Determine if this specific sequence is a Cauchy sequence
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I have the following sequence:
$$a_n =sum_k = 1^n (-1)^b_k 1over k^2 $$
And the hint is that I have to prove that:
$$ 1over k^2 < 1over k-1 - 1over k $$
So assuming $m>n$, I have to prove that:
$$forall epsilon >0, exists N in mathbbN,$$ so that $$ forall m,n > N Rightarrow lvert a_m - a_nrvert < epsilon $$
What I gathered so far:
$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert $
$b_k$ is a sequence of natural numbers $1,2,3.....$, so in absolute value, $(-1)^b_k $ is $1$.
Therefore:
$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert leq sum_k = n+1^m 1over k^2. $
From here on, its not so clear to me as to how to proceed.
What should be my next steps?
calculus sequences-and-series limits cauchy-sequences
edited Dec 9 at 10:51
Jonas Lenz
505212
asked Dec 9 at 10:28
Tegernako
616
add a comment |
up vote
3
down vote
favorite
I have the following sequence:
$$a_n =sum_k = 1^n (-1)^b_k 1over k^2 $$
And the hint is that I have to prove that:
$$ 1over k^2 < 1over k-1 - 1over k $$
So assuming $m>n$, I have to prove that:
$$forall epsilon >0, exists N in mathbbN,$$ so that $$ forall m,n > N Rightarrow lvert a_m - a_nrvert < epsilon $$
What I gathered so far:
$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert $
$b_k$ is a sequence of natural numbers $1,2,3.....$, so in absolute value, $(-1)^b_k $ is $1$.
Therefore:
$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert leq sum_k = n+1^m 1over k^2. $
From here on, its not so clear to me as to how to proceed.
What should be my next steps?
calculus sequences-and-series limits cauchy-sequences
edited Dec 9 at 10:51
Jonas Lenz
505212
asked Dec 9 at 10:28
Tegernako
616
1
Note that (as suggested in one of the edits) the last equality in the last line should be an inequality $leq$.
– Yanko
Dec 9 at 10:50
Just saw it, fixed, thanks a lot :)
– Tegernako
Dec 9 at 10:51
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have the following sequence:
$$a_n =sum_k = 1^n (-1)^b_k 1over k^2 $$
And the hint is that I have to prove that:
$$ 1over k^2 < 1over k-1 - 1over k $$
So assuming $m>n$, I have to prove that:
$$forall epsilon >0, exists N in mathbbN,$$ so that $$ forall m,n > N Rightarrow lvert a_m - a_nrvert < epsilon $$
What I gathered so far:
$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert $
$b_k$ is a sequence of natural numbers $1,2,3.....$, so in absolute value, $(-1)^b_k $ is $1$.
Therefore:
$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert leq sum_k = n+1^m 1over k^2. $
From here on, its not so clear to me as to how to proceed.
What should be my next steps?
calculus sequences-and-series limits cauchy-sequences
edited Dec 9 at 10:51
Jonas Lenz
505212
asked Dec 9 at 10:28
Tegernako
616
I have the following sequence:
$$a_n =sum_k = 1^n (-1)^b_k 1over k^2 $$
And the hint is that I have to prove that:
$$ 1over k^2 < 1over k-1 - 1over k $$
So assuming $m>n$, I have to prove that:
$$forall epsilon >0, exists N in mathbbN,$$ so that $$ forall m,n > N Rightarrow lvert a_m - a_nrvert < epsilon $$
What I gathered so far:
$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert $
$b_k$ is a sequence of natural numbers $1,2,3.....$, so in absolute value, $(-1)^b_k $ is $1$.
Therefore:
$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert leq sum_k = n+1^m 1over k^2. $
From here on, its not so clear to me as to how to proceed.
