If a sequence $s_n$ of real numbers converges to $s>a$ then $s_n>a$Prove that the convergence of the sequence ($s_n$) implies the convergence of ($s_n^3$)If a sequence $s_n$ converges, does $f(s_n)$ converge?Simple sequence convergence question…Conclude that if all but finitely many $s_n$ belong to $[a, b]$, lim $s_n$ belongs to $[a,b]$Simple proof that this sequence converges [verification]Show that if $(sum x_n)$ converges absolutely and $(y_n)$ is bounded then $(sum x_n y_n)$ convergesShow that if all convergent sub-sequences of a sequence $s_n$ converge to 0 and $s_n$ is bounded, then $s_n$ converges to $0$Sequence of divisions such that its upper and lower function limits are equalIf sequence $s_n$ converges, then sequence $$ converges as well(check the logic)Show directly that if $s_n$ is a Cauchy sequence then so is $$. Conclude that $$ converges whenever $s_n$ converges.
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If a sequence $s_n$ of real numbers converges to $s>a$ then $s_n>a$
Prove that the convergence of the sequence ($s_n$) implies the convergence of ($s_n^3$)If a sequence $s_n$ converges, does $f(s_n)$ converge?Simple sequence convergence question…Conclude that if all but finitely many $s_n$ belong to $[a, b]$, lim $s_n$ belongs to $[a,b]$Simple proof that this sequence converges [verification]Show that if $(sum x_n)$ converges absolutely and $(y_n)$ is bounded then $(sum x_n y_n)$ convergesShow that if all convergent sub-sequences of a sequence $s_n$ converge to 0 and $s_n$ is bounded, then $s_n$ converges to $0$Sequence of divisions such that its upper and lower function limits are equalIf sequence $s_n$ converges, then sequence $$ converges as well(check the logic)Show directly that if $s_n$ is a Cauchy sequence then so is $$. Conclude that $$ converges whenever $s_n$ converges.
$begingroup$
Let $s_n$ be a sequence of real numbers that converges to $s>a$. Then there exists an $N>0$ such that $n>Nimplies s_n>a$.
PROOF:
By definition of convergence in sequences we have that for all $varepsilon>0$, there exists $N>0$ such that $n>N$ implies $|s_n-s|<varepsilon$.
Choose $varepsilon=s-a$. Then $|s_n-s|<s-a$. By properties of absolute values and inequalities we have $-s+a<s_n-s<s-a$ which implies $s_n>a$.
Is my proof rigorous, correct? I'd appreciate any comments for improvement.
real-analysis proof-verification convergence
edited yesterday
Shaun
9,334113684
asked yesterday
johnny09johnny09
673624
$endgroup$
add a comment |
$begingroup$
Let $s_n$ be a sequence of real numbers that converges to $s>a$. Then there exists an $N>0$ such that $n>Nimplies s_n>a$.
PROOF:
By definition of convergence in sequences we have that for all $varepsilon>0$, there exists $N>0$ such that $n>N$ implies $|s_n-s|<varepsilon$.
Choose $varepsilon=s-a$. Then $|s_n-s|<s-a$. By properties of absolute values and inequalities we have $-s+a<s_n-s<s-a$ which implies $s_n>a$.
Is my proof rigorous, correct? I'd appreciate any comments for improvement.
real-analysis proof-verification convergence
edited yesterday
Shaun
9,334113684
asked yesterday
johnny09johnny09
673624
$endgroup$
$begingroup$
Looks okay to me, do you know where you used the condition $s>a$ or did you even use it?
$endgroup$
– kingW3
yesterday
$begingroup$
@kingW3 I think implicitly I assumed that $s-a>0$ since $varepsilon>0$. Is that okay to do?
$endgroup$
– johnny09
yesterday
1
$begingroup$
It is, you're able to use $epsilon = s-a$ because $s-a>0iff s>a$. You could mention it explicitly like "Choose $epsilon=s-a>0$" but you don't really need to. I just wanted to check whether you understood that.
$endgroup$
– kingW3
yesterday
add a comment |
$begingroup$
Let $s_n$ be a sequence of real numbers that converges to $s>a$. Then there exists an $N>0$ such that $n>Nimplies s_n>a$.
PROOF:
By definition of convergence in sequences we have that for all $varepsilon>0$, there exists $N>0$ such that $n>N$ implies $|s_n-s|<varepsilon$.
