Convergence or divergence of $ sum_{n=1}^{infty}frac{sqrt{n+2}-sqrt{n}}{n^{3/2}} $: How to argue?
$begingroup$
Does the series
$$
sum_{n=1}^{infty}frac{sqrt{n+2}-sqrt{n}}{n^{3/2}}
$$
converge or diverge?
My attempt was to write the series as
$$
sum_{n=1}^{infty}frac{sqrt{n+2}-sqrt{n}}{n^{3/2}}=sum_{n=1}^{infty}frac{sqrt{n+2}}{n^{3/2}}-sum_{n=1}^{infty}frac{1}{n}
$$
The first series can be estimated from below by the harmonic series:
$$
sum_{n=1}^{infty}frac{sqrt{n+2}}{n^{3/2}}=sum_{n=1}^{infty}frac{(n+2)^{1/2}}{n^{1/2}cdot n}geqslantsum_{n=1}^infty frac{1}{n}=infty
$$
and hence diverges.
The second series is the harmonic series and hence diverges.
Now, I am not sure what the whole thing does.
sequences-and-series convergence
edited Dec 8 '18 at 13:19
Shaun
8,932113681
asked Dec 8 '18 at 13:12
RhjgRhjg
396215
$endgroup$
add a comment |
$begingroup$
Does the series
$$
sum_{n=1}^{infty}frac{sqrt{n+2}-sqrt{n}}{n^{3/2}}
$$
converge or diverge?
My attempt was to write the series as
$$
sum_{n=1}^{infty}frac{sqrt{n+2}-sqrt{n}}{n^{3/2}}=sum_{n=1}^{infty}frac{sqrt{n+2}}{n^{3/2}}-sum_{n=1}^{infty}frac{1}{n}
$$
The first series can be estimated from below by the harmonic series:
$$
sum_{n=1}^{infty}frac{sqrt{n+2}}{n^{3/2}}=sum_{n=1}^{infty}frac{(n+2)^{1/2}}{n^{1/2}cdot n}geqslantsum_{n=1}^infty frac{1}{n}=infty
$$
and hence diverges.
The second series is the harmonic series and hence diverges.
Now, I am not sure what the whole thing does.
sequences-and-series convergence
edited Dec 8 '18 at 13:19
Shaun
8,932113681
asked Dec 8 '18 at 13:12
RhjgRhjg
396215
$endgroup$
add a comment |
$begingroup$
Does the series
$$
sum_{n=1}^{infty}frac{sqrt{n+2}-sqrt{n}}{n^{3/2}}
$$
converge or diverge?
My attempt was to write the series as
$$
sum_{n=1}^{infty}frac{sqrt{n+2}-sqrt{n}}{n^{3/2}}=sum_{n=1}^{infty}frac{sqrt{n+2}}{n^{3/2}}-sum_{n=1}^{infty}frac{1}{n}
$$
The first series can be estimated from below by the harmonic series:
$$
sum_{n=1}^{infty}frac{sqrt{n+2}}{n^{3/2}}=sum_{n=1}^{infty}frac{(n+2)^{1/2}}{n^{1/2}cdot n}geqslantsum_{n=1}^infty frac{1}{n}=infty
$$
and hence diverges.
The second series is the harmonic series and hence diverges.
Now, I am not sure what the whole thing does.
sequences-and-series convergence
edited Dec 8 '18 at 13:19
Shaun
8,932113681
asked Dec 8 '18 at 13:12
RhjgRhjg
396215
$endgroup$
Does the series
$$
sum_{n=1}^{infty}frac{sqrt{n+2}-sqrt{n}}{n^{3/2}}
$$
converge or diverge?
My attempt was to write the series as
$$
sum_{n=1}^{infty}frac{sqrt{n+2}-sqrt{n}}{n^{3/2}}=sum_{n=1}^{infty}frac{sqrt{n+2}}{n^{3/2}}-sum_{n=1}^{infty}frac{1}{n}
$$
The first series can be estimated from below by the harmonic series:
$$
sum_{n=1}^{infty}frac{sqrt{n+2}}{n^{3/2}}=sum_{n=1}^{infty}frac{(n+2)^{1/2}}{n^{1/2}cdot n}geqslantsum_{n=1}^infty frac{1}{n}=infty
$$
and hence diverges.
