divergence of $sum_{n=3}^infty frac{sqrt{n}+2}{n-2}$ verification/ alternative method
I wish to prove divergence of
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2}$$
I wish to do so by comparison, since $ngeq 3$:
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2} > sum_{n=3}^infty frac{1+2}{n-2}>sum_{n=3}^infty frac{3}{n}>sum_{n=3}^infty frac{1}{n} rightarrow infty$$
And the harmonic series is divergent, so if we just remove finitely many terms, we still have that it is divergent, because divergence is determined "in the tail". We have a divergent minorant series and hence the original series diverges to $infty$.
Is this approach fine, or is there some more elegant method, this was about the simplest thing I could think of.
Alternatively we have:
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2} > sum_{n=3}^infty frac{sqrt{n}+2}{n}=sum_{n=3}^infty frac{1}{sqrt{n}}+ frac{2}{n}rightarrow infty$$
real-analysis sequences-and-series proof-verification alternative-proof
asked Nov 29 at 21:50
Wesley Strik
1,486422
add a comment |
I wish to prove divergence of
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2}$$
I wish to do so by comparison, since $ngeq 3$:
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2} > sum_{n=3}^infty frac{1+2}{n-2}>sum_{n=3}^infty frac{3}{n}>sum_{n=3}^infty frac{1}{n} rightarrow infty$$
And the harmonic series is divergent, so if we just remove finitely many terms, we still have that it is divergent, because divergence is determined "in the tail". We have a divergent minorant series and hence the original series diverges to $infty$.
Is this approach fine, or is there some more elegant method, this was about the simplest thing I could think of.
Alternatively we have:
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2} > sum_{n=3}^infty frac{sqrt{n}+2}{n}=sum_{n=3}^infty frac{1}{sqrt{n}}+ frac{2}{n}rightarrow infty$$
real-analysis sequences-and-series proof-verification alternative-proof
asked Nov 29 at 21:50
Wesley Strik
1,486422
3
What's your question?
– user23793
Nov 29 at 21:52
1
The summand is $O(frac{1}{sqrt{n}})$, which diverges
– Alex
Nov 29 at 21:53
My bad. I had the question in my head but did not actually write it out xD
– Wesley Strik
Nov 29 at 21:55
Apparently $frac{1}{sqrt{n}}$ diverges, I was not sure about this, but with that method I could just carry out the division and conclude this immediately I suppose
– Wesley Strik
Nov 29 at 21:56
add a comment |
I wish to prove divergence of
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2}$$
I wish to do so by comparison, since $ngeq 3$:
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2} > sum_{n=3}^infty frac{1+2}{n-2}>sum_{n=3}^infty frac{3}{n}>sum_{n=3}^infty frac{1}{n} rightarrow infty$$
And the harmonic series is divergent, so if we just remove finitely many terms, we still have that it is divergent, because divergence is determined "in the tail". We have a divergent minorant series and hence the original series diverges to $infty$.
Is this approach fine, or is there some more elegant method, this was about the simplest thing I could think of.
Alternatively we have:
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2} > sum_{n=3}^infty frac{sqrt{n}+2}{n}=sum_{n=3}^infty frac{1}{sqrt{n}}+ frac{2}{n}rightarrow infty$$
real-analysis sequences-and-series proof-verification alternative-proof
asked Nov 29 at 21:50
Wesley Strik
1,486422
I wish to prove divergence of
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2}$$
I wish to do so by comparison, since $ngeq 3$:
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2} > sum_{n=3}^infty frac{1+2}{n-2}>sum_{n=3}^infty frac{3}{n}>sum_{n=3}^infty frac{1}{n} rightarrow infty$$
And the harmonic series is divergent, so if we just remove finitely many terms, we still have that it is divergent, because divergence is determined "in the tail". We have a divergent minorant series and hence the original series diverges to $infty$.
Is this approach fine, or is there some more elegant method, this was about the simplest thing I could think of.