What should be my next steps?
calculus sequences-and-series limits cauchy-sequences
calculus sequences-and-series limits cauchy-sequences
edited Dec 9 at 10:51
Jonas Lenz
505212
asked Dec 9 at 10:28
Tegernako
616
edited Dec 9 at 10:51
Jonas Lenz
505212
asked Dec 9 at 10:28
Tegernako
616
edited Dec 9 at 10:51
Jonas Lenz
505212
edited Dec 9 at 10:51
Jonas Lenz
505212
edited Dec 9 at 10:51
Jonas Lenz
505212
505212
asked Dec 9 at 10:28
Tegernako
616
asked Dec 9 at 10:28
Tegernako
616
asked Dec 9 at 10:28
Tegernako
616
616
1
Note that (as suggested in one of the edits) the last equality in the last line should be an inequality $leq$.
– Yanko
Dec 9 at 10:50
Just saw it, fixed, thanks a lot :)
– Tegernako
Dec 9 at 10:51
add a comment |
1
Note that (as suggested in one of the edits) the last equality in the last line should be an inequality $leq$.
– Yanko
Dec 9 at 10:50
Just saw it, fixed, thanks a lot :)
– Tegernako
Dec 9 at 10:51
1
1
Note that (as suggested in one of the edits) the last equality in the last line should be an inequality $leq$.
– Yanko
Dec 9 at 10:50
Note that (as suggested in one of the edits) the last equality in the last line should be an inequality $leq$.
– Yanko
Dec 9 at 10:50
Just saw it, fixed, thanks a lot :)
– Tegernako
Dec 9 at 10:51
Just saw it, fixed, thanks a lot :)
– Tegernako
Dec 9 at 10:51
add a comment |
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
You're almost done. Since $frac1k^2 leq frac1k-1-frac1k$ you have that
$$sum_k = n+1^m 1over k^2leqsum_k=n+1^m left[frac1k-1-frac1kright]$$ This is a telescoping series which is equal to $frac1n-frac1m$. It converges to zero as $n,mrightarrow 0$.
answered Dec 9 at 10:35
Yanko
5,760723
Exactly! Thanks.
– Tegernako
Dec 9 at 10:41
add a comment |
up vote
5
down vote
Now, use the fact that$$sum_k=n+1^mfrac1k^2<sum_k=n+1^mfrac1k-frac1k-1=frac1n-frac1m.$$
answered Dec 9 at 10:36
José Carlos Santos
147k22117218
1
I've edited my answer. Thank you.
– José Carlos Santos
Dec 9 at 10:39
Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
– Tegernako
Dec 9 at 10:40
1
Why do we need the hint? Can't we simply say that as $sum_k=1^infty frac1k^2$ is convergent, we know that that $sum_k=n^infty frac1k^2$ converges to $0$ as $n to infty$ which shows the claim, too.
– Jonas Lenz
Dec 9 at 10:44
1
In order to prove that, for each $varepsilon>0$, $sum_k=n+1^mfrac1k^2<varepsilon$, if $m$ and $n$ are large enough.
– José Carlos Santos
Dec 9 at 10:46
1
@JonasLenz I guess that is how one shows that $sum_k=1^infty frac1k^2$ convergent.
– Yanko
Dec 9 at 10:49
|
show 1 more comment
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
You're almost done. Since $frac1k^2 leq frac1k-1-frac1k$ you have that
$$sum_k = n+1^m 1over k^2leqsum_k=n+1^m left[frac1k-1-frac1kright]$$ This is a telescoping series which is equal to $frac1n-frac1m$. It converges to zero as $n,mrightarrow 0$.
answered Dec 9 at 10:35
Yanko
5,760723
Exactly! Thanks.
– Tegernako
Dec 9 at 10:41
add a comment |
up vote
5
down vote
accepted
You're almost done. Since $frac1k^2 leq frac1k-1-frac1k$ you have that
$$sum_k = n+1^m 1over k^2leqsum_k=n+1^m left[frac1k-1-frac1kright]$$ This is a telescoping series which is equal to $frac1n-frac1m$. It converges to zero as $n,mrightarrow 0$.
answered Dec 9 at 10:35
Yanko
5,760723
Exactly! Thanks.