The second series is the harmonic series and hence diverges.
Now, I am not sure what the whole thing does.
sequences-and-series convergence
sequences-and-series convergence
edited Dec 8 '18 at 13:19
Shaun
8,932113681
asked Dec 8 '18 at 13:12
RhjgRhjg
396215
edited Dec 8 '18 at 13:19
Shaun
8,932113681
asked Dec 8 '18 at 13:12
RhjgRhjg
396215
edited Dec 8 '18 at 13:19
Shaun
8,932113681
edited Dec 8 '18 at 13:19
Shaun
8,932113681
edited Dec 8 '18 at 13:19
Shaun
8,932113681
8,932113681
asked Dec 8 '18 at 13:12
RhjgRhjg
396215
asked Dec 8 '18 at 13:12
RhjgRhjg
396215
asked Dec 8 '18 at 13:12
RhjgRhjg
396215
396215
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Your attempt is wrong because you can't separate the series into a difference of divergent series. You got an indeterminate form $infty - infty$.
You could try to rationalize the numerator:
$$(sqrt{n+2}-sqrt{n})frac{sqrt{n+2}+sqrt{n}}{sqrt{n+2}+sqrt{n}}=frac{2}{sqrt{n+2}+sqrt{n}}$$
Then the series becomes
$$sum_{n=1}^{infty}frac{2}{(sqrt{n+2}+sqrt{n})n^{3/2}}$$
You can use comparison test to conclude.
answered Dec 8 '18 at 13:19
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
$begingroup$
I suggest to do not give full answer suddenly when possible, since it can be not useful for the site nor for the asker who could be interested to solve it by themself. In such latter cases the use of "spoiler" comand can be useful).
$endgroup$
– gimusi
Dec 8 '18 at 13:21
$begingroup$
Using "spoiler" command is a good idea
$endgroup$
– Lorenzo B.
Dec 8 '18 at 13:23
1
$begingroup$
This wasn't and isn't a full answer. It leaves the determination of convergence/divergence to the asker. It merely suggests rationalizing the numerator, and displays the result upon doing so. I think it is a fine answer. Nice work, @LorenzoB.
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– amWhy
Dec 8 '18 at 13:30
1
$begingroup$
@Rhjg, yes you are
$endgroup$
– Lorenzo B.
Dec 8 '18 at 13:38
1
$begingroup$
Thank you very much for your help.
$endgroup$
– Rhjg
Dec 8 '18 at 13:39
|
show 3 more comments
$begingroup$
HINT
We can more effectively use that
$$sqrt{n+2}-sqrt{n}=frac{2}{sqrt{n+2}+sqrt{n}}$$
answered Dec 8 '18 at 13:14
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
You can use the estimate:
$$frac{sqrt{n+2}-sqrt{n}}{n^{3/2}}<frac1{n^2} iff sqrt{n}(sqrt{n+2}-sqrt{n})<1 iff \
sqrt{n(n+2)}<n+1 iff n^2+2n<n^2+2n+1.$$
answered Dec 8 '18 at 13:45
farruhotafarruhota
19.8k2738
$endgroup$
add a comment |
$begingroup$
Rewriting :
$2=(n+2) -n=$
$(sqrt{n+2}-√n)(sqrt{n+2}+√n) gt$
$sqrt{n+2}-√n$, since
$sqrt{n+2}+√n >1$.
Hence
$dfrac{sqrt{n+2}-√n}{n^{3/2}} lt dfrac{2}{n^{3/2}}.$
Use comparison test.
answered Dec 8 '18 at 13:55
Peter SzilasPeter Szilas
11k2821
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add a comment |
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Note that $sqrt{n+2}-sqrt nto 0$ as $ntoinfty$, so $;dfrac{sqrt{n+2}-sqrt{n}}{n^{3/2}}=obiggl(dfrac 1{n^{3/2}}biggr)$, and the latter series converges.
answered Dec 8 '18 at 13:33
BernardBernard
119k740113
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That's a nice way but why $sqrt{n+2}-sqrt nto 0$?
