Determine the convergence of $sum (sqrt{n+1}) -(sqrt{n})$
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1
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I tried using the conjugate of $sum (sqrt{n+1}) -(sqrt{n})$, which leads to
$sum frac{1}{(sqrt{n+1})+(sqrt{n})}$,
and tried to comparing it with
$sum frac{1}{(sqrt{n+1})+(sqrt{n})}geq sum frac{1}{2(sqrt{n+1})}$,
which I do not know the further steps to this. Can anyone help me out?
real-analysis sequences-and-series convergence
edited Nov 16 at 15:58
José Carlos Santos
142k20112208
asked Nov 16 at 15:45
Jeffry Santosa
92
add a comment |
up vote
1
down vote
favorite
I tried using the conjugate of $sum (sqrt{n+1}) -(sqrt{n})$, which leads to
$sum frac{1}{(sqrt{n+1})+(sqrt{n})}$,
and tried to comparing it with
$sum frac{1}{(sqrt{n+1})+(sqrt{n})}geq sum frac{1}{2(sqrt{n+1})}$,
which I do not know the further steps to this. Can anyone help me out?
real-analysis sequences-and-series convergence
edited Nov 16 at 15:58
José Carlos Santos
142k20112208
asked Nov 16 at 15:45
Jeffry Santosa
92
And this (up to a positive factor) dominates the harmonic series.
– user1551
Nov 16 at 15:47
Almost everything cancels straight away, without using the conjugate. $sqrt2-sqrt1+sqrt3-sqrt2=sqrt3-sqrt1$
– Empy2
Nov 16 at 15:54
This is an example of a "telescoping" series
– user25959
Nov 16 at 16:11
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I tried using the conjugate of $sum (sqrt{n+1}) -(sqrt{n})$, which leads to
$sum frac{1}{(sqrt{n+1})+(sqrt{n})}$,
and tried to comparing it with
$sum frac{1}{(sqrt{n+1})+(sqrt{n})}geq sum frac{1}{2(sqrt{n+1})}$,
which I do not know the further steps to this. Can anyone help me out?
real-analysis sequences-and-series convergence
edited Nov 16 at 15:58
José Carlos Santos
142k20112208
asked Nov 16 at 15:45
Jeffry Santosa
92
I tried using the conjugate of $sum (sqrt{n+1}) -(sqrt{n})$, which leads to
$sum frac{1}{(sqrt{n+1})+(sqrt{n})}$,
and tried to comparing it with
$sum frac{1}{(sqrt{n+1})+(sqrt{n})}geq sum frac{1}{2(sqrt{n+1})}$,
which I do not know the further steps to this. Can anyone help me out?
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
edited Nov 16 at 15:58
José Carlos Santos
142k20112208
asked Nov 16 at 15:45
Jeffry Santosa
92
edited Nov 16 at 15:58
José Carlos Santos
142k20112208
asked Nov 16 at 15:45
Jeffry Santosa
92
edited Nov 16 at 15:58
José Carlos Santos
142k20112208
edited Nov 16 at 15:58
José Carlos Santos
142k20112208
edited Nov 16 at 15:58
José Carlos Santos
142k20112208
142k20112208
asked Nov 16 at 15:45
Jeffry Santosa
92
asked Nov 16 at 15:45
Jeffry Santosa
92
asked Nov 16 at 15:45
Jeffry Santosa
92
92
And this (up to a positive factor) dominates the harmonic series.
– user1551
Nov 16 at 15:47
Almost everything cancels straight away, without using the conjugate. $sqrt2-sqrt1+sqrt3-sqrt2=sqrt3-sqrt1$
– Empy2
Nov 16 at 15:54
This is an example of a "telescoping" series
– user25959
Nov 16 at 16:11
add a comment |
And this (up to a positive factor) dominates the harmonic series.
– user1551
Nov 16 at 15:47
Almost everything cancels straight away, without using the conjugate. $sqrt2-sqrt1+sqrt3-sqrt2=sqrt3-sqrt1$
– Empy2
Nov 16 at 15:54
This is an example of a "telescoping" series
– user25959
Nov 16 at 16:11
And this (up to a positive factor) dominates the harmonic series.