Choose $varepsilon=s-a$. Then $|s_n-s|<s-a$. By properties of absolute values and inequalities we have $-s+a<s_n-s<s-a$ which implies $s_n>a$.
Is my proof rigorous, correct? I'd appreciate any comments for improvement.
real-analysis proof-verification convergence
edited yesterday
Shaun
9,334113684
asked yesterday
johnny09johnny09
673624
$endgroup$
Let $s_n$ be a sequence of real numbers that converges to $s>a$. Then there exists an $N>0$ such that $n>Nimplies s_n>a$.
PROOF:
By definition of convergence in sequences we have that for all $varepsilon>0$, there exists $N>0$ such that $n>N$ implies $|s_n-s|<varepsilon$.
Choose $varepsilon=s-a$. Then $|s_n-s|<s-a$. By properties of absolute values and inequalities we have $-s+a<s_n-s<s-a$ which implies $s_n>a$.
Is my proof rigorous, correct? I'd appreciate any comments for improvement.
real-analysis proof-verification convergence
real-analysis proof-verification convergence
edited yesterday
Shaun
9,334113684
asked yesterday
johnny09johnny09
673624
edited yesterday
Shaun
9,334113684
asked yesterday
johnny09johnny09
673624
edited yesterday
Shaun
9,334113684
edited yesterday
Shaun
9,334113684
edited yesterday
Shaun
9,334113684
9,334113684
asked yesterday
johnny09johnny09
673624
asked yesterday
johnny09johnny09
673624
asked yesterday
johnny09johnny09
673624
673624
$begingroup$
Looks okay to me, do you know where you used the condition $s>a$ or did you even use it?
$endgroup$
– kingW3
yesterday
$begingroup$
@kingW3 I think implicitly I assumed that $s-a>0$ since $varepsilon>0$. Is that okay to do?
$endgroup$
– johnny09
yesterday
1
$begingroup$
It is, you're able to use $epsilon = s-a$ because $s-a>0iff s>a$. You could mention it explicitly like "Choose $epsilon=s-a>0$" but you don't really need to. I just wanted to check whether you understood that.
$endgroup$
– kingW3
yesterday
add a comment |
$begingroup$
Looks okay to me, do you know where you used the condition $s>a$ or did you even use it?
$endgroup$
– kingW3
yesterday
$begingroup$
@kingW3 I think implicitly I assumed that $s-a>0$ since $varepsilon>0$. Is that okay to do?
$endgroup$
– johnny09
yesterday
1
$begingroup$
It is, you're able to use $epsilon = s-a$ because $s-a>0iff s>a$. You could mention it explicitly like "Choose $epsilon=s-a>0$" but you don't really need to. I just wanted to check whether you understood that.
$endgroup$
– kingW3
yesterday
$begingroup$
Looks okay to me, do you know where you used the condition $s>a$ or did you even use it?
$endgroup$
– kingW3
yesterday
$begingroup$
Looks okay to me, do you know where you used the condition $s>a$ or did you even use it?
$endgroup$
– kingW3
yesterday
$begingroup$
@kingW3 I think implicitly I assumed that $s-a>0$ since $varepsilon>0$. Is that okay to do?
$endgroup$
– johnny09
yesterday
$begingroup$
@kingW3 I think implicitly I assumed that $s-a>0$ since $varepsilon>0$. Is that okay to do?
$endgroup$
– johnny09
yesterday
1
1
$begingroup$
It is, you're able to use $epsilon = s-a$ because $s-a>0iff s>a$. You could mention it explicitly like "Choose $epsilon=s-a>0$" but you don't really need to. I just wanted to check whether you understood that.
$endgroup$
– kingW3
yesterday
$begingroup$
It is, you're able to use $epsilon = s-a$ because $s-a>0iff s>a$. You could mention it explicitly like "Choose $epsilon=s-a>0$" but you don't really need to. I just wanted to check whether you understood that.
$endgroup$
– kingW3
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I would just add, after “Choose $varepsilon=s-a$”, that there is a $Ninmathbb N$ such that $n>Nimplieslvert s_n-srvert<varepsilon$.