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– gimusi
Dec 8 '18 at 13:37
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That is standard in high school (and your answer shows how it is proved).
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– Bernard
Dec 8 '18 at 13:42
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Yes but it seems that the asker ignore that otherwise probably he wouldn't have posted the question. In my opinion at the present state you answer is not complete regarding a crucial point. Moreover once we know it goes to zero we could simply use comparison test. Please consider my observations in a constructive way.
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– gimusi
Dec 8 '18 at 13:46
1
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@gimusi : I think our answers are more or less complementary. They do not aim the same audience, and I wanted to propose an as concise as possible answer for those who already have some familiarity with the basics of limits.
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– Bernard
Dec 8 '18 at 13:57
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That's ansolutely fine my aim was only to suggest what I thought could make some improvements with that for the asker. But that's of corse all up to you and the answer is fine also in that way. Bye
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– gimusi
Dec 8 '18 at 14:08
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your attempt is wrong because you can't separate the series into a difference of divergent series. You got an indeterminate form $infty - infty$.
You could try to rationalize the numerator:
$$(sqrt{n+2}-sqrt{n})frac{sqrt{n+2}+sqrt{n}}{sqrt{n+2}+sqrt{n}}=frac{2}{sqrt{n+2}+sqrt{n}}$$
Then the series becomes
$$sum_{n=1}^{infty}frac{2}{(sqrt{n+2}+sqrt{n})n^{3/2}}$$
You can use comparison test to conclude.
answered Dec 8 '18 at 13:19
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
$begingroup$
I suggest to do not give full answer suddenly when possible, since it can be not useful for the site nor for the asker who could be interested to solve it by themself. In such latter cases the use of "spoiler" comand can be useful).
$endgroup$
– gimusi
Dec 8 '18 at 13:21
$begingroup$
Using "spoiler" command is a good idea
$endgroup$
– Lorenzo B.
Dec 8 '18 at 13:23
1
$begingroup$
This wasn't and isn't a full answer. It leaves the determination of convergence/divergence to the asker. It merely suggests rationalizing the numerator, and displays the result upon doing so. I think it is a fine answer. Nice work, @LorenzoB.
$endgroup$
– amWhy
Dec 8 '18 at 13:30
1
$begingroup$
@Rhjg, yes you are
$endgroup$
– Lorenzo B.
Dec 8 '18 at 13:38
1
$begingroup$
Thank you very much for your help.
$endgroup$
– Rhjg
Dec 8 '18 at 13:39
|
show 3 more comments
$begingroup$
Your attempt is wrong because you can't separate the series into a difference of divergent series. You got an indeterminate form $infty - infty$.
You could try to rationalize the numerator:
$$(sqrt{n+2}-sqrt{n})frac{sqrt{n+2}+sqrt{n}}{sqrt{n+2}+sqrt{n}}=frac{2}{sqrt{n+2}+sqrt{n}}$$
Then the series becomes
$$sum_{n=1}^{infty}frac{2}{(sqrt{n+2}+sqrt{n})n^{3/2}}$$
You can use comparison test to conclude.
answered Dec 8 '18 at 13:19
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
$begingroup$
I suggest to do not give full answer suddenly when possible, since it can be not useful for the site nor for the asker who could be interested to solve it by themself. In such latter cases the use of "spoiler" comand can be useful).
$endgroup$
– gimusi
Dec 8 '18 at 13:21
$begingroup$
Using "spoiler" command is a good idea
$endgroup$
– Lorenzo B.
Dec 8 '18 at 13:23
1
$begingroup$
This wasn't and isn't a full answer. It leaves the determination of convergence/divergence to the asker. It merely suggests rationalizing the numerator, and displays the result upon doing so. I think it is a fine answer. Nice work, @LorenzoB.