Alternatively we have:
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2} > sum_{n=3}^infty frac{sqrt{n}+2}{n}=sum_{n=3}^infty frac{1}{sqrt{n}}+ frac{2}{n}rightarrow infty$$
real-analysis sequences-and-series proof-verification alternative-proof
real-analysis sequences-and-series proof-verification alternative-proof
asked Nov 29 at 21:50
Wesley Strik
1,486422
asked Nov 29 at 21:50
Wesley Strik
1,486422
edited Nov 29 at 22:00
asked Nov 29 at 21:50
Wesley Strik
1,486422
asked Nov 29 at 21:50
Wesley Strik
1,486422
asked Nov 29 at 21:50
Wesley Strik
1,486422
1,486422
3
What's your question?
– user23793
Nov 29 at 21:52
1
The summand is $O(frac{1}{sqrt{n}})$, which diverges
– Alex
Nov 29 at 21:53
My bad. I had the question in my head but did not actually write it out xD
– Wesley Strik
Nov 29 at 21:55
Apparently $frac{1}{sqrt{n}}$ diverges, I was not sure about this, but with that method I could just carry out the division and conclude this immediately I suppose
– Wesley Strik
Nov 29 at 21:56
add a comment |
3
What's your question?
– user23793
Nov 29 at 21:52
1
The summand is $O(frac{1}{sqrt{n}})$, which diverges
– Alex
Nov 29 at 21:53
My bad. I had the question in my head but did not actually write it out xD
– Wesley Strik
Nov 29 at 21:55
Apparently $frac{1}{sqrt{n}}$ diverges, I was not sure about this, but with that method I could just carry out the division and conclude this immediately I suppose
– Wesley Strik
Nov 29 at 21:56
3
3
What's your question?
– user23793
Nov 29 at 21:52
What's your question?
– user23793
Nov 29 at 21:52
1
1
The summand is $O(frac{1}{sqrt{n}})$, which diverges
– Alex
Nov 29 at 21:53
The summand is $O(frac{1}{sqrt{n}})$, which diverges
– Alex
Nov 29 at 21:53
My bad. I had the question in my head but did not actually write it out xD
– Wesley Strik
Nov 29 at 21:55
My bad. I had the question in my head but did not actually write it out xD
– Wesley Strik
Nov 29 at 21:55
Apparently $frac{1}{sqrt{n}}$ diverges, I was not sure about this, but with that method I could just carry out the division and conclude this immediately I suppose
– Wesley Strik
Nov 29 at 21:56
Apparently $frac{1}{sqrt{n}}$ diverges, I was not sure about this, but with that method I could just carry out the division and conclude this immediately I suppose
– Wesley Strik
Nov 29 at 21:56
add a comment |
3 Answers
3
active
oldest
votes
Not clear what your question is, but your answer is correct.
Your approach is fine. Comparison test would be the proper test to use.
One can also show that $$sum _{n=3}^{infty :}frac{sqrt{n}+2}{n-2}ge sum _{n=3}^{infty :}frac{sqrt{n}+2}{n}$$
and show that the rightmost sum is diverging via the integral test.
answered Nov 29 at 21:53
K Split X
4,12911031
That's a nice alternative, we haven't practised a lot with this yet, but this method would work very well :)
– Wesley Strik
Nov 29 at 22:01
Right now I am still trying to figure out which method would be best suited for which situation, I still feel I'm just throwing all I know at a question most of the time - but I'm getting better.
– Wesley Strik
Nov 29 at 22:02
Generally if you can't evaluate it directly, but you can "visualize" the series diverges, then this test is good.
– K Split X
Nov 29 at 22:03
Intuiton can be developed, you're right ;) "seeing it" and a proof are sometimes closer than it seems.
– Wesley Strik
Nov 29 at 22:05
add a comment |
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2}=sum_{n=3}^infty frac1{sqrt{n}-2}$$ and the terms are of order $n^{-1/2}$.
answered Nov 29 at 22:14
Yves Daoust
124k671221
add a comment |
Since
$$frac{sqrt{n}+2}{n-2} sim frac {sqrt n}n=frac1{sqrt n}$$
the series diverges by limit comparison test with $sum frac1{sqrt n}$.
answered Nov 29 at 22:07
gimusi
1
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Not clear what your question is, but your answer is correct.
Your approach is fine. Comparison test would be the proper test to use.