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
Something like: "Let $varepsilon=s-a>0$. Then, by definition of limit of sequence, there exists $N in mathbbN$ such that $|s_n-s|<varepsilon=s-a$ whenever $n>N$. By properties of absolute values and inequalities we have $-s+a<s_n-s$ which implies $s_n>a$ for $n>N$."
$endgroup$
– md2perpe
yesterday
$begingroup$
That would be just fine!
$endgroup$
– José Carlos Santos
yesterday
add a comment |
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1 Answer
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1 Answer
1
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oldest
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oldest
votes
$begingroup$
I would just add, after “Choose $varepsilon=s-a$”, that there is a $Ninmathbb N$ such that $n>Nimplieslvert s_n-srvert<varepsilon$.
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
Something like: "Let $varepsilon=s-a>0$. Then, by definition of limit of sequence, there exists $N in mathbbN$ such that $|s_n-s|<varepsilon=s-a$ whenever $n>N$. By properties of absolute values and inequalities we have $-s+a<s_n-s$ which implies $s_n>a$ for $n>N$."
$endgroup$
– md2perpe
yesterday
$begingroup$
That would be just fine!
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
I would just add, after “Choose $varepsilon=s-a$”, that there is a $Ninmathbb N$ such that $n>Nimplieslvert s_n-srvert<varepsilon$.
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
Something like: "Let $varepsilon=s-a>0$. Then, by definition of limit of sequence, there exists $N in mathbbN$ such that $|s_n-s|<varepsilon=s-a$ whenever $n>N$. By properties of absolute values and inequalities we have $-s+a<s_n-s$ which implies $s_n>a$ for $n>N$."
$endgroup$
– md2perpe
yesterday
$begingroup$
That would be just fine!
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
I would just add, after “Choose $varepsilon=s-a$”, that there is a $Ninmathbb N$ such that $n>Nimplieslvert s_n-srvert<varepsilon$.
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
I would just add, after “Choose $varepsilon=s-a$”, that there is a $Ninmathbb N$ such that $n>Nimplieslvert s_n-srvert<varepsilon$.
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
$begingroup$
Something like: "Let $varepsilon=s-a>0$. Then, by definition of limit of sequence, there exists $N in mathbbN$ such that $|s_n-s|<varepsilon=s-a$ whenever $n>N$. By properties of absolute values and inequalities we have $-s+a<s_n-s$ which implies $s_n>a$ for $n>N$."
$endgroup$
– md2perpe
yesterday
$begingroup$
That would be just fine!
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
Something like: "Let $varepsilon=s-a>0$. Then, by definition of limit of sequence, there exists $N in mathbbN$ such that $|s_n-s|<varepsilon=s-a$ whenever $n>N$. By properties of absolute values and inequalities we have $-s+a<s_n-s$ which implies $s_n>a$ for $n>N$."
$endgroup$
– md2perpe
yesterday
$begingroup$
That would be just fine!
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
Something like: "Let $varepsilon=s-a>0$. Then, by definition of limit of sequence, there exists $N in mathbbN$ such that $|s_n-s|<varepsilon=s-a$ whenever $n>N$. By properties of absolute values and inequalities we have $-s+a<s_n-s$ which implies $s_n>a$ for $n>N$."
$endgroup$
– md2perpe
yesterday
$begingroup$
Something like: "Let $varepsilon=s-a>0$. Then, by definition of limit of sequence, there exists $N in mathbbN$ such that $|s_n-s|<varepsilon=s-a$ whenever $n>N$. By properties of absolute values and inequalities we have $-s+a<s_n-s$ which implies $s_n>a$ for $n>N$."
$endgroup$
– md2perpe
yesterday
$begingroup$
That would be just fine!
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
That would be just fine!
$endgroup$
– José Carlos Santos
yesterday
add a comment |
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$begingroup$
Looks okay to me, do you know where you used the condition $s>a$ or did you even use it?
$endgroup$
– kingW3
yesterday
$begingroup$
@kingW3 I think implicitly I assumed that $s-a>0$ since $varepsilon>0$. Is that okay to do?
$endgroup$
– johnny09
yesterday
1
$begingroup$
It is, you're able to use $epsilon = s-a$ because $s-a>0iff s>a$. You could mention it explicitly like "Choose $epsilon=s-a>0$" but you don't really need to. I just wanted to check whether you understood that.
$endgroup$
– kingW3
yesterday