Sequence diverging to infiinity definition question
$begingroup$
Let $a_n$ be a real sequence
I know a definition for a sequence to diverge to infinity, that being: if for all m ∈ real numbers, exists N ∈ natural numbers :
$a_n > m , n≥N$
Is the following definition a proper definition for a real sequence to diverge to infinity as well?
For all m ∈ real numbers, exists $a_n$ ∈ real numbers :
$a_n > m $
Otherwise, I would like some clarification, please.
sequences-and-series
asked Jan 7 at 22:47
ValVal
557
$endgroup$
add a comment |
$begingroup$
Let $a_n$ be a real sequence
I know a definition for a sequence to diverge to infinity, that being: if for all m ∈ real numbers, exists N ∈ natural numbers :
$a_n > m , n≥N$
Is the following definition a proper definition for a real sequence to diverge to infinity as well?
For all m ∈ real numbers, exists $a_n$ ∈ real numbers :
$a_n > m $
Otherwise, I would like some clarification, please.
sequences-and-series
asked Jan 7 at 22:47
ValVal
557
$endgroup$
1
$begingroup$
No. Consider $a_n$ such that $a_{2n}=0$ and $a_{2n+1}=n$ for $n in mathbb{N}$. Then $a_n$ does not converge to $infty$, but satisfies your "definition".
$endgroup$
– greelious
Jan 7 at 22:54
add a comment |
$begingroup$
Let $a_n$ be a real sequence
I know a definition for a sequence to diverge to infinity, that being: if for all m ∈ real numbers, exists N ∈ natural numbers :
$a_n > m , n≥N$
Is the following definition a proper definition for a real sequence to diverge to infinity as well?
For all m ∈ real numbers, exists $a_n$ ∈ real numbers :
$a_n > m $
Otherwise, I would like some clarification, please.
sequences-and-series
asked Jan 7 at 22:47
ValVal
557
$endgroup$
Let $a_n$ be a real sequence
I know a definition for a sequence to diverge to infinity, that being: if for all m ∈ real numbers, exists N ∈ natural numbers :
$a_n > m , n≥N$
Is the following definition a proper definition for a real sequence to diverge to infinity as well?
For all m ∈ real numbers, exists $a_n$ ∈ real numbers :
$a_n > m $
Otherwise, I would like some clarification, please.
sequences-and-series
sequences-and-series
asked Jan 7 at 22:47
ValVal
557
asked Jan 7 at 22:47
ValVal
557
asked Jan 7 at 22:47
ValVal
557
asked Jan 7 at 22:47
ValVal
557
asked Jan 7 at 22:47
ValVal
557
557
1
$begingroup$
No. Consider $a_n$ such that $a_{2n}=0$ and $a_{2n+1}=n$ for $n in mathbb{N}$. Then $a_n$ does not converge to $infty$, but satisfies your "definition".
$endgroup$
– greelious
Jan 7 at 22:54
add a comment |
1
$begingroup$
No. Consider $a_n$ such that $a_{2n}=0$ and $a_{2n+1}=n$ for $n in mathbb{N}$. Then $a_n$ does not converge to $infty$, but satisfies your "definition".
$endgroup$
– greelious
Jan 7 at 22:54
1
1
$begingroup$
No. Consider $a_n$ such that $a_{2n}=0$ and $a_{2n+1}=n$ for $n in mathbb{N}$. Then $a_n$ does not converge to $infty$, but satisfies your "definition".
$endgroup$
– greelious
Jan 7 at 22:54
$begingroup$
No. Consider $a_n$ such that $a_{2n}=0$ and $a_{2n+1}=n$ for $n in mathbb{N}$. Then $a_n$ does not converge to $infty$, but satisfies your "definition".
$endgroup$
– greelious
Jan 7 at 22:54
add a comment |
1 Answer
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$begingroup$
No take the sequence
$$
a_n=(-1)^nn.
$$
Under your proposed definition this sequence diverges to infinity (but it clearly does not).
answered Jan 7 at 22:54
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
add a comment |
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$begingroup$
No take the sequence
$$
a_n=(-1)^nn.
$$
Under your proposed definition this sequence diverges to infinity (but it clearly does not).
answered Jan 7 at 22:54
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
add a comment |
$begingroup$
No take the sequence
$$
a_n=(-1)^nn.
$$
Under your proposed definition this sequence diverges to infinity (but it clearly does not).
answered Jan 7 at 22:54
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
add a comment |
$begingroup$
No take the sequence
$$
a_n=(-1)^nn.
$$
Under your proposed definition this sequence diverges to infinity (but it clearly does not).
answered Jan 7 at 22:54
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
No take the sequence
$$
a_n=(-1)^nn.
$$
Under your proposed definition this sequence diverges to infinity (but it clearly does not).
answered Jan 7 at 22:54
Foobaz JohnFoobaz John
22.9k41552
answered Jan 7 at 22:54
Foobaz JohnFoobaz John
22.9k41552
answered Jan 7 at 22:54
Foobaz JohnFoobaz John
22.9k41552
answered Jan 7 at 22:54
Foobaz JohnFoobaz John
22.9k41552
22.9k41552
add a comment |
add a comment |
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$begingroup$
No. Consider $a_n$ such that $a_{2n}=0$ and $a_{2n+1}=n$ for $n in mathbb{N}$. Then $a_n$ does not converge to $infty$, but satisfies your "definition".
$endgroup$
– greelious
Jan 7 at 22:54