– user1551
Nov 16 at 15:47
And this (up to a positive factor) dominates the harmonic series.
– user1551
Nov 16 at 15:47
Almost everything cancels straight away, without using the conjugate. $sqrt2-sqrt1+sqrt3-sqrt2=sqrt3-sqrt1$
– Empy2
Nov 16 at 15:54
Almost everything cancels straight away, without using the conjugate. $sqrt2-sqrt1+sqrt3-sqrt2=sqrt3-sqrt1$
– Empy2
Nov 16 at 15:54
This is an example of a "telescoping" series
– user25959
Nov 16 at 16:11
This is an example of a "telescoping" series
– user25959
Nov 16 at 16:11
add a comment |
4 Answers
4
active
oldest
votes
up vote
4
down vote
$$sum_{i=1}^nsqrt{i+1}-sqrt i=sqrt{n+1}-1to infty$$therefore the series is divergent.
answered Nov 16 at 16:03
Mostafa Ayaz
13k3735
add a comment |
up vote
1
down vote
Since $(forall ninmathbb{N}):sqrt{n+1}leqslant n+1$, you have that $(forall ninmathbb{N}):frac1{2sqrt{n+1}}geqslantfrac1{2(n+1)}$. So…
answered Nov 16 at 15:49
José Carlos Santos
142k20112208
And I just can say that since the Harmonic Series diverge, then $frac{1}{2(n+1)}$, which closely resembles the harmonic series, also diverge, and yada yada yada... Am I correct?
– Jeffry Santosa
Nov 16 at 16:23
@JeffrySantosa Yes, that's it. Well, it depends upon the “yada yada yada”, but my guess is that you will be able to do it.
– José Carlos Santos
Nov 16 at 16:30
add a comment |
up vote
1
down vote
Even easier(without resorting to the harmonic series): note that $1 geq frac{1}{sqrt{n}}, frac{1}{2} geq frac{1}{sqrt{n}}, ..., frac{1}{sqrt{n-1}} geq frac{1}{sqrt{n}} and frac{1}{sqrt{n}} geq frac{1}{sqrt{n}}$, which when added up give $sum_{i=1}^{n}frac{1}{sqrt{i}} geq nfrac{1}{sqrt{n}}=sqrt{n}$ which clearly tends to infinity
answered Nov 16 at 16:00
Sorin Tirc
76210
add a comment |
up vote
0
down vote
You are almost done, simply note that
$$frac{1}{2sqrt{n+1}}ge frac1{2(n+1)}$$
which diverges by harmonic series.
answered Nov 16 at 15:50
gimusi
88.5k74394
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
$$sum_{i=1}^nsqrt{i+1}-sqrt i=sqrt{n+1}-1to infty$$therefore the series is divergent.
answered Nov 16 at 16:03
Mostafa Ayaz
13k3735
add a comment |
up vote
4
down vote
$$sum_{i=1}^nsqrt{i+1}-sqrt i=sqrt{n+1}-1to infty$$therefore the series is divergent.
answered Nov 16 at 16:03
Mostafa Ayaz
13k3735
add a comment |
up vote
4
down vote
up vote
4
down vote
$$sum_{i=1}^nsqrt{i+1}-sqrt i=sqrt{n+1}-1to infty$$therefore the series is divergent.
answered Nov 16 at 16:03
Mostafa Ayaz
13k3735
$$sum_{i=1}^nsqrt{i+1}-sqrt i=sqrt{n+1}-1to infty$$therefore the series is divergent.
answered Nov 16 at 16:03
Mostafa Ayaz
13k3735
answered Nov 16 at 16:03
Mostafa Ayaz
13k3735
answered Nov 16 at 16:03
Mostafa Ayaz
13k3735
answered Nov 16 at 16:03
Mostafa Ayaz
13k3735
13k3735
add a comment |
add a comment |
up vote
1
down vote
Since $(forall ninmathbb{N}):sqrt{n+1}leqslant n+1$, you have that $(forall ninmathbb{N}):frac1{2sqrt{n+1}}geqslantfrac1{2(n+1)}$. So…
answered Nov 16 at 15:49
José Carlos Santos
142k20112208
And I just can say that since the Harmonic Series diverge, then $frac{1}{2(n+1)}$, which closely resembles the harmonic series, also diverge, and yada yada yada... Am I correct?