– Tegernako
Dec 9 at 10:41
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
You're almost done. Since $frac1k^2 leq frac1k-1-frac1k$ you have that
$$sum_k = n+1^m 1over k^2leqsum_k=n+1^m left[frac1k-1-frac1kright]$$ This is a telescoping series which is equal to $frac1n-frac1m$. It converges to zero as $n,mrightarrow 0$.
answered Dec 9 at 10:35
Yanko
5,760723
You're almost done. Since $frac1k^2 leq frac1k-1-frac1k$ you have that
$$sum_k = n+1^m 1over k^2leqsum_k=n+1^m left[frac1k-1-frac1kright]$$ This is a telescoping series which is equal to $frac1n-frac1m$. It converges to zero as $n,mrightarrow 0$.
answered Dec 9 at 10:35
Yanko
5,760723
edited Dec 9 at 10:51
answered Dec 9 at 10:35
Yanko
5,760723
answered Dec 9 at 10:35
Yanko
5,760723
answered Dec 9 at 10:35
Yanko
5,760723
5,760723
Exactly! Thanks.
– Tegernako
Dec 9 at 10:41
add a comment |
Exactly! Thanks.
– Tegernako
Dec 9 at 10:41
Exactly! Thanks.
– Tegernako
Dec 9 at 10:41
Exactly! Thanks.
– Tegernako
Dec 9 at 10:41
add a comment |
up vote
5
down vote
Now, use the fact that$$sum_k=n+1^mfrac1k^2<sum_k=n+1^mfrac1k-frac1k-1=frac1n-frac1m.$$
answered Dec 9 at 10:36
José Carlos Santos
147k22117218
1
I've edited my answer. Thank you.
– José Carlos Santos
Dec 9 at 10:39
Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
– Tegernako
Dec 9 at 10:40
1
Why do we need the hint? Can't we simply say that as $sum_k=1^infty frac1k^2$ is convergent, we know that that $sum_k=n^infty frac1k^2$ converges to $0$ as $n to infty$ which shows the claim, too.
– Jonas Lenz
Dec 9 at 10:44
1
In order to prove that, for each $varepsilon>0$, $sum_k=n+1^mfrac1k^2<varepsilon$, if $m$ and $n$ are large enough.
– José Carlos Santos
Dec 9 at 10:46
1
@JonasLenz I guess that is how one shows that $sum_k=1^infty frac1k^2$ convergent.
– Yanko
Dec 9 at 10:49
|
show 1 more comment
up vote
5
down vote
Now, use the fact that$$sum_k=n+1^mfrac1k^2<sum_k=n+1^mfrac1k-frac1k-1=frac1n-frac1m.$$
answered Dec 9 at 10:36
José Carlos Santos
147k22117218
1
I've edited my answer. Thank you.
– José Carlos Santos
Dec 9 at 10:39
Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
– Tegernako
Dec 9 at 10:40
1
Why do we need the hint? Can't we simply say that as $sum_k=1^infty frac1k^2$ is convergent, we know that that $sum_k=n^infty frac1k^2$ converges to $0$ as $n to infty$ which shows the claim, too.
– Jonas Lenz
Dec 9 at 10:44
1
In order to prove that, for each $varepsilon>0$, $sum_k=n+1^mfrac1k^2<varepsilon$, if $m$ and $n$ are large enough.
– José Carlos Santos
Dec 9 at 10:46
1
@JonasLenz I guess that is how one shows that $sum_k=1^infty frac1k^2$ convergent.
