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 '18 at 21:50
Wesley StrikWesley Strik
1,635423
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 '18 at 21:50
Wesley StrikWesley Strik
1,635423
3
What's your question?
– user23793
Nov 29 '18 at 21:52
1
The summand is $O(frac{1}{sqrt{n}})$, which diverges
– Alex
Nov 29 '18 at 21:53
My bad. I had the question in my head but did not actually write it out xD
– Wesley Strik
Nov 29 '18 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 '18 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 '18 at 21:50
Wesley StrikWesley Strik
1,635423
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 '18 at 21:50
Wesley StrikWesley Strik
1,635423
asked Nov 29 '18 at 21:50
Wesley StrikWesley Strik
1,635423
edited Nov 29 '18 at 22:00
Wesley Strik
asked Nov 29 '18 at 21:50
Wesley StrikWesley Strik
1,635423
asked Nov 29 '18 at 21:50
Wesley StrikWesley Strik
1,635423
asked Nov 29 '18 at 21:50
Wesley StrikWesley Strik
1,635423
1,635423
3
What's your question?
– user23793
Nov 29 '18 at 21:52
1
The summand is $O(frac{1}{sqrt{n}})$, which diverges
– Alex
Nov 29 '18 at 21:53
My bad. I had the question in my head but did not actually write it out xD
– Wesley Strik
Nov 29 '18 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 '18 at 21:56
add a comment |
3
What's your question?
– user23793
Nov 29 '18 at 21:52
1
The summand is $O(frac{1}{sqrt{n}})$, which diverges
– Alex
Nov 29 '18 at 21:53
My bad. I had the question in my head but did not actually write it out xD
– Wesley Strik
Nov 29 '18 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 '18 at 21:56
3
3
What's your question?
– user23793
Nov 29 '18 at 21:52
What's your question?
– user23793
Nov 29 '18 at 21:52
1
1
The summand is $O(frac{1}{sqrt{n}})$, which diverges
– Alex
Nov 29 '18 at 21:53
The summand is $O(frac{1}{sqrt{n}})$, which diverges
– Alex
Nov 29 '18 at 21:53
My bad. I had the question in my head but did not actually write it out xD
– Wesley Strik
Nov 29 '18 at 21:55
My bad. I had the question in my head but did not actually write it out xD
– Wesley Strik
Nov 29 '18 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 '18 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 '18 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 '18 at 21:53
K Split XK Split X
4,21111031
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 '18 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 '18 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 '18 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 '18 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 '18 at 22:14
Yves DaoustYves Daoust
124k671222
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 '18 at 22:07
gimusigimusi
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 '18 at 21:53
K Split XK Split X
4,21111031
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 '18 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 '18 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 '18 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 '18 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 '18 at 21:53
K Split XK Split X
4,21111031
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 '18 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 '18 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 '18 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 '18 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 '18 at 21:53
K Split XK Split X
4,21111031
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 '18 at 21:53
K Split XK Split X
4,21111031
answered Nov 29 '18 at 21:53
K Split XK Split X
4,21111031
answered Nov 29 '18 at 21:53
K Split XK Split X
4,21111031
answered Nov 29 '18 at 21:53
K Split XK Split X
4,21111031
4,21111031
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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 22:14
Yves DaoustYves Daoust
124k671222
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 '18 at 22:14
Yves DaoustYves Daoust
124k671222
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 '18 at 22:14
Yves DaoustYves Daoust
124k671222
$$sum_{n=3}^infty frac{sqrt{n}+2}{n-2}=sum_{n=3}^infty frac1{sqrt{n}-2}$$ and the terms are of order $n^{-1/2}$.
answered Nov 29 '18 at 22:14
Yves DaoustYves Daoust
124k671222
answered Nov 29 '18 at 22:14
Yves DaoustYves Daoust
124k671222
answered Nov 29 '18 at 22:14
Yves DaoustYves Daoust
124k671222
answered Nov 29 '18 at 22:14
Yves DaoustYves Daoust
124k671222
124k671222
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 '18 at 22:07
gimusigimusi
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 '18 at 22:07
gimusigimusi
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 '18 at 22:07
gimusigimusi
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 '18 at 22:07
gimusigimusi
1
answered Nov 29 '18 at 22:07
gimusigimusi
1
answered Nov 29 '18 at 22:07
gimusigimusi
1
answered Nov 29 '18 at 22:07
gimusigimusi
1
1
add a comment |
add a comment |
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3
What's your question?
– user23793
Nov 29 '18 at 21:52
1
The summand is $O(frac{1}{sqrt{n}})$, which diverges
– Alex
Nov 29 '18 at 21:53
My bad. I had the question in my head but did not actually write it out xD
– Wesley Strik
Nov 29 '18 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 '18 at 21:56