Proof for sequence $a_n$ with accumulation points $0$ and $2$
I tried answering the following question, do my arguments make sense and are they correct?
Suppose we have a sequence $a_n$ with positive terms and accumulation points $0$ and $2$.
We consider the new sequence, $nin mathbb{N}_+$:
$$ b_n= frac{n+a_n}{n cdot a_n}$$
I am now asked to answer the questions:
(a) give an accumulation point of $b_n$
Well, notice that:
$$ b_n= frac{n+a_n}{n cdot a_n}=frac{1}{a_n} + frac{1}{n}$$
In the limit for large $n$, we have that $frac{1}{n} rightarrow 0$. We notice that $a_n$ has a subsequence that converges to $2$, then certainly we have that $b_n$ has a subsequence $b_{n_j}$:
$$ b_{n_j}=frac{1}{a_{n_j}} + frac{1}{{n_j}}rightarrow frac{1}{2}+0$$
So $frac{1}{2}$ is an accumulation point.
(b) Is $b_n$ bounded?
Notice that we have subsequence $a_{n_k}$ that converges to $0$, we now consider $b_{n_k}$, such that:
$$ b_{n_k}=frac{1}{a_{n_k}} + frac{1}{n_k}rightarrow infty $$
Because the sequence $b_n$ has a subsequence that is divergent to $infty$, it cannot be bounded $square$.
real-analysis proof-verification
asked Nov 28 at 17:42
Wesley Strik
1,524422
add a comment |
I tried answering the following question, do my arguments make sense and are they correct?
Suppose we have a sequence $a_n$ with positive terms and accumulation points $0$ and $2$.
We consider the new sequence, $nin mathbb{N}_+$:
$$ b_n= frac{n+a_n}{n cdot a_n}$$
I am now asked to answer the questions:
(a) give an accumulation point of $b_n$
Well, notice that:
$$ b_n= frac{n+a_n}{n cdot a_n}=frac{1}{a_n} + frac{1}{n}$$
In the limit for large $n$, we have that $frac{1}{n} rightarrow 0$. We notice that $a_n$ has a subsequence that converges to $2$, then certainly we have that $b_n$ has a subsequence $b_{n_j}$:
$$ b_{n_j}=frac{1}{a_{n_j}} + frac{1}{{n_j}}rightarrow frac{1}{2}+0$$
So $frac{1}{2}$ is an accumulation point.
(b) Is $b_n$ bounded?
Notice that we have subsequence $a_{n_k}$ that converges to $0$, we now consider $b_{n_k}$, such that:
$$ b_{n_k}=frac{1}{a_{n_k}} + frac{1}{n_k}rightarrow infty $$
Because the sequence $b_n$ has a subsequence that is divergent to $infty$, it cannot be bounded $square$.
real-analysis proof-verification
asked Nov 28 at 17:42
Wesley Strik
1,524422
1
You cannot leave $frac 1n$ as is when taking subsequences. Otherwise, the arguments are fine.
– Gabriel Romon
Nov 28 at 17:52
add a comment |
I tried answering the following question, do my arguments make sense and are they correct?
Suppose we have a sequence $a_n$ with positive terms and accumulation points $0$ and $2$.
We consider the new sequence, $nin mathbb{N}_+$:
$$ b_n= frac{n+a_n}{n cdot a_n}$$
I am now asked to answer the questions:
(a) give an accumulation point of $b_n$
Well, notice that:
$$ b_n= frac{n+a_n}{n cdot a_n}=frac{1}{a_n} + frac{1}{n}$$
In the limit for large $n$, we have that $frac{1}{n} rightarrow 0$. We notice that $a_n$ has a subsequence that converges to $2$, then certainly we have that $b_n$ has a subsequence $b_{n_j}$:
$$ b_{n_j}=frac{1}{a_{n_j}} + frac{1}{{n_j}}rightarrow frac{1}{2}+0$$
So $frac{1}{2}$ is an accumulation point.
(b) Is $b_n$ bounded?
Notice that we have subsequence $a_{n_k}$ that converges to $0$, we now consider $b_{n_k}$, such that:
$$ b_{n_k}=frac{1}{a_{n_k}} + frac{1}{n_k}rightarrow infty $$
Because the sequence $b_n$ has a subsequence that is divergent to $infty$, it cannot be bounded $square$.
real-analysis proof-verification
asked Nov 28 at 17:42
Wesley Strik
1,524422
I tried answering the following question, do my arguments make sense and are they correct?
Suppose we have a sequence $a_n$ with positive terms and accumulation points $0$ and $2$.