$endgroup$
– amWhy
Dec 8 '18 at 13:30
1
$begingroup$
@Rhjg, yes you are
$endgroup$
– Lorenzo B.
Dec 8 '18 at 13:38
1
$begingroup$
Thank you very much for your help.
$endgroup$
– Rhjg
Dec 8 '18 at 13:39
|
show 3 more comments
$begingroup$
Your attempt is wrong because you can't separate the series into a difference of divergent series. You got an indeterminate form $infty - infty$.
You could try to rationalize the numerator:
$$(sqrt{n+2}-sqrt{n})frac{sqrt{n+2}+sqrt{n}}{sqrt{n+2}+sqrt{n}}=frac{2}{sqrt{n+2}+sqrt{n}}$$
Then the series becomes
$$sum_{n=1}^{infty}frac{2}{(sqrt{n+2}+sqrt{n})n^{3/2}}$$
You can use comparison test to conclude.
answered Dec 8 '18 at 13:19
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
Your attempt is wrong because you can't separate the series into a difference of divergent series. You got an indeterminate form $infty - infty$.
You could try to rationalize the numerator:
$$(sqrt{n+2}-sqrt{n})frac{sqrt{n+2}+sqrt{n}}{sqrt{n+2}+sqrt{n}}=frac{2}{sqrt{n+2}+sqrt{n}}$$
Then the series becomes
$$sum_{n=1}^{infty}frac{2}{(sqrt{n+2}+sqrt{n})n^{3/2}}$$
You can use comparison test to conclude.
answered Dec 8 '18 at 13:19
Lorenzo B.Lorenzo B.
1,8402520
edited Dec 8 '18 at 13:22
answered Dec 8 '18 at 13:19
Lorenzo B.Lorenzo B.
1,8402520
answered Dec 8 '18 at 13:19
Lorenzo B.Lorenzo B.
1,8402520
answered Dec 8 '18 at 13:19
Lorenzo B.Lorenzo B.
1,8402520
1,8402520
$begingroup$
I suggest to do not give full answer suddenly when possible, since it can be not useful for the site nor for the asker who could be interested to solve it by themself. In such latter cases the use of "spoiler" comand can be useful).
$endgroup$
– gimusi
Dec 8 '18 at 13:21
$begingroup$
Using "spoiler" command is a good idea
$endgroup$
– Lorenzo B.
Dec 8 '18 at 13:23
1
$begingroup$
This wasn't and isn't a full answer. It leaves the determination of convergence/divergence to the asker. It merely suggests rationalizing the numerator, and displays the result upon doing so. I think it is a fine answer. Nice work, @LorenzoB.
$endgroup$
– amWhy
Dec 8 '18 at 13:30
1
$begingroup$
@Rhjg, yes you are
$endgroup$
– Lorenzo B.
Dec 8 '18 at 13:38
1
$begingroup$
Thank you very much for your help.
$endgroup$
– Rhjg
Dec 8 '18 at 13:39
|
show 3 more comments
$begingroup$
I suggest to do not give full answer suddenly when possible, since it can be not useful for the site nor for the asker who could be interested to solve it by themself. In such latter cases the use of "spoiler" comand can be useful).
$endgroup$
– gimusi
Dec 8 '18 at 13:21
$begingroup$
Using "spoiler" command is a good idea
$endgroup$
– Lorenzo B.
Dec 8 '18 at 13:23
1
$begingroup$
This wasn't and isn't a full answer. It leaves the determination of convergence/divergence to the asker. It merely suggests rationalizing the numerator, and displays the result upon doing so. I think it is a fine answer. Nice work, @LorenzoB.
$endgroup$
– amWhy
Dec 8 '18 at 13:30
1
$begingroup$
@Rhjg, yes you are
$endgroup$
– Lorenzo B.
Dec 8 '18 at 13:38
1
$begingroup$
Thank you very much for your help.
