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Define a sequence $(a_n)$ as follows $a_1 = 1$ and for $ngeq 1,;a_n+1 = (-1)^nfrac12 left(|a_n| + frac 2right)$
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Define a sequence $(a_n)$ as follows $a_1 = 1$ and for $ngeq 1$
$$a_n+1 = (-1)^nfrac12 left(|a_n| + frac 2right)$$
then what will be it's $limsup (a_n)$ and $liminf (a_n)$ and $sup(a_n)$, $inf(a_n)$.
I tried to determine the nature of $a_2n+1$ and $a_2n$. I got $a_2n+1$ is increasing and $a_2n+1 geq textsomething$ where $a_2n$ is decreasing and $a_2nleq textsomething$. So I could not draw any conclusion.
Can anyone please help me?
sequences-and-series analysis
edited Jul 28 at 17:13
rtybase
8,78221333
asked Jul 27 at 19:18
cmi
6029
add a comment |Â
up vote
2
down vote
favorite
Define a sequence $(a_n)$ as follows $a_1 = 1$ and for $ngeq 1$
$$a_n+1 = (-1)^nfrac12 left(|a_n| + frac 2right)$$
then what will be it's $limsup (a_n)$ and $liminf (a_n)$ and $sup(a_n)$, $inf(a_n)$.
I tried to determine the nature of $a_2n+1$ and $a_2n$. I got $a_2n+1$ is increasing and $a_2n+1 geq textsomething$ where $a_2n$ is decreasing and $a_2nleq textsomething$. So I could not draw any conclusion.
Can anyone please help me?
sequences-and-series analysis
edited Jul 28 at 17:13
rtybase
8,78221333
asked Jul 27 at 19:18
cmi
6029
2
This sequence satisfy $a_2n+1geq sqrt 2$ and decreasing for $ngeq 1$, $a_2nleq -sqrt 2$ and increasing for $ngeq 1$, $limsup a_n = sqrt 2$, $liminf a_n=-sqrt 2$, $sup a_n = frac12 (frac 32+ frac 43)$, $inf a_n=-frac 32$.
– i707107
Jul 27 at 22:01
To prove these, use induction.
– i707107
Jul 27 at 22:02
How $(a_2n)$ becomes increasing sequence for $ngeq 1$? Can you please elaborately explain? I calculated and I got $a_2 = -3/2$ and $a_4 = -201/104$. So $a_2 ge a_4$. @i707107
– cmi
Jul 28 at 1:36
It is not possible for $a_2n$ to be less than $-3/2$. I don't think your $a_4$ is correct.
– i707107
Jul 28 at 2:08
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Define a sequence $(a_n)$ as follows $a_1 = 1$ and for $ngeq 1$
$$a_n+1 = (-1)^nfrac12 left(|a_n| + frac 2right)$$
then what will be it's $limsup (a_n)$ and $liminf (a_n)$ and $sup(a_n)$, $inf(a_n)$.
I tried to determine the nature of $a_2n+1$ and $a_2n$. I got $a_2n+1$ is increasing and $a_2n+1 geq textsomething$ where $a_2n$ is decreasing and $a_2nleq textsomething$. So I could not draw any conclusion.
Can anyone please help me?
sequences-and-series analysis
edited Jul 28 at 17:13
rtybase
8,78221333
asked Jul 27 at 19:18
cmi
6029
Define a sequence $(a_n)$ as follows $a_1 = 1$ and for $ngeq 1$
$$a_n+1 = (-1)^nfrac12 left(|a_n| + frac 2right)$$
then what will be it's $limsup (a_n)$ and $liminf (a_n)$ and $sup(a_n)$, $inf(a_n)$.
I tried to determine the nature of $a_2n+1$ and $a_2n$. I got $a_2n+1$ is increasing and $a_2n+1 geq textsomething$ where $a_2n$ is decreasing and $a_2nleq textsomething$. So I could not draw any conclusion.
Can anyone please help me?
sequences-and-series analysis
edited Jul 28 at 17:13
rtybase
8,78221333
asked Jul 27 at 19:18
cmi
6029
edited Jul 28 at 17:13
rtybase
8,78221333
edited Jul 28 at 17:13
rtybase
8,78221333
edited Jul 28 at 17:13
rtybase
8,78221333
8,78221333
asked Jul 27 at 19:18
cmi
6029
asked Jul 27 at 19:18
cmi
6029
asked Jul 27 at 19:18
cmi
6029
6029
2
This sequence satisfy $a_2n+1geq sqrt 2$ and decreasing for $ngeq 1$, $a_2nleq -sqrt 2$ and increasing for $ngeq 1$, $limsup a_n = sqrt 2$, $liminf a_n=-sqrt 2$, $sup a_n = frac12 (frac 32+ frac 43)$, $inf a_n=-frac 32$.