– Yanko
Dec 9 at 10:49
|
show 1 more comment
up vote
5
down vote
up vote
5
down vote
Now, use the fact that$$sum_k=n+1^mfrac1k^2<sum_k=n+1^mfrac1k-frac1k-1=frac1n-frac1m.$$
answered Dec 9 at 10:36
José Carlos Santos
147k22117218
Now, use the fact that$$sum_k=n+1^mfrac1k^2<sum_k=n+1^mfrac1k-frac1k-1=frac1n-frac1m.$$
answered Dec 9 at 10:36
José Carlos Santos
147k22117218
edited Dec 9 at 10:39
answered Dec 9 at 10:36
José Carlos Santos
147k22117218
answered Dec 9 at 10:36
José Carlos Santos
147k22117218
answered Dec 9 at 10:36
José Carlos Santos
147k22117218
147k22117218
1
I've edited my answer. Thank you.
– José Carlos Santos
Dec 9 at 10:39
Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
– Tegernako
Dec 9 at 10:40
1
Why do we need the hint? Can't we simply say that as $sum_k=1^infty frac1k^2$ is convergent, we know that that $sum_k=n^infty frac1k^2$ converges to $0$ as $n to infty$ which shows the claim, too.
– Jonas Lenz
Dec 9 at 10:44
1
In order to prove that, for each $varepsilon>0$, $sum_k=n+1^mfrac1k^2<varepsilon$, if $m$ and $n$ are large enough.
– José Carlos Santos
Dec 9 at 10:46
1
@JonasLenz I guess that is how one shows that $sum_k=1^infty frac1k^2$ convergent.
– Yanko
Dec 9 at 10:49
|
show 1 more comment
1
I've edited my answer. Thank you.
– José Carlos Santos
Dec 9 at 10:39
Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
– Tegernako
Dec 9 at 10:40
1
Why do we need the hint? Can't we simply say that as $sum_k=1^infty frac1k^2$ is convergent, we know that that $sum_k=n^infty frac1k^2$ converges to $0$ as $n to infty$ which shows the claim, too.
– Jonas Lenz
Dec 9 at 10:44
1
In order to prove that, for each $varepsilon>0$, $sum_k=n+1^mfrac1k^2<varepsilon$, if $m$ and $n$ are large enough.
– José Carlos Santos
Dec 9 at 10:46
1
@JonasLenz I guess that is how one shows that $sum_k=1^infty frac1k^2$ convergent.
– Yanko
Dec 9 at 10:49
1
1
I've edited my answer. Thank you.
– José Carlos Santos
Dec 9 at 10:39
I've edited my answer. Thank you.
– José Carlos Santos
Dec 9 at 10:39
Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
– Tegernako
Dec 9 at 10:40
Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
– Tegernako
Dec 9 at 10:40
1
1
Why do we need the hint? Can't we simply say that as $sum_k=1^infty frac1k^2$ is convergent, we know that that $sum_k=n^infty frac1k^2$ converges to $0$ as $n to infty$ which shows the claim, too.
– Jonas Lenz
Dec 9 at 10:44
Why do we need the hint? Can't we simply say that as $sum_k=1^infty frac1k^2$ is convergent, we know that that $sum_k=n^infty frac1k^2$ converges to $0$ as $n to infty$ which shows the claim, too.
– Jonas Lenz
Dec 9 at 10:44
1
1
In order to prove that, for each $varepsilon>0$, $sum_k=n+1^mfrac1k^2<varepsilon$, if $m$ and $n$ are large enough.
– José Carlos Santos
Dec 9 at 10:46
In order to prove that, for each $varepsilon>0$, $sum_k=n+1^mfrac1k^2<varepsilon$, if $m$ and $n$ are large enough.
– José Carlos Santos
Dec 9 at 10:46
1
1
@JonasLenz I guess that is how one shows that $sum_k=1^infty frac1k^2$ convergent.
– Yanko
Dec 9 at 10:49
@JonasLenz I guess that is how one shows that $sum_k=1^infty frac1k^2$ convergent.
– Yanko
Dec 9 at 10:49
|
show 1 more comment
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1
Note that (as suggested in one of the edits) the last equality in the last line should be an inequality $leq$.
– Yanko
Dec 9 at 10:50
Just saw it, fixed, thanks a lot :)
– Tegernako
Dec 9 at 10:51