How does this series diverge by limit comparison test?
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How does this series diverge by limit comparison test? $$sum_n=1^infty sqrtfracn+4n^4+4$$
I origionally tried using $frac1n^2$ for the comparison, but I'm pretty sure it has to be $fracnn^2$ to properly compare.
sequences-and-series divergent-series
edited Dec 18 at 5:36
Chinnapparaj R
5,2751826
asked Dec 18 at 5:33
Luke D
796
add a comment |
How does this series diverge by limit comparison test? $$sum_n=1^infty sqrtfracn+4n^4+4$$
I origionally tried using $frac1n^2$ for the comparison, but I'm pretty sure it has to be $fracnn^2$ to properly compare.
sequences-and-series divergent-series
edited Dec 18 at 5:36
Chinnapparaj R
5,2751826
asked Dec 18 at 5:33
Luke D
796
add a comment |
How does this series diverge by limit comparison test? $$sum_n=1^infty sqrtfracn+4n^4+4$$
I origionally tried using $frac1n^2$ for the comparison, but I'm pretty sure it has to be $fracnn^2$ to properly compare.
sequences-and-series divergent-series
edited Dec 18 at 5:36
Chinnapparaj R
5,2751826
asked Dec 18 at 5:33
Luke D
796
How does this series diverge by limit comparison test? $$sum_n=1^infty sqrtfracn+4n^4+4$$
I origionally tried using $frac1n^2$ for the comparison, but I'm pretty sure it has to be $fracnn^2$ to properly compare.
sequences-and-series divergent-series
sequences-and-series divergent-series
edited Dec 18 at 5:36
Chinnapparaj R
5,2751826
asked Dec 18 at 5:33
Luke D
796
edited Dec 18 at 5:36
Chinnapparaj R
5,2751826
asked Dec 18 at 5:33
Luke D
796
edited Dec 18 at 5:36
Chinnapparaj R
5,2751826
edited Dec 18 at 5:36
Chinnapparaj R
5,2751826
edited Dec 18 at 5:36
Chinnapparaj R
5,2751826
5,2751826
asked Dec 18 at 5:33
Luke D
796
asked Dec 18 at 5:33
Luke D
796
asked Dec 18 at 5:33
Luke D
796
796
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$fracn+4n^4+4 leq fracn+4nn^4 leq frac5n^3$,
hence, $sqrtfracn+4n^4+4 leq fracsqrt5n^3/2$
so the series converges by comparison with convergent p-series $sum frac1n^3/2$
answered Dec 18 at 5:37
Mustafa Said
2,9411913
Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n
– Luke D
Dec 18 at 5:41
@LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt5$.
– Mustafa Said
Dec 18 at 5:43
This can also be done with the limit comparison test, which is what the OP asked for.
– David
Dec 18 at 5:45
Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
– Luke D
Dec 18 at 5:48
add a comment |
For this sort of thing it is strongly advised to do a rough calculation first. We have
$$sqrtfracn+4n^4+4approxsqrtfracnn^4=frac1n^3/2 ,$$
which suggests comparing with
$$sumfrac1n^3/2 .$$
We have
$$sqrtfracn+4n^4+4bigg/frac1n^3/2=sqrtfracn^4+4n^3n^4+4
=sqrtfrac1+4n^-11+4n^-4to1quadhboxas $ntoinfty$ .$$
Since this limit exists and is finite and not zero, and we know that
$$sumfrac1n^3/2$$
converges, your series converges too. (Doesn't diverge!!!)
answered Dec 18 at 5:44
David
67.7k663126
So the answer our teacher gave us is wrong... thanks for the heads up.
– Luke D
Dec 18 at 5:46
If your teacher said the series diverges, yes, that's wrong.
– David
Dec 18 at 5:47
add a comment |
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2 Answers
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2 Answers
2
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$fracn+4n^4+4 leq fracn+4nn^4 leq frac5n^3$,
hence, $sqrtfracn+4n^4+4 leq fracsqrt5n^3/2$
so the series converges by comparison with convergent p-series $sum frac1n^3/2$
answered Dec 18 at 5:37
Mustafa Said
2,9411913
Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n
– Luke D
Dec 18 at 5:41
@LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt5$.
– Mustafa Said
Dec 18 at 5:43
This can also be done with the limit comparison test, which is what the OP asked for.
– David
Dec 18 at 5:45
Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
– Luke D
Dec 18 at 5:48
add a comment |
$fracn+4n^4+4 leq fracn+4nn^4 leq frac5n^3$,
hence, $sqrtfracn+4n^4+4 leq fracsqrt5n^3/2$
so the series converges by comparison with convergent p-series $sum frac1n^3/2$
answered Dec 18 at 5:37
Mustafa Said
2,9411913
Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n
– Luke D
Dec 18 at 5:41
@LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt5$.
– Mustafa Said
Dec 18 at 5:43
This can also be done with the limit comparison test, which is what the OP asked for.
– David
Dec 18 at 5:45
Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
– Luke D
Dec 18 at 5:48
add a comment |
$fracn+4n^4+4 leq fracn+4nn^4 leq frac5n^3$,
hence, $sqrtfracn+4n^4+4 leq fracsqrt5n^3/2$
so the series converges by comparison with convergent p-series $sum frac1n^3/2$
answered Dec 18 at 5:37
Mustafa Said
2,9411913
$fracn+4n^4+4 leq fracn+4nn^4 leq frac5n^3$,
hence, $sqrtfracn+4n^4+4 leq fracsqrt5n^3/2$
so the series converges by comparison with convergent p-series $sum frac1n^3/2$
answered Dec 18 at 5:37
Mustafa Said
2,9411913
answered Dec 18 at 5:37
Mustafa Said
2,9411913
answered Dec 18 at 5:37
Mustafa Said
2,9411913
answered Dec 18 at 5:37
Mustafa Said
2,9411913
2,9411913
Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n
– Luke D
Dec 18 at 5:41
@LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt5$.