– i707107
Jul 27 at 22:01
To prove these, use induction.
– i707107
Jul 27 at 22:02
How $(a_2n)$ becomes increasing sequence for $ngeq 1$? Can you please elaborately explain? I calculated and I got $a_2 = -3/2$ and $a_4 = -201/104$. So $a_2 ge a_4$. @i707107
– cmi
Jul 28 at 1:36
It is not possible for $a_2n$ to be less than $-3/2$. I don't think your $a_4$ is correct.
– i707107
Jul 28 at 2:08
add a comment |Â
2
This sequence satisfy $a_2n+1geq sqrt 2$ and decreasing for $ngeq 1$, $a_2nleq -sqrt 2$ and increasing for $ngeq 1$, $limsup a_n = sqrt 2$, $liminf a_n=-sqrt 2$, $sup a_n = frac12 (frac 32+ frac 43)$, $inf a_n=-frac 32$.
– i707107
Jul 27 at 22:01
To prove these, use induction.
– i707107
Jul 27 at 22:02
How $(a_2n)$ becomes increasing sequence for $ngeq 1$? Can you please elaborately explain? I calculated and I got $a_2 = -3/2$ and $a_4 = -201/104$. So $a_2 ge a_4$. @i707107
– cmi
Jul 28 at 1:36
It is not possible for $a_2n$ to be less than $-3/2$. I don't think your $a_4$ is correct.
– i707107
Jul 28 at 2:08
2
2
This sequence satisfy $a_2n+1geq sqrt 2$ and decreasing for $ngeq 1$, $a_2nleq -sqrt 2$ and increasing for $ngeq 1$, $limsup a_n = sqrt 2$, $liminf a_n=-sqrt 2$, $sup a_n = frac12 (frac 32+ frac 43)$, $inf a_n=-frac 32$.
– i707107
Jul 27 at 22:01
This sequence satisfy $a_2n+1geq sqrt 2$ and decreasing for $ngeq 1$, $a_2nleq -sqrt 2$ and increasing for $ngeq 1$, $limsup a_n = sqrt 2$, $liminf a_n=-sqrt 2$, $sup a_n = frac12 (frac 32+ frac 43)$, $inf a_n=-frac 32$.
– i707107
Jul 27 at 22:01
To prove these, use induction.
– i707107
Jul 27 at 22:02
To prove these, use induction.
– i707107
Jul 27 at 22:02
How $(a_2n)$ becomes increasing sequence for $ngeq 1$? Can you please elaborately explain? I calculated and I got $a_2 = -3/2$ and $a_4 = -201/104$. So $a_2 ge a_4$. @i707107
– cmi
Jul 28 at 1:36
How $(a_2n)$ becomes increasing sequence for $ngeq 1$? Can you please elaborately explain? I calculated and I got $a_2 = -3/2$ and $a_4 = -201/104$. So $a_2 ge a_4$. @i707107
– cmi
Jul 28 at 1:36
It is not possible for $a_2n$ to be less than $-3/2$. I don't think your $a_4$ is correct.
– i707107
Jul 28 at 2:08
It is not possible for $a_2n$ to be less than $-3/2$. I don't think your $a_4$ is correct.
– i707107
Jul 28 at 2:08
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Hint. Taking absolute value of $$a_n+1 = (-1)^nfrac12 left(|a_n| + frac 2right)Rightarrow
|a_n+1|=left|(-1)^nfrac12 left(|a_n| + frac 2right)right| Rightarrow \
|a_n+1|=frac12 left(|a_n| + frac 2right)$$
Using this question and answers provided we have
$$limlimits_nrightarrowinfty |a_n|=sqrt2$$
and $(|a_n|)_ninmathbbN$ is decreasing.
It shouldn't be difficult to see that
$limlimits_krightarrowinfty a_2k+1=sqrt2$ and
$limlimits_krightarrowinfty a_2k=-sqrt2$, because
- $0<sqrt2<a_2(k+1)+1<a_2k+1$, thus $(a_2k+1)_kinmathbbN$ has a limit.
