Let $(a_n), (b_n),$ be bounded, then prove that $c_n$ converges and give its value.












3












$begingroup$


Could I get some feedback on the following proof, I feel becoming a good mathematician is through constant feedback and improvement of your work, I tried to make it short, but well-readable.




We are given that $a_n$ and $b_n$ are bounded sequences, and also:
$$ (n-1)a_n leq n^2 c_n leq (n+1)b_n$$
Prove that $c_n$ converges and give the value of the limit.




First of all, we know that these sequences are bounded, so surely we have that:
$$L geq a_n land b_n leq U$$
For some lower bound $L$ and some upper bound $U$ in $mathbb{R}$.
We now apply this:
$$ (n-1) L leq (n-1)a_n leq n^2 c_n leq (n+1)b_n leq (n+1)U$$
$$ frac{L}{n}-frac{L}{n^2}=frac{(n-1) L}{n^2} leq c_nleq frac{(n+1) U}{n^2} =frac{U}{n}+frac{U}{n^2}$$
Both these sequence converge to zero, so we have that in the limit:
$$0 leq lim_{nrightarrow infty} c_n leq 0$$
By the squeeze theorem we have that $lim_{nrightarrow infty} c_n =0$






real-analysis sequences-and-series limits proof-verification







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes, it is correct.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 21:29


















3












$begingroup$


Could I get some feedback on the following proof, I feel becoming a good mathematician is through constant feedback and improvement of your work, I tried to make it short, but well-readable.




We are given that $a_n$ and $b_n$ are bounded sequences, and also:
$$ (n-1)a_n leq n^2 c_n leq (n+1)b_n$$
Prove that $c_n$ converges and give the value of the limit.




First of all, we know that these sequences are bounded, so surely we have that:
$$L geq a_n land b_n leq U$$
For some lower bound $L$ and some upper bound $U$ in $mathbb{R}$.
We now apply this:
$$ (n-1) L leq (n-1)a_n leq n^2 c_n leq (n+1)b_n leq (n+1)U$$
$$ frac{L}{n}-frac{L}{n^2}=frac{(n-1) L}{n^2} leq c_nleq frac{(n+1) U}{n^2} =frac{U}{n}+frac{U}{n^2}$$
Both these sequence converge to zero, so we have that in the limit:
$$0 leq lim_{nrightarrow infty} c_n leq 0$$
By the squeeze theorem we have that $lim_{nrightarrow infty} c_n =0$






real-analysis sequences-and-series limits proof-verification







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes, it is correct.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 21:29
















3












3








3





$begingroup$


Could I get some feedback on the following proof, I feel becoming a good mathematician is through constant feedback and improvement of your work, I tried to make it short, but well-readable.




We are given that $a_n$ and $b_n$ are bounded sequences, and also:
$$ (n-1)a_n leq n^2 c_n leq (n+1)b_n$$
Prove that $c_n$ converges and give the value of the limit.




First of all, we know that these sequences are bounded, so surely we have that:
$$L geq a_n land b_n leq U$$
For some lower bound $L$ and some upper bound $U$ in $mathbb{R}$.
We now apply this:
$$ (n-1) L leq (n-1)a_n leq n^2 c_n leq (n+1)b_n leq (n+1)U$$
$$ frac{L}{n}-frac{L}{n^2}=frac{(n-1) L}{n^2} leq c_nleq frac{(n+1) U}{n^2} =frac{U}{n}+frac{U}{n^2}$$
Both these sequence converge to zero, so we have that in the limit:
$$0 leq lim_{nrightarrow infty} c_n leq 0$$
By the squeeze theorem we have that $lim_{nrightarrow infty} c_n =0$






real-analysis sequences-and-series limits proof-verification







share|cite|improve this question









$endgroup$




Could I get some feedback on the following proof, I feel becoming a good mathematician is through constant feedback and improvement of your work, I tried to make it short, but well-readable.




We are given that $a_n$ and $b_n$ are bounded sequences, and also:
$$ (n-1)a_n leq n^2 c_n leq (n+1)b_n$$
Prove that $c_n$ converges and give the value of the limit.




