$limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1 Rightarrow exists c in mathbb{R}: limlimits_{n to infty} {a_n}...
$begingroup$
Let $(a_n)_{nin mathbb{N}}$ be a sequence of real numbers. I was wondering if the follwing implication is true:
$$limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1 Rightarrow exists c in mathbb{R}: limlimits_{n to infty} {a_n} = c$$
Put into words: If $limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1$ then $limlimits_{n to infty} a_{n}$ converges.
My intuition behind $lim_{n to infty} frac{a_n}{a_{n+1}} = 1$ is that at some point $a_n$ and $a_{n+1}$ are alsmost equal. If this is the case $(a_n)_{nin mathbb{N}}$ is a Cauchy sequence and so converges. However I wasn't able to formally prove the statement. So I wondered:
- Is the statement really true?
- (If so, how could you prove it?)
Cheers,
Pascal
limits convergence
$endgroup$
add a comment |
$begingroup$
Let $(a_n)_{nin mathbb{N}}$ be a sequence of real numbers. I was wondering if the follwing implication is true:
$$limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1 Rightarrow exists c in mathbb{R}: limlimits_{n to infty} {a_n} = c$$
Put into words: If $limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1$ then $limlimits_{n to infty} a_{n}$ converges.
My intuition behind $lim_{n to infty} frac{a_n}{a_{n+1}} = 1$ is that at some point $a_n$ and $a_{n+1}$ are alsmost equal. If this is the case $(a_n)_{nin mathbb{N}}$ is a Cauchy sequence and so converges. However I wasn't able to formally prove the statement. So I wondered:
- Is the statement really true?
- (If so, how could you prove it?)
Cheers,
Pascal
limits convergence
$endgroup$
3
$begingroup$
$$ a_n = n.{} $$
$endgroup$
– MisterRiemann
Nov 26 '18 at 22:08
1
$begingroup$
@MisterRiemann Thanks for the counter example. :) I guess I can answer the question myself then.
$endgroup$
– plauer
Nov 26 '18 at 22:11
add a comment |
$begingroup$
Let $(a_n)_{nin mathbb{N}}$ be a sequence of real numbers. I was wondering if the follwing implication is true:
$$limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1 Rightarrow exists c in mathbb{R}: limlimits_{n to infty} {a_n} = c$$
Put into words: If $limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1$ then $limlimits_{n to infty} a_{n}$ converges.
My intuition behind $lim_{n to infty} frac{a_n}{a_{n+1}} = 1$ is that at some point $a_n$ and $a_{n+1}$ are alsmost equal. If this is the case $(a_n)_{nin mathbb{N}}$ is a Cauchy sequence and so converges. However I wasn't able to formally prove the statement. So I wondered:
- Is the statement really true?
- (If so, how could you prove it?)
Cheers,
Pascal
limits convergence
$endgroup$
Let $(a_n)_{nin mathbb{N}}$ be a sequence of real numbers. I was wondering if the follwing implication is true:
$$limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1 Rightarrow exists c in mathbb{R}: limlimits_{n to infty} {a_n} = c$$
Put into words: If $limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1$ then $limlimits_{n to infty} a_{n}$ converges.
My intuition behind $lim_{n to infty} frac{a_n}{a_{n+1}} = 1$ is that at some point $a_n$ and $a_{n+1}$ are alsmost equal. If this is the case $(a_n)_{nin mathbb{N}}$ is a Cauchy sequence and so converges. However I wasn't able to formally prove the statement. So I wondered:
- Is the statement really true?
- (If so, how could you prove it?)
Cheers,
Pascal
limits convergence
limits convergence
asked Nov 26 '18 at 22:07
plauerplauer
305
305
3
$begingroup$
$$ a_n = n.{} $$
$endgroup$
– MisterRiemann
Nov 26 '18 at 22:08
1
$begingroup$
@MisterRiemann Thanks for the counter example. :) I guess I can answer the question myself then.
$endgroup$
– plauer
Nov 26 '18 at 22:11
add a comment |
3
$begingroup$
$$ a_n = n.{} $$
$endgroup$
– MisterRiemann
Nov 26 '18 at 22:08
1
$begingroup$
@MisterRiemann Thanks for the counter example. :) I guess I can answer the question myself then.
$endgroup$
– plauer
Nov 26 '18 at 22:11
3
3
$begingroup$
$$ a_n = n.{} $$
$endgroup$
– MisterRiemann
Nov 26 '18 at 22:08
$begingroup$
$$ a_n = n.{} $$
$endgroup$
– MisterRiemann
Nov 26 '18 at 22:08
1
1
$begingroup$
@MisterRiemann Thanks for the counter example. :) I guess I can answer the question myself then.
$endgroup$
– plauer
Nov 26 '18 at 22:11
$begingroup$
@MisterRiemann Thanks for the counter example. :) I guess I can answer the question myself then.