One can also show that $$sum _{n=3}^{infty :}frac{sqrt{n}+2}{n-2}ge sum _{n=3}^{infty :}frac{sqrt{n}+2}{n}$$
and show that the rightmost sum is diverging via the integral test.
answered Nov 29 at 21:53
K Split X
4,12911031
That's a nice alternative, we haven't practised a lot with this yet, but this method would work very well :)
– Wesley Strik
Nov 29 at 22:01
Right now I am still trying to figure out which method would be best suited for which situation, I still feel I'm just throwing all I know at a question most of the time - but I'm getting better.
– Wesley Strik
Nov 29 at 22:02
Generally if you can't evaluate it directly, but you can "visualize" the series diverges, then this test is good.
– K Split X
Nov 29 at 22:03
Intuiton can be developed, you're right ;) "seeing it" and a proof are sometimes closer than it seems.
– Wesley Strik
Nov 29 at 22:05
add a comment |
Not clear what your question is, but your answer is correct.
Your approach is fine. Comparison test would be the proper test to use.
One can also show that $$sum _{n=3}^{infty :}frac{sqrt{n}+2}{n-2}ge sum _{n=3}^{infty :}frac{sqrt{n}+2}{n}$$
and show that the rightmost sum is diverging via the integral test.
answered Nov 29 at 21:53
K Split X
4,12911031
That's a nice alternative, we haven't practised a lot with this yet, but this method would work very well :)
– Wesley Strik
Nov 29 at 22:01
Right now I am still trying to figure out which method would be best suited for which situation, I still feel I'm just throwing all I know at a question most of the time - but I'm getting better.
– Wesley Strik
Nov 29 at 22:02
Generally if you can't evaluate it directly, but you can "visualize" the series diverges, then this test is good.
– K Split X
Nov 29 at 22:03
Intuiton can be developed, you're right ;) "seeing it" and a proof are sometimes closer than it seems.
– Wesley Strik
Nov 29 at 22:05
add a comment |
Not clear what your question is, but your answer is correct.
Your approach is fine. Comparison test would be the proper test to use.
One can also show that $$sum _{n=3}^{infty :}frac{sqrt{n}+2}{n-2}ge sum _{n=3}^{infty :}frac{sqrt{n}+2}{n}$$
and show that the rightmost sum is diverging via the integral test.
answered Nov 29 at 21:53
K Split X
4,12911031
Not clear what your question is, but your answer is correct.
Your approach is fine. Comparison test would be the proper test to use.
One can also show that $$sum _{n=3}^{infty :}frac{sqrt{n}+2}{n-2}ge sum _{n=3}^{infty :}frac{sqrt{n}+2}{n}$$
and show that the rightmost sum is diverging via the integral test.
answered Nov 29 at 21:53
K Split X
4,12911031
answered Nov 29 at 21:53
K Split X
4,12911031
answered Nov 29 at 21:53
K Split X
4,12911031
answered Nov 29 at 21:53
K Split X
4,12911031
4,12911031
That's a nice alternative, we haven't practised a lot with this yet, but this method would work very well :)
– Wesley Strik
Nov 29 at 22:01
Right now I am still trying to figure out which method would be best suited for which situation, I still feel I'm just throwing all I know at a question most of the time - but I'm getting better.
– Wesley Strik
Nov 29 at 22:02
Generally if you can't evaluate it directly, but you can "visualize" the series diverges, then this test is good.
– K Split X
Nov 29 at 22:03
Intuiton can be developed, you're right ;) "seeing it" and a proof are sometimes closer than it seems.
– Wesley Strik
Nov 29 at 22:05
add a comment |
That's a nice alternative, we haven't practised a lot with this yet, but this method would work very well :)
– Wesley Strik
Nov 29 at 22:01
Right now I am still trying to figure out which method would be best suited for which situation, I still feel I'm just throwing all I know at a question most of the time - but I'm getting better.
– Wesley Strik
Nov 29 at 22:02
Generally if you can't evaluate it directly, but you can "visualize" the series diverges, then this test is good.
– K Split X
Nov 29 at 22:03
Intuiton can be developed, you're right ;) "seeing it" and a proof are sometimes closer than it seems.