– Mustafa Said
Dec 18 at 5:43
This can also be done with the limit comparison test, which is what the OP asked for.
– David
Dec 18 at 5:45
Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
– Luke D
Dec 18 at 5:48
add a comment |
Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n
– Luke D
Dec 18 at 5:41
@LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt5$.
– Mustafa Said
Dec 18 at 5:43
This can also be done with the limit comparison test, which is what the OP asked for.
– David
Dec 18 at 5:45
Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
– Luke D
Dec 18 at 5:48
Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n
– Luke D
Dec 18 at 5:41
Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n
– Luke D
Dec 18 at 5:41
@LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt5$.
– Mustafa Said
Dec 18 at 5:43
@LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt5$.
– Mustafa Said
Dec 18 at 5:43
This can also be done with the limit comparison test, which is what the OP asked for.
– David
Dec 18 at 5:45
This can also be done with the limit comparison test, which is what the OP asked for.
– David
Dec 18 at 5:45
Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
– Luke D
Dec 18 at 5:48
Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
– Luke D
Dec 18 at 5:48
add a comment |
For this sort of thing it is strongly advised to do a rough calculation first. We have
$$sqrtfracn+4n^4+4approxsqrtfracnn^4=frac1n^3/2 ,$$
which suggests comparing with
$$sumfrac1n^3/2 .$$
We have
$$sqrtfracn+4n^4+4bigg/frac1n^3/2=sqrtfracn^4+4n^3n^4+4
=sqrtfrac1+4n^-11+4n^-4to1quadhboxas $ntoinfty$ .$$
Since this limit exists and is finite and not zero, and we know that
$$sumfrac1n^3/2$$
converges, your series converges too. (Doesn't diverge!!!)
answered Dec 18 at 5:44
David
67.7k663126
So the answer our teacher gave us is wrong... thanks for the heads up.
– Luke D
Dec 18 at 5:46
If your teacher said the series diverges, yes, that's wrong.
– David
Dec 18 at 5:47
add a comment |
For this sort of thing it is strongly advised to do a rough calculation first. We have
$$sqrtfracn+4n^4+4approxsqrtfracnn^4=frac1n^3/2 ,$$
which suggests comparing with
$$sumfrac1n^3/2 .$$
We have
$$sqrtfracn+4n^4+4bigg/frac1n^3/2=sqrtfracn^4+4n^3n^4+4
=sqrtfrac1+4n^-11+4n^-4to1quadhboxas $ntoinfty$ .$$
Since this limit exists and is finite and not zero, and we know that
$$sumfrac1n^3/2$$
converges, your series converges too. (Doesn't diverge!!!)
answered Dec 18 at 5:44
David
67.7k663126
So the answer our teacher gave us is wrong... thanks for the heads up.
– Luke D
Dec 18 at 5:46
If your teacher said the series diverges, yes, that's wrong.
– David
Dec 18 at 5:47
add a comment |
For this sort of thing it is strongly advised to do a rough calculation first. We have
$$sqrtfracn+4n^4+4approxsqrtfracnn^4=frac1n^3/2 ,$$
which suggests comparing with
$$sumfrac1n^3/2 .$$
We have
$$sqrtfracn+4n^4+4bigg/frac1n^3/2=sqrtfracn^4+4n^3n^4+4
=sqrtfrac1+4n^-11+4n^-4to1quadhboxas $ntoinfty$ .$$
Since this limit exists and is finite and not zero, and we know that
$$sumfrac1n^3/2$$
converges, your series converges too. (Doesn't diverge!!!)
answered Dec 18 at 5:44
David
67.7k663126
For this sort of thing it is strongly advised to do a rough calculation first. We have
$$sqrtfracn+4n^4+4approxsqrtfracnn^4=frac1n^3/2 ,$$
which suggests comparing with
$$sumfrac1n^3/2 .$$
We have
$$sqrtfracn+4n^4+4bigg/frac1n^3/2=sqrtfracn^4+4n^3n^4+4
=sqrtfrac1+4n^-11+4n^-4to1quadhboxas $ntoinfty$ .$$
Since this limit exists and is finite and not zero, and we know that
$$sumfrac1n^3/2$$
converges, your series converges too. (Doesn't diverge!!!)
answered Dec 18 at 5:44
David
67.7k663126
answered Dec 18 at 5:44
David
67.7k663126
answered Dec 18 at 5:44
David
67.7k663126
answered Dec 18 at 5:44
David
67.7k663126
67.7k663126
So the answer our teacher gave us is wrong... thanks for the heads up.
– Luke D
Dec 18 at 5:46
If your teacher said the series diverges, yes, that's wrong.
– David
Dec 18 at 5:47
add a comment |
So the answer our teacher gave us is wrong... thanks for the heads up.
– Luke D
Dec 18 at 5:46
If your teacher said the series diverges, yes, that's wrong.
– David
Dec 18 at 5:47
So the answer our teacher gave us is wrong... thanks for the heads up.
– Luke D
Dec 18 at 5:46
So the answer our teacher gave us is wrong... thanks for the heads up.
– Luke D
Dec 18 at 5:46
If your teacher said the series diverges, yes, that's wrong.
– David
Dec 18 at 5:47
If your teacher said the series diverges, yes, that's wrong.
– David
Dec 18 at 5:47
add a comment |
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