Concerning the limit of the numerator of a rational function
$begingroup$
I have a basic question about the limits of rational sequences:
Suppose we know that a sequence $(a_{n})$ converges to $0$ and that each $a_n$ is defined by $a_n := frac{b_n}{c_n}$ for two sequences $(b_{n})$ and $(c_{n})$. We further know that $(c_{n})$ goes to infinity, i.e
$$lim_{n rightarrow infty} c_n = infty.$$
Can we deduce from this that $lim_{n rightarrow infty} b_n = 0$ and if yes how?
Edit: I have an additional question: Are we at least able to deduce that $b_{(n)}$ converges?
sequences-and-series limits
asked Dec 17 '18 at 13:10
3nondatur3nondatur
385111
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add a comment |
$begingroup$
I have a basic question about the limits of rational sequences:
Suppose we know that a sequence $(a_{n})$ converges to $0$ and that each $a_n$ is defined by $a_n := frac{b_n}{c_n}$ for two sequences $(b_{n})$ and $(c_{n})$. We further know that $(c_{n})$ goes to infinity, i.e
$$lim_{n rightarrow infty} c_n = infty.$$
Can we deduce from this that $lim_{n rightarrow infty} b_n = 0$ and if yes how?
Edit: I have an additional question: Are we at least able to deduce that $b_{(n)}$ converges?
sequences-and-series limits
asked Dec 17 '18 at 13:10
3nondatur3nondatur
385111
$endgroup$
add a comment |
$begingroup$
I have a basic question about the limits of rational sequences:
Suppose we know that a sequence $(a_{n})$ converges to $0$ and that each $a_n$ is defined by $a_n := frac{b_n}{c_n}$ for two sequences $(b_{n})$ and $(c_{n})$. We further know that $(c_{n})$ goes to infinity, i.e
$$lim_{n rightarrow infty} c_n = infty.$$
Can we deduce from this that $lim_{n rightarrow infty} b_n = 0$ and if yes how?
Edit: I have an additional question: Are we at least able to deduce that $b_{(n)}$ converges?
sequences-and-series limits
asked Dec 17 '18 at 13:10
3nondatur3nondatur
385111
$endgroup$
I have a basic question about the limits of rational sequences:
Suppose we know that a sequence $(a_{n})$ converges to $0$ and that each $a_n$ is defined by $a_n := frac{b_n}{c_n}$ for two sequences $(b_{n})$ and $(c_{n})$. We further know that $(c_{n})$ goes to infinity, i.e
$$lim_{n rightarrow infty} c_n = infty.$$
Can we deduce from this that $lim_{n rightarrow infty} b_n = 0$ and if yes how?
Edit: I have an additional question: Are we at least able to deduce that $b_{(n)}$ converges?
sequences-and-series limits
sequences-and-series limits
asked Dec 17 '18 at 13:10
3nondatur3nondatur
385111
asked Dec 17 '18 at 13:10
3nondatur3nondatur
385111
edited Dec 17 '18 at 14:10
3nondatur
asked Dec 17 '18 at 13:10
3nondatur3nondatur
385111
asked Dec 17 '18 at 13:10
3nondatur3nondatur
385111
asked Dec 17 '18 at 13:10
3nondatur3nondatur
385111
385111
add a comment |
add a comment |
2 Answers
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No.
What about $$left( a_n right)=left( frac{1}{n}right)$$
where $a_n=frac{1}{n}$,
$ b_n=1$ and $ c_n=n$
Edit-Edited to answer your additional question.
Again No.
What about $$left( a_n right)=left( frac{n}{n^2}right)$$
where $a_n=frac{n}{n^2}$,
$ b_n=n$ and $ c_n=n^2$
answered Dec 17 '18 at 13:13
Rakesh BhattRakesh Bhatt
952114
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$begingroup$
No, that’s not true. For instance, for $b_n = n$ and $c_n = n^2$, we have
$$lim_{n to infty}frac{b_n}{c_n} = 0$$
while both sequences tend to $infty$. More generally, when both $b_n$ and $c_n$ are polynomials with $c_n$ having the higher degree, the limit tends to $0$, even though both polynomials may tend to $infty$.
Edit: Also no, the example above once again shows this. Both polynomials clearly diverge to $infty$, but since the denominator has a greater degree, it diverges faster. Hence, the limit tends to $0$.
answered Dec 17 '18 at 13:22
KM101KM101
5,9131523
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
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votes
$begingroup$
No.
What about $$left( a_n right)=left( frac{1}{n}right)$$
where $a_n=frac{1}{n}$,
$ b_n=1$ and $ c_n=n$
Edit-Edited to answer your additional question.
Again No.
What about $$left( a_n right)=left( frac{n}{n^2}right)$$
where $a_n=frac{n}{n^2}$,
$ b_n=n$ and $ c_n=n^2$
answered Dec 17 '18 at 13:13
Rakesh BhattRakesh Bhatt
952114
$endgroup$
add a comment |
$begingroup$
No.