$endgroup$
– plauer
Nov 26 '18 at 22:11
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
As noticed by MisterRiemann $a_n=n$ is a first counterexample but also $a_n=n^2$ works or $a_n=log n$ and so on.
Therefore unfortunately your guess is definitely not true!
As a remark, other common myths on limits are:
1) $a_n to infty implies a_{n+1}ge a_n$
2) $a_n to L implies a_n to L^+ quad lor quad a_n to L^-$
3) $a_n to 0^+ implies a_{n+1}le a_n$
4) $a_n$ bounded $implies a_n to L$
$endgroup$
add a comment |
$begingroup$
Thanks to MisterRiemann for the counterexample of $a_n = n$.
$$limlimits_{n to infty} frac{a_n}{a_{n+1}} = limlimits_{n to infty} frac{n}{{n+1}} = 1$$
but:
$$limlimits_{n to infty} {a_n} =limlimits_{n to infty} n = infty Rightarrow nexists c in mathbb{R}: limlimits_{n to infty} {a_n} = c $$
$endgroup$
add a comment |
$begingroup$
The conclusion is wrong even for bounded sequences.
An example is $a_ n = 2 + sin(log(n))$ which is not convergent (it oscillates between $1$ and $3$). But
$$
leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert
= leftlvert frac{sin(log(n+1))- sin(log(n))}{2 + sin(log(n))} rightrvert \
le leftlvert sin(log(n+1))- sin(log(n)) rightrvert
= leftlvert frac{cos(log(x_n))}{x_n}rightrvert
$$
for some $x_n in (n, n+1)$, using the mean value theorem. It follows that
$$
leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert le frac{1}{n} to 0 , ,
$$
i.e. $frac{a_{n+1}}{a_n} to 1$.
Roughly speaking, $(a_n)$ oscillates, but with decreasing frequency,
so that the ratio of successive sequence elements approaches one.
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As noticed by MisterRiemann $a_n=n$ is a first counterexample but also $a_n=n^2$ works or $a_n=log n$ and so on.
Therefore unfortunately your guess is definitely not true!
As a remark, other common myths on limits are:
1) $a_n to infty implies a_{n+1}ge a_n$
2) $a_n to L implies a_n to L^+ quad lor quad a_n to L^-$
3) $a_n to 0^+ implies a_{n+1}le a_n$
4) $a_n$ bounded $implies a_n to L$
$endgroup$
add a comment |
$begingroup$
As noticed by MisterRiemann $a_n=n$ is a first counterexample but also $a_n=n^2$ works or $a_n=log n$ and so on.
Therefore unfortunately your guess is definitely not true!
As a remark, other common myths on limits are:
1) $a_n to infty implies a_{n+1}ge a_n$
2) $a_n to L implies a_n to L^+ quad lor quad a_n to L^-$
3) $a_n to 0^+ implies a_{n+1}le a_n$
4) $a_n$ bounded $implies a_n to L$
$endgroup$
add a comment |
$begingroup$
As noticed by MisterRiemann $a_n=n$ is a first counterexample but also $a_n=n^2$ works or $a_n=log n$ and so on.
Therefore unfortunately your guess is definitely not true!
As a remark, other common myths on limits are:
1) $a_n to infty implies a_{n+1}ge a_n$
2) $a_n to L implies a_n to L^+ quad lor quad a_n to L^-$
3) $a_n to 0^+ implies a_{n+1}le a_n$
4) $a_n$ bounded $implies a_n to L$
$endgroup$
As noticed by MisterRiemann $a_n=n$ is a first counterexample but also $a_n=n^2$ works or $a_n=log n$ and so on.
Therefore unfortunately your guess is definitely not true!
As a remark, other common myths on limits are:
1) $a_n to infty implies a_{n+1}ge a_n$
2) $a_n to L implies a_n to L^+ quad lor quad a_n to L^-$
3) $a_n to 0^+ implies a_{n+1}le a_n$
4) $a_n$ bounded $implies a_n to L$
edited Nov 26 '18 at 22:34
answered Nov 26 '18 at 22:12
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
$begingroup$
Thanks to MisterRiemann for the counterexample of $a_n = n$.
$$limlimits_{n to infty} frac{a_n}{a_{n+1}} = limlimits_{n to infty} frac{n}{{n+1}} = 1$$
but:
$$limlimits_{n to infty} {a_n} =limlimits_{n to infty} n = infty Rightarrow nexists c in mathbb{R}: limlimits_{n to infty} {a_n} = c $$
$endgroup$
add a comment |
$begingroup$
Thanks to MisterRiemann for the counterexample of $a_n = n$.
$$limlimits_{n to infty} frac{a_n}{a_{n+1}} = limlimits_{n to infty} frac{n}{{n+1}} = 1$$
but:
$$limlimits_{n to infty} {a_n} =limlimits_{n to infty} n = infty Rightarrow nexists c in mathbb{R}: limlimits_{n to infty} {a_n} = c $$
$endgroup$
add a comment |
$begingroup$
Thanks to MisterRiemann for the counterexample of $a_n = n$.
