For a convergent sequence $limsup b_n = lim b_n$
$begingroup$
Happy new year!
I wish to prove the following result to practice with $limsup:
$
For a convergent sequence$ (b_n)$ we know $limsup b_n = lim b_n$
If $b_n$ is convergent, the set of accumulation points is definitely nonempty as it must contain the limit of $b_n$ and we know that $b_n$ is bounded. Therefore if $V$ is the set of accumulation points of $(b_n)$, we know that $limsup(b_n)=sup(V)$.
Intuitively and by previous exercises I know that since limits are unique, there must be precisely one accumulation point and therefore $V$ contains one element, namely $lim(b_n)$. We now conclude that:
$$limsup b_n = lim b_n $$
Was this argument convincing? Was I a bit sloppy, can I make this argument more precise, is it correct?
After this question I will attempt to prove a similar result for a divergent sequence $(b_n)to infty$ then $limsup b_n = infty$.
Edit/addition: $limsup b_n = infty$ follows immediately from unboundedness (from above) of a divergent sequence and the definition of $limsup$, which is nice.
real-analysis proof-verification limsup-and-liminf
asked Jan 2 at 15:33
Wesley StrikWesley Strik
2,194424
$endgroup$
add a comment |
$begingroup$
Happy new year!
I wish to prove the following result to practice with $limsup:
$
For a convergent sequence$ (b_n)$ we know $limsup b_n = lim b_n$
If $b_n$ is convergent, the set of accumulation points is definitely nonempty as it must contain the limit of $b_n$ and we know that $b_n$ is bounded. Therefore if $V$ is the set of accumulation points of $(b_n)$, we know that $limsup(b_n)=sup(V)$.
Intuitively and by previous exercises I know that since limits are unique, there must be precisely one accumulation point and therefore $V$ contains one element, namely $lim(b_n)$. We now conclude that:
$$limsup b_n = lim b_n $$
Was this argument convincing? Was I a bit sloppy, can I make this argument more precise, is it correct?
After this question I will attempt to prove a similar result for a divergent sequence $(b_n)to infty$ then $limsup b_n = infty$.
Edit/addition: $limsup b_n = infty$ follows immediately from unboundedness (from above) of a divergent sequence and the definition of $limsup$, which is nice.
real-analysis proof-verification limsup-and-liminf
asked Jan 2 at 15:33
Wesley StrikWesley Strik
2,194424
$endgroup$
$begingroup$
What is $sup b_n$?
$endgroup$
– El borito
Jan 2 at 15:41
1
$begingroup$
Oh, my bad, that's a typo
$endgroup$
– Wesley Strik
Jan 2 at 15:44
1
$begingroup$
Your argument is correct.
$endgroup$
– Song
Jan 2 at 17:21
add a comment |
$begingroup$
Happy new year!
I wish to prove the following result to practice with $limsup:
$
For a convergent sequence$ (b_n)$ we know $limsup b_n = lim b_n$
If $b_n$ is convergent, the set of accumulation points is definitely nonempty as it must contain the limit of $b_n$ and we know that $b_n$ is bounded. Therefore if $V$ is the set of accumulation points of $(b_n)$, we know that $limsup(b_n)=sup(V)$.
Intuitively and by previous exercises I know that since limits are unique, there must be precisely one accumulation point and therefore $V$ contains one element, namely $lim(b_n)$. We now conclude that:
$$limsup b_n = lim b_n $$
Was this argument convincing? Was I a bit sloppy, can I make this argument more precise, is it correct?
After this question I will attempt to prove a similar result for a divergent sequence $(b_n)to infty$ then $limsup b_n = infty$.
Edit/addition: $limsup b_n = infty$ follows immediately from unboundedness (from above) of a divergent sequence and the definition of $limsup$, which is nice.
real-analysis proof-verification limsup-and-liminf
asked Jan 2 at 15:33
Wesley StrikWesley Strik
2,194424
$endgroup$
Happy new year!
I wish to prove the following result to practice with $limsup:
$
For a convergent sequence$ (b_n)$ we know $limsup b_n = lim b_n$
If $b_n$ is convergent, the set of accumulation points is definitely nonempty as it must contain the limit of $b_n$ and we know that $b_n$ is bounded. Therefore if $V$ is the set of accumulation points of $(b_n)$, we know that $limsup(b_n)=sup(V)$.
Intuitively and by previous exercises I know that since limits are unique, there must be precisely one accumulation point and therefore $V$ contains one element, namely $lim(b_n)$. We now conclude that:
$$limsup b_n = lim b_n $$
Was this argument convincing? Was I a bit sloppy, can I make this argument more precise, is it correct?
After this question I will attempt to prove a similar result for a divergent sequence $(b_n)to infty$ then $limsup b_n = infty$.
Edit/addition: $limsup b_n = infty$ follows immediately from unboundedness (from above) of a divergent sequence and the definition of $limsup$, which is nice.
real-analysis proof-verification limsup-and-liminf
real-analysis proof-verification limsup-and-liminf
asked Jan 2 at 15:33
Wesley StrikWesley Strik
2,194424
asked Jan 2 at 15:33
Wesley StrikWesley Strik
2,194424
edited Jan 2 at 15:44
Wesley Strik
asked Jan 2 at 15:33
Wesley StrikWesley Strik
2,194424
asked Jan 2 at 15:33
Wesley StrikWesley Strik
2,194424
asked Jan 2 at 15:33
Wesley StrikWesley Strik
2,194424
2,194424
$begingroup$
What is $sup b_n$?
$endgroup$
– El borito
Jan 2 at 15:41
1
$begingroup$
Oh, my bad, that's a typo
$endgroup$
– Wesley Strik
Jan 2 at 15:44
1
$begingroup$
Your argument is correct.
$endgroup$
– Song
Jan 2 at 17:21
add a comment |
$begingroup$
What is $sup b_n$?
$endgroup$
– El borito
Jan 2 at 15:41
1
$begingroup$
Oh, my bad, that's a typo
$endgroup$
– Wesley Strik
Jan 2 at 15:44
1
$begingroup$
Your argument is correct.
$endgroup$
– Song
Jan 2 at 17:21
$begingroup$
What is $sup b_n$?
$endgroup$
– El borito
Jan 2 at 15:41
$begingroup$
What is $sup b_n$?
$endgroup$
– El borito
Jan 2 at 15:41
1
1
$begingroup$
Oh, my bad, that's a typo
$endgroup$
– Wesley Strik
Jan 2 at 15:44
$begingroup$
Oh, my bad, that's a typo
$endgroup$
– Wesley Strik
Jan 2 at 15:44
1
1
$begingroup$
Your argument is correct.
$endgroup$
– Song
Jan 2 at 17:21
$begingroup$
Your argument is correct.
$endgroup$
– Song
Jan 2 at 17:21
add a comment |
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$begingroup$
What is $sup b_n$?
$endgroup$
– El borito
Jan 2 at 15:41
1
$begingroup$
Oh, my bad, that's a typo
$endgroup$
– Wesley Strik
Jan 2 at 15:44
1
$begingroup$
Your argument is correct.
$endgroup$
– Song
Jan 2 at 17:21