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a monotonic sequence $x_n$ in $mathbbR$ is convergent iff the sequence $(x_n)^2$ is convergent
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a monotonic sequence $x_n$ in $mathbbR$ is convergent iff the sequence $(x_n)^2$ is convergent.
Forward Implications is easy, nothing to worry about that. I am confused about backward implication For backward Implication my attempt.
Attempt(Please check if this reasoning is True)
Case 1:-
$x_n$ is Increasing.
We have two possibilities further.
Either all $x_n<0$ or $x_n >0$ for n>N, where N is some natural number.
When all $x_n<0$ , we have it bounded above by 0. Hence its convergent.
Now if $x_n >0$ for n>N, where N is some natural number. Then because $(x_n)^2$ is convergent implies $(x_n)^2$ is bounded above. Hence
$x_n$ is also bounded above. Hence $x_n$ is convergent.
Simmilar for when $x_n$ is decreasing.
If its not correct, Tell me the solution
real-analysis
asked Jul 26 at 23:09
Rakesh Bhatt
638112
add a comment |Â
up vote
2
down vote
favorite
a monotonic sequence $x_n$ in $mathbbR$ is convergent iff the sequence $(x_n)^2$ is convergent.
Forward Implications is easy, nothing to worry about that. I am confused about backward implication For backward Implication my attempt.
Attempt(Please check if this reasoning is True)
Case 1:-
$x_n$ is Increasing.
We have two possibilities further.
Either all $x_n<0$ or $x_n >0$ for n>N, where N is some natural number.
When all $x_n<0$ , we have it bounded above by 0. Hence its convergent.
Now if $x_n >0$ for n>N, where N is some natural number. Then because $(x_n)^2$ is convergent implies $(x_n)^2$ is bounded above. Hence
$x_n$ is also bounded above. Hence $x_n$ is convergent.
Simmilar for when $x_n$ is decreasing.
If its not correct, Tell me the solution
real-analysis
asked Jul 26 at 23:09
Rakesh Bhatt
638112
1
This almost works. You need to prove that $x_n$ is bounded in the case that $x_n>0$ eventually. You say it, but don’t prove it.
– Clayton
Jul 26 at 23:20
1
This argument is perfect.
– user578878
Jul 26 at 23:43
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
a monotonic sequence $x_n$ in $mathbbR$ is convergent iff the sequence $(x_n)^2$ is convergent.
Forward Implications is easy, nothing to worry about that. I am confused about backward implication For backward Implication my attempt.
Attempt(Please check if this reasoning is True)
Case 1:-
$x_n$ is Increasing.
We have two possibilities further.
Either all $x_n<0$ or $x_n >0$ for n>N, where N is some natural number.
When all $x_n<0$ , we have it bounded above by 0. Hence its convergent.
Now if $x_n >0$ for n>N, where N is some natural number. Then because $(x_n)^2$ is convergent implies $(x_n)^2$ is bounded above. Hence
$x_n$ is also bounded above. Hence $x_n$ is convergent.
Simmilar for when $x_n$ is decreasing.
If its not correct, Tell me the solution
real-analysis
asked Jul 26 at 23:09
Rakesh Bhatt
638112
a monotonic sequence $x_n$ in $mathbbR$ is convergent iff the sequence $(x_n)^2$ is convergent.
Forward Implications is easy, nothing to worry about that. I am confused about backward implication For backward Implication my attempt.
Attempt(Please check if this reasoning is True)
Case 1:-
$x_n$ is Increasing.
We have two possibilities further.
Either all $x_n<0$ or $x_n >0$ for n>N, where N is some natural number.
When all $x_n<0$ , we have it bounded above by 0. Hence its convergent.
Now if $x_n >0$ for n>N, where N is some natural number. Then because $(x_n)^2$ is convergent implies $(x_n)^2$ is bounded above. Hence
$x_n$ is also bounded above. Hence $x_n$ is convergent.
Simmilar for when $x_n$ is decreasing.
If its not correct, Tell me the solution
real-analysis
asked Jul 26 at 23:09
Rakesh Bhatt
638112
asked Jul 26 at 23:09
Rakesh Bhatt
638112
asked Jul 26 at 23:09
Rakesh Bhatt
638112
asked Jul 26 at 23:09
Rakesh Bhatt
638112
638112
1
This almost works. You need to prove that $x_n$ is bounded in the case that $x_n>0$ eventually. You say it, but don’t prove it.
– Clayton
Jul 26 at 23:20
1
This argument is perfect.
– user578878
Jul 26 at 23:43
add a comment |Â
1
This almost works. You need to prove that $x_n$ is bounded in the case that $x_n>0$ eventually. You say it, but don’t prove it.
– Clayton
Jul 26 at 23:20
1
This argument is perfect.
– user578878
Jul 26 at 23:43
1
1
This almost works. You need to prove that $x_n$ is bounded in the case that $x_n>0$ eventually. You say it, but don’t prove it.
– Clayton
Jul 26 at 23:20
This almost works. You need to prove that $x_n$ is bounded in the case that $x_n>0$ eventually. You say it, but don’t prove it.
– Clayton
Jul 26 at 23:20
1
1
This argument is perfect.
– user578878
Jul 26 at 23:43
This argument is perfect.
– user578878
Jul 26 at 23:43
add a comment |Â
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1
This almost works. You need to prove that $x_n$ is bounded in the case that $x_n>0$ eventually. You say it, but don’t prove it.
– Clayton
Jul 26 at 23:20
1
This argument is perfect.
– user578878
Jul 26 at 23:43