$endgroup$
– Rhjg
Dec 8 '18 at 13:39
$begingroup$
I suggest to do not give full answer suddenly when possible, since it can be not useful for the site nor for the asker who could be interested to solve it by themself. In such latter cases the use of "spoiler" comand can be useful).
$endgroup$
– gimusi
Dec 8 '18 at 13:21
$begingroup$
I suggest to do not give full answer suddenly when possible, since it can be not useful for the site nor for the asker who could be interested to solve it by themself. In such latter cases the use of "spoiler" comand can be useful).
$endgroup$
– gimusi
Dec 8 '18 at 13:21
$begingroup$
Using "spoiler" command is a good idea
$endgroup$
– Lorenzo B.
Dec 8 '18 at 13:23
$begingroup$
Using "spoiler" command is a good idea
$endgroup$
– Lorenzo B.
Dec 8 '18 at 13:23
1
1
$begingroup$
This wasn't and isn't a full answer. It leaves the determination of convergence/divergence to the asker. It merely suggests rationalizing the numerator, and displays the result upon doing so. I think it is a fine answer. Nice work, @LorenzoB.
$endgroup$
– amWhy
Dec 8 '18 at 13:30
$begingroup$
This wasn't and isn't a full answer. It leaves the determination of convergence/divergence to the asker. It merely suggests rationalizing the numerator, and displays the result upon doing so. I think it is a fine answer. Nice work, @LorenzoB.
$endgroup$
– amWhy
Dec 8 '18 at 13:30
1
1
$begingroup$
@Rhjg, yes you are
$endgroup$
– Lorenzo B.
Dec 8 '18 at 13:38
$begingroup$
@Rhjg, yes you are
$endgroup$
– Lorenzo B.
Dec 8 '18 at 13:38
1
1
$begingroup$
Thank you very much for your help.
$endgroup$
– Rhjg
Dec 8 '18 at 13:39
$begingroup$
Thank you very much for your help.
$endgroup$
– Rhjg
Dec 8 '18 at 13:39
|
show 3 more comments
$begingroup$
HINT
We can more effectively use that
$$sqrt{n+2}-sqrt{n}=frac{2}{sqrt{n+2}+sqrt{n}}$$
answered Dec 8 '18 at 13:14
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
HINT
We can more effectively use that
$$sqrt{n+2}-sqrt{n}=frac{2}{sqrt{n+2}+sqrt{n}}$$
answered Dec 8 '18 at 13:14
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
HINT
We can more effectively use that
$$sqrt{n+2}-sqrt{n}=frac{2}{sqrt{n+2}+sqrt{n}}$$
answered Dec 8 '18 at 13:14
gimusigimusi
92.8k84494
$endgroup$
HINT
We can more effectively use that
$$sqrt{n+2}-sqrt{n}=frac{2}{sqrt{n+2}+sqrt{n}}$$
answered Dec 8 '18 at 13:14
gimusigimusi
92.8k84494
answered Dec 8 '18 at 13:14
gimusigimusi
92.8k84494
answered Dec 8 '18 at 13:14
gimusigimusi
92.8k84494
answered Dec 8 '18 at 13:14
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
$begingroup$
You can use the estimate:
$$frac{sqrt{n+2}-sqrt{n}}{n^{3/2}}<frac1{n^2} iff sqrt{n}(sqrt{n+2}-sqrt{n})<1 iff \
sqrt{n(n+2)}<n+1 iff n^2+2n<n^2+2n+1.$$