- $a_2k<a_2(k+1)<-sqrt2<0$ (because $|a_2k|>|a_2(k+1)|$), thus $(a_2k)_kinmathbbN$ has a limit.
In both cases we can use the fact that $limlimits_ntoinftya_k=a Rightarrow limlimits_ntoinfty|a_k|=|a|$ to conclude (e.g. proof by contradiction) that:
- $limlimits_krightarrowinfty a_2k+1=sqrt2$ and
- $limlimits_krightarrowinfty a_2k=-sqrt2$
answered Jul 27 at 22:55
rtybase
8,78221333
I can't see the last line. Can you please elaborate little bit?
– cmi
Jul 29 at 11:08
@cmi updated ... let me know if you need more details!
– rtybase
Jul 29 at 11:39
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint. Taking absolute value of $$a_n+1 = (-1)^nfrac12 left(|a_n| + frac 2right)Rightarrow
|a_n+1|=left|(-1)^nfrac12 left(|a_n| + frac 2right)right| Rightarrow \
|a_n+1|=frac12 left(|a_n| + frac 2right)$$
Using this question and answers provided we have
$$limlimits_nrightarrowinfty |a_n|=sqrt2$$
and $(|a_n|)_ninmathbbN$ is decreasing.
It shouldn't be difficult to see that
$limlimits_krightarrowinfty a_2k+1=sqrt2$ and
$limlimits_krightarrowinfty a_2k=-sqrt2$, because
- $0<sqrt2<a_2(k+1)+1<a_2k+1$, thus $(a_2k+1)_kinmathbbN$ has a limit.
- $a_2k<a_2(k+1)<-sqrt2<0$ (because $|a_2k|>|a_2(k+1)|$), thus $(a_2k)_kinmathbbN$ has a limit.
In both cases we can use the fact that $limlimits_ntoinftya_k=a Rightarrow limlimits_ntoinfty|a_k|=|a|$ to conclude (e.g. proof by contradiction) that:
- $limlimits_krightarrowinfty a_2k+1=sqrt2$ and
- $limlimits_krightarrowinfty a_2k=-sqrt2$
answered Jul 27 at 22:55
rtybase
8,78221333
I can't see the last line. Can you please elaborate little bit?
– cmi
Jul 29 at 11:08
@cmi updated ... let me know if you need more details!
– rtybase
Jul 29 at 11:39
add a comment |Â
up vote
1
down vote
Hint. Taking absolute value of $$a_n+1 = (-1)^nfrac12 left(|a_n| + frac 2right)Rightarrow
|a_n+1|=left|(-1)^nfrac12 left(|a_n| + frac 2right)right| Rightarrow \
|a_n+1|=frac12 left(|a_n| + frac 2right)$$
Using this question and answers provided we have
$$limlimits_nrightarrowinfty |a_n|=sqrt2$$
and $(|a_n|)_ninmathbbN$ is decreasing.
It shouldn't be difficult to see that
$limlimits_krightarrowinfty a_2k+1=sqrt2$ and
$limlimits_krightarrowinfty a_2k=-sqrt2$, because
- $0<sqrt2<a_2(k+1)+1<a_2k+1$, thus $(a_2k+1)_kinmathbbN$ has a limit.
- $a_2k<a_2(k+1)<-sqrt2<0$ (because $|a_2k|>|a_2(k+1)|$), thus $(a_2k)_kinmathbbN$ has a limit.
In both cases we can use the fact that $limlimits_ntoinftya_k=a Rightarrow limlimits_ntoinfty|a_k|=|a|$ to conclude (e.g. proof by contradiction) that:
- $limlimits_krightarrowinfty a_2k+1=sqrt2$ and
- $limlimits_krightarrowinfty a_2k=-sqrt2$
answered Jul 27 at 22:55
rtybase
8,78221333
I can't see the last line. Can you please elaborate little bit?
– cmi
Jul 29 at 11:08
@cmi updated ... let me know if you need more details!
– rtybase
Jul 29 at 11:39
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint. Taking absolute value of $$a_n+1 = (-1)^nfrac12 left(|a_n| + frac 2right)Rightarrow
|a_n+1|=left|(-1)^nfrac12 left(|a_n| + frac 2right)right| Rightarrow \
|a_n+1|=frac12 left(|a_n| + frac 2right)$$
Using this question and answers provided we have
$$limlimits_nrightarrowinfty |a_n|=sqrt2$$
and $(|a_n|)_ninmathbbN$ is decreasing.