$$limlimits_{n to infty} frac{a_n}{a_{n+1}} = limlimits_{n to infty} frac{n}{{n+1}} = 1$$
but:
$$limlimits_{n to infty} {a_n} =limlimits_{n to infty} n = infty Rightarrow nexists c in mathbb{R}: limlimits_{n to infty} {a_n} = c $$
$endgroup$
Thanks to MisterRiemann for the counterexample of $a_n = n$.
$$limlimits_{n to infty} frac{a_n}{a_{n+1}} = limlimits_{n to infty} frac{n}{{n+1}} = 1$$
but:
$$limlimits_{n to infty} {a_n} =limlimits_{n to infty} n = infty Rightarrow nexists c in mathbb{R}: limlimits_{n to infty} {a_n} = c $$
answered Nov 26 '18 at 22:19
plauerplauer
305
305
add a comment |
add a comment |
$begingroup$
The conclusion is wrong even for bounded sequences.
An example is $a_ n = 2 + sin(log(n))$ which is not convergent (it oscillates between $1$ and $3$). But
$$
leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert
= leftlvert frac{sin(log(n+1))- sin(log(n))}{2 + sin(log(n))} rightrvert \
le leftlvert sin(log(n+1))- sin(log(n)) rightrvert
= leftlvert frac{cos(log(x_n))}{x_n}rightrvert
$$
for some $x_n in (n, n+1)$, using the mean value theorem. It follows that
$$
leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert le frac{1}{n} to 0 , ,
$$
i.e. $frac{a_{n+1}}{a_n} to 1$.
Roughly speaking, $(a_n)$ oscillates, but with decreasing frequency,
so that the ratio of successive sequence elements approaches one.
$endgroup$
add a comment |
$begingroup$
The conclusion is wrong even for bounded sequences.
An example is $a_ n = 2 + sin(log(n))$ which is not convergent (it oscillates between $1$ and $3$). But
$$
leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert
= leftlvert frac{sin(log(n+1))- sin(log(n))}{2 + sin(log(n))} rightrvert \
le leftlvert sin(log(n+1))- sin(log(n)) rightrvert
= leftlvert frac{cos(log(x_n))}{x_n}rightrvert
$$
for some $x_n in (n, n+1)$, using the mean value theorem. It follows that
$$
leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert le frac{1}{n} to 0 , ,
$$
i.e. $frac{a_{n+1}}{a_n} to 1$.
Roughly speaking, $(a_n)$ oscillates, but with decreasing frequency,
so that the ratio of successive sequence elements approaches one.
$endgroup$
add a comment |
$begingroup$
The conclusion is wrong even for bounded sequences.
An example is $a_ n = 2 + sin(log(n))$ which is not convergent (it oscillates between $1$ and $3$). But
$$
leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert
= leftlvert frac{sin(log(n+1))- sin(log(n))}{2 + sin(log(n))} rightrvert \
le leftlvert sin(log(n+1))- sin(log(n)) rightrvert
= leftlvert frac{cos(log(x_n))}{x_n}rightrvert
$$
for some $x_n in (n, n+1)$, using the mean value theorem. It follows that
$$
leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert le frac{1}{n} to 0 , ,
$$
i.e. $frac{a_{n+1}}{a_n} to 1$.
Roughly speaking, $(a_n)$ oscillates, but with decreasing frequency,
so that the ratio of successive sequence elements approaches one.
$endgroup$
The conclusion is wrong even for bounded sequences.
An example is $a_ n = 2 + sin(log(n))$ which is not convergent (it oscillates between $1$ and $3$). But
$$
leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert
= leftlvert frac{sin(log(n+1))- sin(log(n))}{2 + sin(log(n))} rightrvert \
le leftlvert sin(log(n+1))- sin(log(n)) rightrvert
= leftlvert frac{cos(log(x_n))}{x_n}rightrvert
$$
for some $x_n in (n, n+1)$, using the mean value theorem. It follows that
$$
leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert le frac{1}{n} to 0 , ,
$$
i.e. $frac{a_{n+1}}{a_n} to 1$.
Roughly speaking, $(a_n)$ oscillates, but with decreasing frequency,
so that the ratio of successive sequence elements approaches one.
edited Nov 27 '18 at 13:18
answered Nov 27 '18 at 12:58
Martin RMartin R
27.9k33255
27.9k33255
add a comment |
add a comment |
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3
$begingroup$
$$ a_n = n.{} $$
$endgroup$
– MisterRiemann
Nov 26 '18 at 22:08
1
$begingroup$
@MisterRiemann Thanks for the counter example. :) I guess I can answer the question myself then.
$endgroup$
– plauer
Nov 26 '18 at 22:11