First of all, we know that these sequences are bounded, so surely we have that:
$$L geq a_n land b_n leq U$$
For some lower bound $L$ and some upper bound $U$ in $mathbb{R}$.
We now apply this:
$$ (n-1) L leq (n-1)a_n leq n^2 c_n leq (n+1)b_n leq (n+1)U$$
$$ frac{L}{n}-frac{L}{n^2}=frac{(n-1) L}{n^2} leq c_nleq frac{(n+1) U}{n^2} =frac{U}{n}+frac{U}{n^2}$$
Both these sequence converge to zero, so we have that in the limit:
$$0 leq lim_{nrightarrow infty} c_n leq 0$$
By the squeeze theorem we have that $lim_{nrightarrow infty} c_n =0$






real-analysis sequences-and-series limits proof-verification




real-analysis sequences-and-series limits proof-verification






share|cite|improve this question













share|cite|improve this question











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asked Nov 29 '18 at 21:15









Wesley StrikWesley Strik

1,653423




1,653423








  • 1




    $begingroup$
    Yes, it is correct.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 21:29
















  • 1




    $begingroup$
    Yes, it is correct.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 21:29










1




1




$begingroup$
Yes, it is correct.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 21:29






$begingroup$
Yes, it is correct.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 21:29












1 Answer
1






active

oldest

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1












$begingroup$

Absolutely fine proof. You may want to find if the given condition can be relaxed. For example, if $n^2$ is replaced by $n^alpha$ in the above proof, which values would work? $n = n^1$ would not work : try to find a counterexample.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice variation on the problem :)
    $endgroup$
    – Wesley Strik
    Nov 29 '18 at 23:36










  • $begingroup$
    You are welcome!
    $endgroup$
    – астон вілла олоф мэллбэрг
    Nov 30 '18 at 1:20











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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









1












$begingroup$

Absolutely fine proof. You may want to find if the given condition can be relaxed. For example, if $n^2$ is replaced by $n^alpha$ in the above proof, which values would work? $n = n^1$ would not work : try to find a counterexample.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice variation on the problem :)
    $endgroup$
    – Wesley Strik
    Nov 29 '18 at 23:36










  • $begingroup$
    You are welcome!
    $endgroup$
    – астон вілла олоф мэллбэрг
    Nov 30 '18 at 1:20
















1












$begingroup$

Absolutely fine proof. You may want to find if the given condition can be relaxed. For example, if $n^2$ is replaced by $n^alpha$ in the above proof, which values would work? $n = n^1$ would not work : try to find a counterexample.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice variation on the problem :)
    $endgroup$
    – Wesley Strik
    Nov 29 '18 at 23:36










  • $begingroup$
    You are welcome!
    $endgroup$
    – астон вілла олоф мэллбэрг
    Nov 30 '18 at 1:20














1












1








1





$begingroup$

Absolutely fine proof. You may want to find if the given condition can be relaxed. For example, if $n^2$ is replaced by $n^alpha$ in the above proof, which values would work? $n = n^1$ would not work : try to find a counterexample.






share|cite|improve this answer









$endgroup$



Absolutely fine proof. You may want to find if the given condition can be relaxed. For example, if $n^2$ is replaced by $n^alpha$ in the above proof, which values would work? $n = n^1$ would not work : try to find a counterexample.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 29 '18 at 23:22









астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

37.5k33376




37.5k33376












  • $begingroup$
    Nice variation on the problem :)
    $endgroup$
    – Wesley Strik
    Nov 29 '18 at 23:36










  • $begingroup$
    You are welcome!
    $endgroup$
    – астон вілла олоф мэллбэрг
    Nov 30 '18 at 1:20


















  • $begingroup$
    Nice variation on the problem :)
    $endgroup$
    – Wesley Strik
    Nov 29 '18 at 23:36










  • $begingroup$
    You are welcome!
    $endgroup$
    – астон вілла олоф мэллбэрг
    Nov 30 '18 at 1:20
















$begingroup$
Nice variation on the problem :)
$endgroup$
– Wesley Strik
Nov 29 '18 at 23:36




$begingroup$
Nice variation on the problem :)
$endgroup$
– Wesley Strik
Nov 29 '18 at 23:36












$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 1:20




$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 1:20


















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