It shouldn't be difficult to see that
$limlimits_krightarrowinfty a_2k+1=sqrt2$ and
$limlimits_krightarrowinfty a_2k=-sqrt2$, because
- $0<sqrt2<a_2(k+1)+1<a_2k+1$, thus $(a_2k+1)_kinmathbbN$ has a limit.
- $a_2k<a_2(k+1)<-sqrt2<0$ (because $|a_2k|>|a_2(k+1)|$), thus $(a_2k)_kinmathbbN$ has a limit.
In both cases we can use the fact that $limlimits_ntoinftya_k=a Rightarrow limlimits_ntoinfty|a_k|=|a|$ to conclude (e.g. proof by contradiction) that:
- $limlimits_krightarrowinfty a_2k+1=sqrt2$ and
- $limlimits_krightarrowinfty a_2k=-sqrt2$
answered Jul 27 at 22:55
rtybase
8,78221333
Hint. Taking absolute value of $$a_n+1 = (-1)^nfrac12 left(|a_n| + frac 2right)Rightarrow
|a_n+1|=left|(-1)^nfrac12 left(|a_n| + frac 2right)right| Rightarrow \
|a_n+1|=frac12 left(|a_n| + frac 2right)$$
Using this question and answers provided we have
$$limlimits_nrightarrowinfty |a_n|=sqrt2$$
and $(|a_n|)_ninmathbbN$ is decreasing.
It shouldn't be difficult to see that
$limlimits_krightarrowinfty a_2k+1=sqrt2$ and
$limlimits_krightarrowinfty a_2k=-sqrt2$, because
- $0<sqrt2<a_2(k+1)+1<a_2k+1$, thus $(a_2k+1)_kinmathbbN$ has a limit.
- $a_2k<a_2(k+1)<-sqrt2<0$ (because $|a_2k|>|a_2(k+1)|$), thus $(a_2k)_kinmathbbN$ has a limit.
In both cases we can use the fact that $limlimits_ntoinftya_k=a Rightarrow limlimits_ntoinfty|a_k|=|a|$ to conclude (e.g. proof by contradiction) that:
- $limlimits_krightarrowinfty a_2k+1=sqrt2$ and
- $limlimits_krightarrowinfty a_2k=-sqrt2$
answered Jul 27 at 22:55
rtybase
8,78221333
edited Jul 29 at 11:39
answered Jul 27 at 22:55
rtybase
8,78221333
answered Jul 27 at 22:55
rtybase
8,78221333
answered Jul 27 at 22:55
rtybase
8,78221333
8,78221333
I can't see the last line. Can you please elaborate little bit?
– cmi
Jul 29 at 11:08
@cmi updated ... let me know if you need more details!
– rtybase
Jul 29 at 11:39
add a comment |Â
I can't see the last line. Can you please elaborate little bit?
– cmi
Jul 29 at 11:08
@cmi updated ... let me know if you need more details!
– rtybase
Jul 29 at 11:39
I can't see the last line. Can you please elaborate little bit?
– cmi
Jul 29 at 11:08
I can't see the last line. Can you please elaborate little bit?
– cmi
Jul 29 at 11:08
@cmi updated ... let me know if you need more details!
– rtybase
Jul 29 at 11:39
@cmi updated ... let me know if you need more details!
– rtybase
Jul 29 at 11:39
add a comment |Â
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2
This sequence satisfy $a_2n+1geq sqrt 2$ and decreasing for $ngeq 1$, $a_2nleq -sqrt 2$ and increasing for $ngeq 1$, $limsup a_n = sqrt 2$, $liminf a_n=-sqrt 2$, $sup a_n = frac12 (frac 32+ frac 43)$, $inf a_n=-frac 32$.
– i707107
Jul 27 at 22:01
To prove these, use induction.
– i707107
Jul 27 at 22:02
How $(a_2n)$ becomes increasing sequence for $ngeq 1$? Can you please elaborately explain? I calculated and I got $a_2 = -3/2$ and $a_4 = -201/104$. So $a_2 ge a_4$. @i707107
– cmi
Jul 28 at 1:36
It is not possible for $a_2n$ to be less than $-3/2$. I don't think your $a_4$ is correct.
– i707107
Jul 28 at 2:08