down vote
Is one of your "diverges" supposed to be "converges"?
Your sum is essentially
$sum_{n=1}^{infty} (-1)^nn
$
or
$-1+2-3+4-5+6...$
with partial sums
$-1,1,-2,2,-3,3,...$.
This diverges.
In what sense can the series converge?
If you are looking at Cesaro sums,
the even ones are zero
and the odd ones
go to -1/2,
so these do not converge.
answered Nov 22 at 6:34
marty cohen
71.9k547126
add a comment |
up vote
0
down vote
up vote
0
down vote
Is one of your "diverges" supposed to be "converges"?
Your sum is essentially
$sum_{n=1}^{infty} (-1)^nn
$
or
$-1+2-3+4-5+6...$
with partial sums
$-1,1,-2,2,-3,3,...$.
This diverges.
In what sense can the series converge?
If you are looking at Cesaro sums,
the even ones are zero
and the odd ones
go to -1/2,
so these do not converge.
answered Nov 22 at 6:34
marty cohen
71.9k547126
Is one of your "diverges" supposed to be "converges"?
Your sum is essentially
$sum_{n=1}^{infty} (-1)^nn
$
or
$-1+2-3+4-5+6...$
with partial sums
$-1,1,-2,2,-3,3,...$.
This diverges.
In what sense can the series converge?
If you are looking at Cesaro sums,
the even ones are zero
and the odd ones
go to -1/2,
so these do not converge.
answered Nov 22 at 6:34
marty cohen
71.9k547126
answered Nov 22 at 6:34
marty cohen
71.9k547126
answered Nov 22 at 6:34
marty cohen
71.9k547126
answered Nov 22 at 6:34
marty cohen
71.9k547126
71.9k547126
add a comment |
add a comment |
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I think you have a fundamental misunderstanding. First, the limit comparison is only going to work if all of the terms are positive (you might take absolute values, but this changes the problem. In some cases, this change means the difference between convergence and divergence.) Second, the limit of the sequence converges, but the limit of the sum doesn’t converge... At any rate, you compared with $b_n=n$. $sum b_n$ diverges (you might need to think about the statement of the tests to wrap your head around what I mean.)
– Clayton
Nov 22 at 6:21