– Jeffry Santosa
Nov 16 at 16:23
@JeffrySantosa Yes, that's it. Well, it depends upon the “yada yada yada”, but my guess is that you will be able to do it.
– José Carlos Santos
Nov 16 at 16:30
add a comment |
up vote
1
down vote
Since $(forall ninmathbb{N}):sqrt{n+1}leqslant n+1$, you have that $(forall ninmathbb{N}):frac1{2sqrt{n+1}}geqslantfrac1{2(n+1)}$. So…
answered Nov 16 at 15:49
José Carlos Santos
142k20112208
And I just can say that since the Harmonic Series diverge, then $frac{1}{2(n+1)}$, which closely resembles the harmonic series, also diverge, and yada yada yada... Am I correct?
– Jeffry Santosa
Nov 16 at 16:23
@JeffrySantosa Yes, that's it. Well, it depends upon the “yada yada yada”, but my guess is that you will be able to do it.
– José Carlos Santos
Nov 16 at 16:30
add a comment |
up vote
1
down vote
up vote
1
down vote
Since $(forall ninmathbb{N}):sqrt{n+1}leqslant n+1$, you have that $(forall ninmathbb{N}):frac1{2sqrt{n+1}}geqslantfrac1{2(n+1)}$. So…
answered Nov 16 at 15:49
José Carlos Santos
142k20112208
Since $(forall ninmathbb{N}):sqrt{n+1}leqslant n+1$, you have that $(forall ninmathbb{N}):frac1{2sqrt{n+1}}geqslantfrac1{2(n+1)}$. So…
answered Nov 16 at 15:49
José Carlos Santos
142k20112208
answered Nov 16 at 15:49
José Carlos Santos
142k20112208
answered Nov 16 at 15:49
José Carlos Santos
142k20112208
answered Nov 16 at 15:49
José Carlos Santos
142k20112208
142k20112208
And I just can say that since the Harmonic Series diverge, then $frac{1}{2(n+1)}$, which closely resembles the harmonic series, also diverge, and yada yada yada... Am I correct?
– Jeffry Santosa
Nov 16 at 16:23
@JeffrySantosa Yes, that's it. Well, it depends upon the “yada yada yada”, but my guess is that you will be able to do it.
– José Carlos Santos
Nov 16 at 16:30
add a comment |
And I just can say that since the Harmonic Series diverge, then $frac{1}{2(n+1)}$, which closely resembles the harmonic series, also diverge, and yada yada yada... Am I correct?
– Jeffry Santosa
Nov 16 at 16:23
@JeffrySantosa Yes, that's it. Well, it depends upon the “yada yada yada”, but my guess is that you will be able to do it.
– José Carlos Santos
Nov 16 at 16:30
And I just can say that since the Harmonic Series diverge, then $frac{1}{2(n+1)}$, which closely resembles the harmonic series, also diverge, and yada yada yada... Am I correct?
– Jeffry Santosa
Nov 16 at 16:23
And I just can say that since the Harmonic Series diverge, then $frac{1}{2(n+1)}$, which closely resembles the harmonic series, also diverge, and yada yada yada... Am I correct?
– Jeffry Santosa
Nov 16 at 16:23
@JeffrySantosa Yes, that's it. Well, it depends upon the “yada yada yada”, but my guess is that you will be able to do it.
– José Carlos Santos
Nov 16 at 16:30
@JeffrySantosa Yes, that's it. Well, it depends upon the “yada yada yada”, but my guess is that you will be able to do it.