What about $$left( a_n right)=left( frac{1}{n}right)$$
where $a_n=frac{1}{n}$,
$ b_n=1$ and $ c_n=n$
Edit-Edited to answer your additional question.
Again No.
What about $$left( a_n right)=left( frac{n}{n^2}right)$$
where $a_n=frac{n}{n^2}$,
$ b_n=n$ and $ c_n=n^2$
answered Dec 17 '18 at 13:13
Rakesh BhattRakesh Bhatt
952114
$endgroup$
add a comment |
$begingroup$
No.
What about $$left( a_n right)=left( frac{1}{n}right)$$
where $a_n=frac{1}{n}$,
$ b_n=1$ and $ c_n=n$
Edit-Edited to answer your additional question.
Again No.
What about $$left( a_n right)=left( frac{n}{n^2}right)$$
where $a_n=frac{n}{n^2}$,
$ b_n=n$ and $ c_n=n^2$
answered Dec 17 '18 at 13:13
Rakesh BhattRakesh Bhatt
952114
$endgroup$
No.
What about $$left( a_n right)=left( frac{1}{n}right)$$
where $a_n=frac{1}{n}$,
$ b_n=1$ and $ c_n=n$
Edit-Edited to answer your additional question.
Again No.
What about $$left( a_n right)=left( frac{n}{n^2}right)$$
where $a_n=frac{n}{n^2}$,
$ b_n=n$ and $ c_n=n^2$
answered Dec 17 '18 at 13:13
Rakesh BhattRakesh Bhatt
952114
edited Dec 17 '18 at 13:58
answered Dec 17 '18 at 13:13
Rakesh BhattRakesh Bhatt
952114
answered Dec 17 '18 at 13:13
Rakesh BhattRakesh Bhatt
952114
answered Dec 17 '18 at 13:13
Rakesh BhattRakesh Bhatt
952114
952114
add a comment |
add a comment |
$begingroup$
No, that’s not true. For instance, for $b_n = n$ and $c_n = n^2$, we have
$$lim_{n to infty}frac{b_n}{c_n} = 0$$
while both sequences tend to $infty$. More generally, when both $b_n$ and $c_n$ are polynomials with $c_n$ having the higher degree, the limit tends to $0$, even though both polynomials may tend to $infty$.
Edit: Also no, the example above once again shows this. Both polynomials clearly diverge to $infty$, but since the denominator has a greater degree, it diverges faster. Hence, the limit tends to $0$.
answered Dec 17 '18 at 13:22
KM101KM101
5,9131523
$endgroup$
add a comment |
$begingroup$
No, that’s not true. For instance, for $b_n = n$ and $c_n = n^2$, we have
$$lim_{n to infty}frac{b_n}{c_n} = 0$$
while both sequences tend to $infty$. More generally, when both $b_n$ and $c_n$ are polynomials with $c_n$ having the higher degree, the limit tends to $0$, even though both polynomials may tend to $infty$.
Edit: Also no, the example above once again shows this. Both polynomials clearly diverge to $infty$, but since the denominator has a greater degree, it diverges faster. Hence, the limit tends to $0$.
answered Dec 17 '18 at 13:22
KM101KM101
5,9131523
$endgroup$
add a comment |
$begingroup$
No, that’s not true. For instance, for $b_n = n$ and $c_n = n^2$, we have
$$lim_{n to infty}frac{b_n}{c_n} = 0$$
while both sequences tend to $infty$. More generally, when both $b_n$ and $c_n$ are polynomials with $c_n$ having the higher degree, the limit tends to $0$, even though both polynomials may tend to $infty$.
Edit: Also no, the example above once again shows this. Both polynomials clearly diverge to $infty$, but since the denominator has a greater degree, it diverges faster. Hence, the limit tends to $0$.
answered Dec 17 '18 at 13:22
KM101KM101
5,9131523
$endgroup$
No, that’s not true. For instance, for $b_n = n$ and $c_n = n^2$, we have
$$lim_{n to infty}frac{b_n}{c_n} = 0$$
while both sequences tend to $infty$. More generally, when both $b_n$ and $c_n$ are polynomials with $c_n$ having the higher degree, the limit tends to $0$, even though both polynomials may tend to $infty$.
Edit: Also no, the example above once again shows this. Both polynomials clearly diverge to $infty$, but since the denominator has a greater degree, it diverges faster. Hence, the limit tends to $0$.
answered Dec 17 '18 at 13:22
KM101KM101
5,9131523
edited Dec 17 '18 at 14:01
answered Dec 17 '18 at 13:22
KM101KM101
5,9131523
answered Dec 17 '18 at 13:22
KM101KM101
5,9131523
answered Dec 17 '18 at 13:22
KM101KM101
5,9131523
5,9131523
add a comment |
add a comment |
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