answered Dec 8 '18 at 13:45
farruhotafarruhota
19.8k2738
$endgroup$
add a comment |
$begingroup$
You can use the estimate:
$$frac{sqrt{n+2}-sqrt{n}}{n^{3/2}}<frac1{n^2} iff sqrt{n}(sqrt{n+2}-sqrt{n})<1 iff \
sqrt{n(n+2)}<n+1 iff n^2+2n<n^2+2n+1.$$
answered Dec 8 '18 at 13:45
farruhotafarruhota
19.8k2738
$endgroup$
add a comment |
$begingroup$
You can use the estimate:
$$frac{sqrt{n+2}-sqrt{n}}{n^{3/2}}<frac1{n^2} iff sqrt{n}(sqrt{n+2}-sqrt{n})<1 iff \
sqrt{n(n+2)}<n+1 iff n^2+2n<n^2+2n+1.$$
answered Dec 8 '18 at 13:45
farruhotafarruhota
19.8k2738
$endgroup$
You can use the estimate:
$$frac{sqrt{n+2}-sqrt{n}}{n^{3/2}}<frac1{n^2} iff sqrt{n}(sqrt{n+2}-sqrt{n})<1 iff \
sqrt{n(n+2)}<n+1 iff n^2+2n<n^2+2n+1.$$
answered Dec 8 '18 at 13:45
farruhotafarruhota
19.8k2738
answered Dec 8 '18 at 13:45
farruhotafarruhota
19.8k2738
answered Dec 8 '18 at 13:45
farruhotafarruhota
19.8k2738
answered Dec 8 '18 at 13:45
farruhotafarruhota
19.8k2738
19.8k2738
add a comment |
add a comment |
$begingroup$
Rewriting :
$2=(n+2) -n=$
$(sqrt{n+2}-√n)(sqrt{n+2}+√n) gt$
$sqrt{n+2}-√n$, since
$sqrt{n+2}+√n >1$.
Hence
$dfrac{sqrt{n+2}-√n}{n^{3/2}} lt dfrac{2}{n^{3/2}}.$
Use comparison test.
answered Dec 8 '18 at 13:55
Peter SzilasPeter Szilas
11k2821
$endgroup$
add a comment |
$begingroup$
Rewriting :
$2=(n+2) -n=$
$(sqrt{n+2}-√n)(sqrt{n+2}+√n) gt$
$sqrt{n+2}-√n$, since
$sqrt{n+2}+√n >1$.
Hence
$dfrac{sqrt{n+2}-√n}{n^{3/2}} lt dfrac{2}{n^{3/2}}.$
Use comparison test.
answered Dec 8 '18 at 13:55
Peter SzilasPeter Szilas
11k2821
$endgroup$
add a comment |
$begingroup$
Rewriting :
$2=(n+2) -n=$
$(sqrt{n+2}-√n)(sqrt{n+2}+√n) gt$
$sqrt{n+2}-√n$, since
$sqrt{n+2}+√n >1$.
Hence
$dfrac{sqrt{n+2}-√n}{n^{3/2}} lt dfrac{2}{n^{3/2}}.$
Use comparison test.
answered Dec 8 '18 at 13:55
Peter SzilasPeter Szilas
11k2821
$endgroup$
Rewriting :
$2=(n+2) -n=$
$(sqrt{n+2}-√n)(sqrt{n+2}+√n) gt$
$sqrt{n+2}-√n$, since
$sqrt{n+2}+√n >1$.
Hence
$dfrac{sqrt{n+2}-√n}{n^{3/2}} lt dfrac{2}{n^{3/2}}.$
Use comparison test.
answered Dec 8 '18 at 13:55
Peter SzilasPeter Szilas
11k2821
edited Dec 8 '18 at 14:00
answered Dec 8 '18 at 13:55
Peter SzilasPeter Szilas
11k2821
answered Dec 8 '18 at 13:55
Peter SzilasPeter Szilas
11k2821
answered Dec 8 '18 at 13:55
Peter SzilasPeter Szilas
11k2821
11k2821
add a comment |
add a comment |
$begingroup$
Note that $sqrt{n+2}-sqrt nto 0$ as $ntoinfty$, so $;dfrac{sqrt{n+2}-sqrt{n}}{n^{3/2}}=obiggl(dfrac 1{n^{3/2}}biggr)$, and the latter series converges.
answered Dec 8 '18 at 13:33
BernardBernard
119k740113
$endgroup$
$begingroup$
That's a nice way but why $sqrt{n+2}-sqrt nto 0$?
$endgroup$
– gimusi
Dec 8 '18 at 13:37
$begingroup$
That is standard in high school (and your answer shows how it is proved).