We consider the new sequence, $nin mathbb{N}_+$:
$$ b_n= frac{n+a_n}{n cdot a_n}$$
I am now asked to answer the questions:
(a) give an accumulation point of $b_n$
Well, notice that:
$$ b_n= frac{n+a_n}{n cdot a_n}=frac{1}{a_n} + frac{1}{n}$$
In the limit for large $n$, we have that $frac{1}{n} rightarrow 0$. We notice that $a_n$ has a subsequence that converges to $2$, then certainly we have that $b_n$ has a subsequence $b_{n_j}$:
$$ b_{n_j}=frac{1}{a_{n_j}} + frac{1}{{n_j}}rightarrow frac{1}{2}+0$$
So $frac{1}{2}$ is an accumulation point.
(b) Is $b_n$ bounded?
Notice that we have subsequence $a_{n_k}$ that converges to $0$, we now consider $b_{n_k}$, such that:
$$ b_{n_k}=frac{1}{a_{n_k}} + frac{1}{n_k}rightarrow infty $$
Because the sequence $b_n$ has a subsequence that is divergent to $infty$, it cannot be bounded $square$.
real-analysis proof-verification
real-analysis proof-verification
asked Nov 28 at 17:42
Wesley Strik
1,524422
asked Nov 28 at 17:42
Wesley Strik
1,524422
edited Nov 28 at 19:32
asked Nov 28 at 17:42
Wesley Strik
1,524422
asked Nov 28 at 17:42
Wesley Strik
1,524422
asked Nov 28 at 17:42
Wesley Strik
1,524422
1,524422
1
You cannot leave $frac 1n$ as is when taking subsequences. Otherwise, the arguments are fine.
– Gabriel Romon
Nov 28 at 17:52
add a comment |
1
You cannot leave $frac 1n$ as is when taking subsequences. Otherwise, the arguments are fine.
– Gabriel Romon
Nov 28 at 17:52
1
1
You cannot leave $frac 1n$ as is when taking subsequences. Otherwise, the arguments are fine.
– Gabriel Romon
Nov 28 at 17:52
You cannot leave $frac 1n$ as is when taking subsequences. Otherwise, the arguments are fine.
– Gabriel Romon
Nov 28 at 17:52
add a comment |
1 Answer
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The argumentation seems to be correct.
I would just caution you to be a little more careful with notation.
How is $a'_n$ being labeled? Do the subindices coincide when passing to $b'_n$?
answered Nov 28 at 17:53
Jorge Fernández
75k1189190
Is this better?
– Wesley Strik
Nov 28 at 19:32
1
I think it is better !
– Jorge Fernández
Nov 29 at 0:19
add a comment |
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1 Answer
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1 Answer
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active
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The argumentation seems to be correct.
I would just caution you to be a little more careful with notation.
How is $a'_n$ being labeled? Do the subindices coincide when passing to $b'_n$?
answered Nov 28 at 17:53
Jorge Fernández
75k1189190
Is this better?
– Wesley Strik
Nov 28 at 19:32
1
I think it is better !
– Jorge Fernández
Nov 29 at 0:19
add a comment |
The argumentation seems to be correct.
I would just caution you to be a little more careful with notation.
How is $a'_n$ being labeled? Do the subindices coincide when passing to $b'_n$?
answered Nov 28 at 17:53
Jorge Fernández
75k1189190
Is this better?
– Wesley Strik
Nov 28 at 19:32
1
I think it is better !
– Jorge Fernández
Nov 29 at 0:19
add a comment |
The argumentation seems to be correct.
I would just caution you to be a little more careful with notation.
How is $a'_n$ being labeled? Do the subindices coincide when passing to $b'_n$?
answered Nov 28 at 17:53
Jorge Fernández
75k1189190
The argumentation seems to be correct.
I would just caution you to be a little more careful with notation.
How is $a'_n$ being labeled? Do the subindices coincide when passing to $b'_n$?
answered Nov 28 at 17:53
Jorge Fernández
75k1189190
answered Nov 28 at 17:53
Jorge Fernández
75k1189190
answered Nov 28 at 17:53
Jorge Fernández
75k1189190
answered Nov 28 at 17:53
Jorge Fernández
75k1189190
75k1189190
Is this better?
– Wesley Strik
Nov 28 at 19:32
1
I think it is better !
– Jorge Fernández
Nov 29 at 0:19
add a comment |
Is this better?
– Wesley Strik
Nov 28 at 19:32
1
I think it is better !
– Jorge Fernández
Nov 29 at 0:19
Is this better?
– Wesley Strik
Nov 28 at 19:32
Is this better?
– Wesley Strik
Nov 28 at 19:32
1
1
I think it is better !
– Jorge Fernández
Nov 29 at 0:19
I think it is better !
– Jorge Fernández
Nov 29 at 0:19
add a comment |
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You cannot leave $frac 1n$ as is when taking subsequences. Otherwise, the arguments are fine.
– Gabriel Romon
Nov 28 at 17:52