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Definition of limit: $forall n>N$ or $forall n geq N$?
Clash Royale CLAN TAG#URR8PPP
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My question is about the definition of limit.
Definition: The number $a$ is said to be the limit of the sequence $x_n$ if $forall epsilon > 0$, $exists N in mathbbN$ such that $forall n > N$, we have
$$|x_n - a| < epsilon.$$
This definition is in the book "The fundamentals of Mathematical Analysis - Fikhtengol'ts". But in "Principles of Mathematical Analysis - Walter Rudin", he uses the condition $n geq N$. Is there any difference in the definition of limit when we use the conditions $n>N$ and $n geq N$?
calculus real-analysis sequences-and-series limits definition
edited Jul 31 at 12:58
psmears
69949
asked Jul 31 at 11:04
Minh
727
add a comment |Â
up vote
2
down vote
favorite
My question is about the definition of limit.
Definition: The number $a$ is said to be the limit of the sequence $x_n$ if $forall epsilon > 0$, $exists N in mathbbN$ such that $forall n > N$, we have
$$|x_n - a| < epsilon.$$
This definition is in the book "The fundamentals of Mathematical Analysis - Fikhtengol'ts". But in "Principles of Mathematical Analysis - Walter Rudin", he uses the condition $n geq N$. Is there any difference in the definition of limit when we use the conditions $n>N$ and $n geq N$?
calculus real-analysis sequences-and-series limits definition
edited Jul 31 at 12:58
psmears
69949
asked Jul 31 at 11:04
Minh
727
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
My question is about the definition of limit.
Definition: The number $a$ is said to be the limit of the sequence $x_n$ if $forall epsilon > 0$, $exists N in mathbbN$ such that $forall n > N$, we have
$$|x_n - a| < epsilon.$$
This definition is in the book "The fundamentals of Mathematical Analysis - Fikhtengol'ts". But in "Principles of Mathematical Analysis - Walter Rudin", he uses the condition $n geq N$. Is there any difference in the definition of limit when we use the conditions $n>N$ and $n geq N$?
calculus real-analysis sequences-and-series limits definition
edited Jul 31 at 12:58
psmears
69949
asked Jul 31 at 11:04
Minh
727
My question is about the definition of limit.
Definition: The number $a$ is said to be the limit of the sequence $x_n$ if $forall epsilon > 0$, $exists N in mathbbN$ such that $forall n > N$, we have
$$|x_n - a| < epsilon.$$
This definition is in the book "The fundamentals of Mathematical Analysis - Fikhtengol'ts". But in "Principles of Mathematical Analysis - Walter Rudin", he uses the condition $n geq N$. Is there any difference in the definition of limit when we use the conditions $n>N$ and $n geq N$?
calculus real-analysis sequences-and-series limits definition
edited Jul 31 at 12:58
psmears
69949
asked Jul 31 at 11:04
Minh
727
edited Jul 31 at 12:58
psmears
69949
edited Jul 31 at 12:58
psmears
69949
edited Jul 31 at 12:58
psmears
69949
69949
asked Jul 31 at 11:04
Minh
727
asked Jul 31 at 11:04
Minh
727
asked Jul 31 at 11:04
Minh
727
727
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
11
down vote
accepted
No, since $n>Niff ngeqslant N+1$.
answered Jul 31 at 11:09
José Carlos Santos
112k1696172
add a comment |Â
up vote
6
down vote
No. There are no differences ! Try a proof !
answered Jul 31 at 11:06
Fred
37k1237
add a comment |Â
up vote
4
down vote
No, because both of them means the same: the inequality
$|x_n-a|<epsilon$
holds true for every natural number except finitely many.
answered Jul 31 at 11:38
MGy
413
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
No, since $n>Niff ngeqslant N+1$.
answered Jul 31 at 11:09
José Carlos Santos
112k1696172
add a comment |Â
up vote
11
down vote
accepted
No, since $n>Niff ngeqslant N+1$.
answered Jul 31 at 11:09
José Carlos Santos
112k1696172
add a comment |Â
up vote
11
down vote
accepted
up vote
11
down vote
accepted
No, since $n>Niff ngeqslant N+1$.
answered Jul 31 at 11:09
José Carlos Santos
112k1696172
No, since $n>Niff ngeqslant N+1$.
answered Jul 31 at 11:09
José Carlos Santos
112k1696172
answered Jul 31 at 11:09
José Carlos Santos
112k1696172
answered Jul 31 at 11:09
José Carlos Santos
112k1696172
answered Jul 31 at 11:09
José Carlos Santos
112k1696172
112k1696172
add a comment |Â
add a comment |Â
up vote
6
down vote
No. There are no differences ! Try a proof !
answered Jul 31 at 11:06
Fred
37k1237
add a comment |Â
up vote
6
down vote
No. There are no differences ! Try a proof !
answered Jul 31 at 11:06
Fred
37k1237
add a comment |Â
up vote
6
down vote
up vote
6
down vote
No. There are no differences ! Try a proof !
answered Jul 31 at 11:06
Fred
37k1237
No. There are no differences ! Try a proof !
answered Jul 31 at 11:06
Fred
37k1237
answered Jul 31 at 11:06
Fred
37k1237
answered Jul 31 at 11:06
Fred
37k1237
answered Jul 31 at 11:06
Fred
37k1237
37k1237
add a comment |Â
add a comment |Â
up vote
4
down vote
No, because both of them means the same: the inequality
$|x_n-a|<epsilon$
holds true for every natural number except finitely many.
answered Jul 31 at 11:38
MGy
413
add a comment |Â
up vote
4
down vote
No, because both of them means the same: the inequality
$|x_n-a|<epsilon$
holds true for every natural number except finitely many.
answered Jul 31 at 11:38
MGy
413
add a comment |Â
up vote
4
down vote
up vote
4
down vote
No, because both of them means the same: the inequality
$|x_n-a|<epsilon$
holds true for every natural number except finitely many.
answered Jul 31 at 11:38
MGy
413
No, because both of them means the same: the inequality
$|x_n-a|<epsilon$
holds true for every natural number except finitely many.
answered Jul 31 at 11:38
MGy
413
answered Jul 31 at 11:38
MGy
413
answered Jul 31 at 11:38
MGy
413
answered Jul 31 at 11:38
MGy
413
413
add a comment |Â
add a comment |Â
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