How does this series diverge by limit comparison test?
How does this series diverge by limit comparison test? $$sum_{n=1}^infty sqrt{frac{n+4}{n^4+4}}$$
I origionally tried using $frac{1}{n^2}$ for the comparison, but I'm pretty sure it has to be $frac{n}{n^2}$ to properly compare.
sequences-and-series divergent-series
edited Dec 18 at 5:36
Chinnapparaj R
5,2251826
asked Dec 18 at 5:33
Luke D
796
add a comment |
How does this series diverge by limit comparison test? $$sum_{n=1}^infty sqrt{frac{n+4}{n^4+4}}$$
I origionally tried using $frac{1}{n^2}$ for the comparison, but I'm pretty sure it has to be $frac{n}{n^2}$ to properly compare.
sequences-and-series divergent-series
edited Dec 18 at 5:36
Chinnapparaj R
5,2251826
asked Dec 18 at 5:33
Luke D
796
add a comment |
How does this series diverge by limit comparison test? $$sum_{n=1}^infty sqrt{frac{n+4}{n^4+4}}$$
I origionally tried using $frac{1}{n^2}$ for the comparison, but I'm pretty sure it has to be $frac{n}{n^2}$ to properly compare.
sequences-and-series divergent-series
edited Dec 18 at 5:36
Chinnapparaj R
5,2251826
asked Dec 18 at 5:33
Luke D
796
How does this series diverge by limit comparison test? $$sum_{n=1}^infty sqrt{frac{n+4}{n^4+4}}$$
I origionally tried using $frac{1}{n^2}$ for the comparison, but I'm pretty sure it has to be $frac{n}{n^2}$ to properly compare.
sequences-and-series divergent-series
sequences-and-series divergent-series
edited Dec 18 at 5:36
Chinnapparaj R
5,2251826
asked Dec 18 at 5:33
Luke D
796
edited Dec 18 at 5:36
Chinnapparaj R
5,2251826
asked Dec 18 at 5:33
Luke D
796
edited Dec 18 at 5:36
Chinnapparaj R
5,2251826
edited Dec 18 at 5:36
Chinnapparaj R
5,2251826
edited Dec 18 at 5:36
Chinnapparaj R
5,2251826
5,2251826
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
$frac{n+4}{n^4+4} leq frac{n+4n}{n^4} leq frac{5}{n^3}$,
hence, $sqrt{frac{n+4}{n^4+4}} leq frac{sqrt{5}}{n^{3/2}}$
so the series converges by comparison with convergent p-series $sum frac{1}{n^{3/2}}$
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 $c = sqrt{5}$.
– 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
$$sqrt{frac{n+4}{n^4+4}}approxsqrt{frac{n}{n^4}}=frac1{n^{3/2}} ,$$
which suggests comparing with
$$sumfrac1{n^{3/2}} .$$
We have
$$sqrt{frac{n+4}{n^4+4}}bigg/frac1{n^{3/2}}=sqrt{frac{n^4+4n^3}{n^4+4}}
=sqrt{frac{1+4n^{-1}}{1+4n^{-4}}}to1quadhbox{as $ntoinfty$} .$$
Since this limit exists and is finite and not zero, and we know that
$$sumfrac1{n^{3/2}}$$
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
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$frac{n+4}{n^4+4} leq frac{n+4n}{n^4} leq frac{5}{n^3}$,
hence, $sqrt{frac{n+4}{n^4+4}} leq frac{sqrt{5}}{n^{3/2}}$
so the series converges by comparison with convergent p-series $sum frac{1}{n^{3/2}}$
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 $c = sqrt{5}$.
– 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 |
$frac{n+4}{n^4+4} leq frac{n+4n}{n^4} leq frac{5}{n^3}$,
hence, $sqrt{frac{n+4}{n^4+4}} leq frac{sqrt{5}}{n^{3/2}}$
so the series converges by comparison with convergent p-series $sum frac{1}{n^{3/2}}$
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 $c = sqrt{5}$.
– 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 |
$frac{n+4}{n^4+4} leq frac{n+4n}{n^4} leq frac{5}{n^3}$,
hence, $sqrt{frac{n+4}{n^4+4}} leq frac{sqrt{5}}{n^{3/2}}$
so the series converges by comparison with convergent p-series $sum frac{1}{n^{3/2}}$
answered Dec 18 at 5:37
Mustafa Said
2,9411913
$frac{n+4}{n^4+4} leq frac{n+4n}{n^4} leq frac{5}{n^3}$,
hence, $sqrt{frac{n+4}{n^4+4}} leq frac{sqrt{5}}{n^{3/2}}$
so the series converges by comparison with convergent p-series $sum frac{1}{n^{3/2}}$
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 $c = sqrt{5}$.
– 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 $c = sqrt{5}$.
– 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 $c = sqrt{5}$.
– Mustafa Said
Dec 18 at 5:43
@LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt{5}$.
– 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
$$sqrt{frac{n+4}{n^4+4}}approxsqrt{frac{n}{n^4}}=frac1{n^{3/2}} ,$$
which suggests comparing with
$$sumfrac1{n^{3/2}} .$$
We have
$$sqrt{frac{n+4}{n^4+4}}bigg/frac1{n^{3/2}}=sqrt{frac{n^4+4n^3}{n^4+4}}
=sqrt{frac{1+4n^{-1}}{1+4n^{-4}}}to1quadhbox{as $ntoinfty$} .$$
Since this limit exists and is finite and not zero, and we know that
$$sumfrac1{n^{3/2}}$$
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
$$sqrt{frac{n+4}{n^4+4}}approxsqrt{frac{n}{n^4}}=frac1{n^{3/2}} ,$$
which suggests comparing with
$$sumfrac1{n^{3/2}} .$$
We have
$$sqrt{frac{n+4}{n^4+4}}bigg/frac1{n^{3/2}}=sqrt{frac{n^4+4n^3}{n^4+4}}
=sqrt{frac{1+4n^{-1}}{1+4n^{-4}}}to1quadhbox{as $ntoinfty$} .$$
Since this limit exists and is finite and not zero, and we know that
$$sumfrac1{n^{3/2}}$$
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
$$sqrt{frac{n+4}{n^4+4}}approxsqrt{frac{n}{n^4}}=frac1{n^{3/2}} ,$$
which suggests comparing with
$$sumfrac1{n^{3/2}} .$$
We have
$$sqrt{frac{n+4}{n^4+4}}bigg/frac1{n^{3/2}}=sqrt{frac{n^4+4n^3}{n^4+4}}
=sqrt{frac{1+4n^{-1}}{1+4n^{-4}}}to1quadhbox{as $ntoinfty$} .$$
Since this limit exists and is finite and not zero, and we know that
$$sumfrac1{n^{3/2}}$$
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
$$sqrt{frac{n+4}{n^4+4}}approxsqrt{frac{n}{n^4}}=frac1{n^{3/2}} ,$$
which suggests comparing with
$$sumfrac1{n^{3/2}} .$$
We have
$$sqrt{frac{n+4}{n^4+4}}bigg/frac1{n^{3/2}}=sqrt{frac{n^4+4n^3}{n^4+4}}
=sqrt{frac{1+4n^{-1}}{1+4n^{-4}}}to1quadhbox{as $ntoinfty$} .$$
Since this limit exists and is finite and not zero, and we know that
$$sumfrac1{n^{3/2}}$$
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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