– José Carlos Santos
Nov 16 at 16:30
add a comment |
up vote
1
down vote
Even easier(without resorting to the harmonic series): note that $1 geq frac{1}{sqrt{n}}, frac{1}{2} geq frac{1}{sqrt{n}}, ..., frac{1}{sqrt{n-1}} geq frac{1}{sqrt{n}} and frac{1}{sqrt{n}} geq frac{1}{sqrt{n}}$, which when added up give $sum_{i=1}^{n}frac{1}{sqrt{i}} geq nfrac{1}{sqrt{n}}=sqrt{n}$ which clearly tends to infinity
answered Nov 16 at 16:00
Sorin Tirc
76210
add a comment |
up vote
1
down vote
Even easier(without resorting to the harmonic series): note that $1 geq frac{1}{sqrt{n}}, frac{1}{2} geq frac{1}{sqrt{n}}, ..., frac{1}{sqrt{n-1}} geq frac{1}{sqrt{n}} and frac{1}{sqrt{n}} geq frac{1}{sqrt{n}}$, which when added up give $sum_{i=1}^{n}frac{1}{sqrt{i}} geq nfrac{1}{sqrt{n}}=sqrt{n}$ which clearly tends to infinity
answered Nov 16 at 16:00
Sorin Tirc
76210
add a comment |
up vote
1
down vote
up vote
1
down vote
Even easier(without resorting to the harmonic series): note that $1 geq frac{1}{sqrt{n}}, frac{1}{2} geq frac{1}{sqrt{n}}, ..., frac{1}{sqrt{n-1}} geq frac{1}{sqrt{n}} and frac{1}{sqrt{n}} geq frac{1}{sqrt{n}}$, which when added up give $sum_{i=1}^{n}frac{1}{sqrt{i}} geq nfrac{1}{sqrt{n}}=sqrt{n}$ which clearly tends to infinity
answered Nov 16 at 16:00
Sorin Tirc
76210
Even easier(without resorting to the harmonic series): note that $1 geq frac{1}{sqrt{n}}, frac{1}{2} geq frac{1}{sqrt{n}}, ..., frac{1}{sqrt{n-1}} geq frac{1}{sqrt{n}} and frac{1}{sqrt{n}} geq frac{1}{sqrt{n}}$, which when added up give $sum_{i=1}^{n}frac{1}{sqrt{i}} geq nfrac{1}{sqrt{n}}=sqrt{n}$ which clearly tends to infinity
answered Nov 16 at 16:00
Sorin Tirc
76210
answered Nov 16 at 16:00
Sorin Tirc
76210
answered Nov 16 at 16:00
Sorin Tirc
76210
answered Nov 16 at 16:00
Sorin Tirc
76210
76210
add a comment |
add a comment |
up vote
0
down vote
You are almost done, simply note that
$$frac{1}{2sqrt{n+1}}ge frac1{2(n+1)}$$
which diverges by harmonic series.
answered Nov 16 at 15:50
gimusi
88.5k74394
add a comment |
up vote
0
down vote
You are almost done, simply note that
$$frac{1}{2sqrt{n+1}}ge frac1{2(n+1)}$$
which diverges by harmonic series.
answered Nov 16 at 15:50
gimusi
88.5k74394
add a comment |
up vote
0
down vote
up vote
0
down vote
You are almost done, simply note that
$$frac{1}{2sqrt{n+1}}ge frac1{2(n+1)}$$
which diverges by harmonic series.
answered Nov 16 at 15:50
gimusi
88.5k74394
You are almost done, simply note that
$$frac{1}{2sqrt{n+1}}ge frac1{2(n+1)}$$
which diverges by harmonic series.
answered Nov 16 at 15:50
gimusi
88.5k74394
edited Nov 19 at 19:00
answered Nov 16 at 15:50
gimusi
88.5k74394
answered Nov 16 at 15:50
gimusi
88.5k74394
answered Nov 16 at 15:50
gimusi
88.5k74394
88.5k74394
add a comment |
add a comment |
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And this (up to a positive factor) dominates the harmonic series.
– user1551
Nov 16 at 15:47
Almost everything cancels straight away, without using the conjugate. $sqrt2-sqrt1+sqrt3-sqrt2=sqrt3-sqrt1$
– Empy2
Nov 16 at 15:54
This is an example of a "telescoping" series
– user25959
Nov 16 at 16:11