– Wesley Strik
Nov 29 at 22:05
That's a nice alternative, we haven't practised a lot with this yet, but this method would work very well :)
– Wesley Strik
Nov 29 at 22:01
That's a nice alternative, we haven't practised a lot with this yet, but this method would work very well :)
– Wesley Strik
Nov 29 at 22:01
Right now I am still trying to figure out which method would be best suited for which situation, I still feel I'm just throwing all I know at a question most of the time - but I'm getting better.
– Wesley Strik
Nov 29 at 22:02
Right now I am still trying to figure out which method would be best suited for which situation, I still feel I'm just throwing all I know at a question most of the time - but I'm getting better.
– Wesley Strik
Nov 29 at 22:02
Generally if you can't evaluate it directly, but you can "visualize" the series diverges, then this test is good.
– K Split X
Nov 29 at 22:03
Generally if you can't evaluate it directly, but you can "visualize" the series diverges, then this test is good.
– K Split X
Nov 29 at 22:03
Intuiton can be developed, you're right ;) "seeing it" and a proof are sometimes closer than it seems.
– Wesley Strik
Nov 29 at 22:05
Intuiton can be developed, you're right ;) "seeing it" and a proof are sometimes closer than it seems.
– Wesley Strik
Nov 29 at 22:05
add a comment |
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2}=sum_{n=3}^infty frac1{sqrt{n}-2}$$ and the terms are of order $n^{-1/2}$.
answered Nov 29 at 22:14
Yves Daoust
124k671221
add a comment |
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2}=sum_{n=3}^infty frac1{sqrt{n}-2}$$ and the terms are of order $n^{-1/2}$.
answered Nov 29 at 22:14
Yves Daoust
124k671221
add a comment |
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2}=sum_{n=3}^infty frac1{sqrt{n}-2}$$ and the terms are of order $n^{-1/2}$.
answered Nov 29 at 22:14
Yves Daoust
124k671221
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2}=sum_{n=3}^infty frac1{sqrt{n}-2}$$ and the terms are of order $n^{-1/2}$.
answered Nov 29 at 22:14
Yves Daoust
124k671221
answered Nov 29 at 22:14
Yves Daoust
124k671221
answered Nov 29 at 22:14
Yves Daoust
124k671221
answered Nov 29 at 22:14
Yves Daoust
124k671221
124k671221
add a comment |
add a comment |
Since
$$frac{sqrt{n}+2}{n-2} sim frac {sqrt n}n=frac1{sqrt n}$$
the series diverges by limit comparison test with $sum frac1{sqrt n}$.
answered Nov 29 at 22:07
gimusi
1
add a comment |
Since
$$frac{sqrt{n}+2}{n-2} sim frac {sqrt n}n=frac1{sqrt n}$$
the series diverges by limit comparison test with $sum frac1{sqrt n}$.
answered Nov 29 at 22:07
gimusi
1
add a comment |
Since
$$frac{sqrt{n}+2}{n-2} sim frac {sqrt n}n=frac1{sqrt n}$$
the series diverges by limit comparison test with $sum frac1{sqrt n}$.
answered Nov 29 at 22:07
gimusi
1
Since
$$frac{sqrt{n}+2}{n-2} sim frac {sqrt n}n=frac1{sqrt n}$$
the series diverges by limit comparison test with $sum frac1{sqrt n}$.
answered Nov 29 at 22:07
gimusi
1
answered Nov 29 at 22:07
gimusi
1
answered Nov 29 at 22:07
gimusi
1
answered Nov 29 at 22:07
gimusi
1
1
add a comment |
add a comment |
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3
What's your question?
– user23793
Nov 29 at 21:52
1
The summand is $O(frac{1}{sqrt{n}})$, which diverges
– Alex
Nov 29 at 21:53
My bad. I had the question in my head but did not actually write it out xD
– Wesley Strik
Nov 29 at 21:55
Apparently $frac{1}{sqrt{n}}$ diverges, I was not sure about this, but with that method I could just carry out the division and conclude this immediately I suppose
– Wesley Strik
Nov 29 at 21:56