$endgroup$
– Bernard
Dec 8 '18 at 13:42
$begingroup$
Yes but it seems that the asker ignore that otherwise probably he wouldn't have posted the question. In my opinion at the present state you answer is not complete regarding a crucial point. Moreover once we know it goes to zero we could simply use comparison test. Please consider my observations in a constructive way.
$endgroup$
– gimusi
Dec 8 '18 at 13:46
1
$begingroup$
@gimusi : I think our answers are more or less complementary. They do not aim the same audience, and I wanted to propose an as concise as possible answer for those who already have some familiarity with the basics of limits.
$endgroup$
– Bernard
Dec 8 '18 at 13:57
$begingroup$
That's ansolutely fine my aim was only to suggest what I thought could make some improvements with that for the asker. But that's of corse all up to you and the answer is fine also in that way. Bye
$endgroup$
– gimusi
Dec 8 '18 at 14:08
add a comment |
$begingroup$
Note that $sqrt{n+2}-sqrt nto 0$ as $ntoinfty$, so $;dfrac{sqrt{n+2}-sqrt{n}}{n^{3/2}}=obiggl(dfrac 1{n^{3/2}}biggr)$, and the latter series converges.
answered Dec 8 '18 at 13:33
BernardBernard
119k740113
$endgroup$
$begingroup$
That's a nice way but why $sqrt{n+2}-sqrt nto 0$?
$endgroup$
– gimusi
Dec 8 '18 at 13:37
$begingroup$
That is standard in high school (and your answer shows how it is proved).
$endgroup$
– Bernard
Dec 8 '18 at 13:42
$begingroup$
Yes but it seems that the asker ignore that otherwise probably he wouldn't have posted the question. In my opinion at the present state you answer is not complete regarding a crucial point. Moreover once we know it goes to zero we could simply use comparison test. Please consider my observations in a constructive way.
$endgroup$
– gimusi
Dec 8 '18 at 13:46
1
$begingroup$
@gimusi : I think our answers are more or less complementary. They do not aim the same audience, and I wanted to propose an as concise as possible answer for those who already have some familiarity with the basics of limits.
$endgroup$
– Bernard
Dec 8 '18 at 13:57
$begingroup$
That's ansolutely fine my aim was only to suggest what I thought could make some improvements with that for the asker. But that's of corse all up to you and the answer is fine also in that way. Bye
$endgroup$
– gimusi
Dec 8 '18 at 14:08
add a comment |
$begingroup$
Note that $sqrt{n+2}-sqrt nto 0$ as $ntoinfty$, so $;dfrac{sqrt{n+2}-sqrt{n}}{n^{3/2}}=obiggl(dfrac 1{n^{3/2}}biggr)$, and the latter series converges.
answered Dec 8 '18 at 13:33
BernardBernard
119k740113
$endgroup$
Note that $sqrt{n+2}-sqrt nto 0$ as $ntoinfty$, so $;dfrac{sqrt{n+2}-sqrt{n}}{n^{3/2}}=obiggl(dfrac 1{n^{3/2}}biggr)$, and the latter series converges.
answered Dec 8 '18 at 13:33
BernardBernard
119k740113
answered Dec 8 '18 at 13:33
BernardBernard
119k740113
answered Dec 8 '18 at 13:33
BernardBernard
119k740113
answered Dec 8 '18 at 13:33
BernardBernard
119k740113
119k740113
$begingroup$
That's a nice way but why $sqrt{n+2}-sqrt nto 0$?
$endgroup$
– gimusi
Dec 8 '18 at 13:37
$begingroup$
That is standard in high school (and your answer shows how it is proved).
$endgroup$
– Bernard
Dec 8 '18 at 13:42
$begingroup$
Yes but it seems that the asker ignore that otherwise probably he wouldn't have posted the question. In my opinion at the present state you answer is not complete regarding a crucial point. Moreover once we know it goes to zero we could simply use comparison test. Please consider my observations in a constructive way.
