Proof for sequence $a_n$ with accumulation points $0$ and $2$












1












$begingroup$


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







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$endgroup$








  • 1




    $begingroup$
    You cannot leave $frac 1n$ as is when taking subsequences. Otherwise, the arguments are fine.
    $endgroup$
    – Gabriel Romon
    Nov 28 '18 at 17:52
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You cannot leave $frac 1n$ as is when taking subsequences. Otherwise, the arguments are fine.
    $endgroup$
    – Gabriel Romon
    Nov 28 '18 at 17:52














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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edited Nov 28 '18 at 19:32







Wesley Strik

















asked Nov 28 '18 at 17:42









Wesley StrikWesley Strik

1,635423




1,635423








  • 1




    $begingroup$
    You cannot leave $frac 1n$ as is when taking subsequences. Otherwise, the arguments are fine.
    $endgroup$
    – Gabriel Romon
    Nov 28 '18 at 17:52














  • 1




    $begingroup$
    You cannot leave $frac 1n$ as is when taking subsequences. Otherwise, the arguments are fine.
    $endgroup$
    – Gabriel Romon
    Nov 28 '18 at 17:52








1




1




$begingroup$
You cannot leave $frac 1n$ as is when taking subsequences. Otherwise, the arguments are fine.
$endgroup$
– Gabriel Romon
Nov 28 '18 at 17:52




$begingroup$
You cannot leave $frac 1n$ as is when taking subsequences. Otherwise, the arguments are fine.
$endgroup$
– Gabriel Romon
Nov 28 '18 at 17:52










1 Answer
1






active

oldest

votes


















2












$begingroup$

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 $b'_n$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is this better?
    $endgroup$
    – Wesley Strik
    Nov 28 '18 at 19:32






  • 1




    $begingroup$
    I think it is better !
    $endgroup$
    – Jorge Fernández
    Nov 29 '18 at 0:19











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

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 $b'_n$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is this better?
    $endgroup$
    – Wesley Strik
    Nov 28 '18 at 19:32






  • 1




    $begingroup$
    I think it is better !
    $endgroup$
    – Jorge Fernández
    Nov 29 '18 at 0:19
















2












$begingroup$

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 $b'_n$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is this better?
    $endgroup$
    – Wesley Strik
    Nov 28 '18 at 19:32






  • 1




    $begingroup$
    I think it is better !
    $endgroup$
    – Jorge Fernández
    Nov 29 '18 at 0:19














2












2








2





$begingroup$

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 $b'_n$?






share|cite|improve this answer









$endgroup$



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 $b'_n$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 28 '18 at 17:53









Jorge FernándezJorge Fernández

75.1k1190191




75.1k1190191












  • $begingroup$
    Is this better?
    $endgroup$
    – Wesley Strik
    Nov 28 '18 at 19:32






  • 1




    $begingroup$
    I think it is better !
    $endgroup$
    – Jorge Fernández
    Nov 29 '18 at 0:19


















  • $begingroup$
    Is this better?
    $endgroup$
    – Wesley Strik
    Nov 28 '18 at 19:32






  • 1




    $begingroup$
    I think it is better !
    $endgroup$
    – Jorge Fernández
    Nov 29 '18 at 0:19
















$begingroup$
Is this better?
$endgroup$
– Wesley Strik
Nov 28 '18 at 19:32




$begingroup$
Is this better?
$endgroup$
– Wesley Strik
Nov 28 '18 at 19:32




1




1




$begingroup$
I think it is better !
$endgroup$
– Jorge Fernández
Nov 29 '18 at 0:19




$begingroup$
I think it is better !
$endgroup$
– Jorge Fernández
Nov 29 '18 at 0:19


















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