$endgroup$
– gimusi
Dec 8 '18 at 13:46
1
$begingroup$
@gimusi : I think our answers are more or less complementary. They do not aim the same audience, and I wanted to propose an as concise as possible answer for those who already have some familiarity with the basics of limits.
$endgroup$
– Bernard
Dec 8 '18 at 13:57
$begingroup$
That's ansolutely fine my aim was only to suggest what I thought could make some improvements with that for the asker. But that's of corse all up to you and the answer is fine also in that way. Bye
$endgroup$
– gimusi
Dec 8 '18 at 14:08
add a comment |
$begingroup$
That's a nice way but why $sqrt{n+2}-sqrt nto 0$?
$endgroup$
– gimusi
Dec 8 '18 at 13:37
$begingroup$
That is standard in high school (and your answer shows how it is proved).
$endgroup$
– Bernard
Dec 8 '18 at 13:42
$begingroup$
Yes but it seems that the asker ignore that otherwise probably he wouldn't have posted the question. In my opinion at the present state you answer is not complete regarding a crucial point. Moreover once we know it goes to zero we could simply use comparison test. Please consider my observations in a constructive way.
$endgroup$
– gimusi
Dec 8 '18 at 13:46
1
$begingroup$
@gimusi : I think our answers are more or less complementary. They do not aim the same audience, and I wanted to propose an as concise as possible answer for those who already have some familiarity with the basics of limits.
$endgroup$
– Bernard
Dec 8 '18 at 13:57
$begingroup$
That's ansolutely fine my aim was only to suggest what I thought could make some improvements with that for the asker. But that's of corse all up to you and the answer is fine also in that way. Bye
$endgroup$
– gimusi
Dec 8 '18 at 14:08
$begingroup$
That's a nice way but why $sqrt{n+2}-sqrt nto 0$?
$endgroup$
– gimusi
Dec 8 '18 at 13:37
$begingroup$
That's a nice way but why $sqrt{n+2}-sqrt nto 0$?
$endgroup$
– gimusi
Dec 8 '18 at 13:37
$begingroup$
That is standard in high school (and your answer shows how it is proved).
$endgroup$
– Bernard
Dec 8 '18 at 13:42
$begingroup$
That is standard in high school (and your answer shows how it is proved).
$endgroup$
– Bernard
Dec 8 '18 at 13:42
$begingroup$
Yes but it seems that the asker ignore that otherwise probably he wouldn't have posted the question. In my opinion at the present state you answer is not complete regarding a crucial point. Moreover once we know it goes to zero we could simply use comparison test. Please consider my observations in a constructive way.
$endgroup$
– gimusi
Dec 8 '18 at 13:46
$begingroup$
Yes but it seems that the asker ignore that otherwise probably he wouldn't have posted the question. In my opinion at the present state you answer is not complete regarding a crucial point. Moreover once we know it goes to zero we could simply use comparison test. Please consider my observations in a constructive way.
$endgroup$
– gimusi
Dec 8 '18 at 13:46
1
1
$begingroup$
@gimusi : I think our answers are more or less complementary. They do not aim the same audience, and I wanted to propose an as concise as possible answer for those who already have some familiarity with the basics of limits.
$endgroup$
– Bernard
Dec 8 '18 at 13:57
$begingroup$
@gimusi : I think our answers are more or less complementary. They do not aim the same audience, and I wanted to propose an as concise as possible answer for those who already have some familiarity with the basics of limits.
$endgroup$
– Bernard
Dec 8 '18 at 13:57
$begingroup$
That's ansolutely fine my aim was only to suggest what I thought could make some improvements with that for the asker. But that's of corse all up to you and the answer is fine also in that way. Bye
$endgroup$
– gimusi
Dec 8 '18 at 14:08
$begingroup$
That's ansolutely fine my aim was only to suggest what I thought could make some improvements with that for the asker. But that's of corse all up to you and the answer is fine also in that way. Bye
$endgroup$
– gimusi
Dec 8 '18 at 14:08
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