Mixing
Finding limit of $m(1-frac1m+1)^m^2-m-1$ as $mrightarrow infty$
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Intuitively I feel like the limit is going to 0. But while proving if I take log then it is coming infinity. Please solve the problem.
calculus limits
asked Aug 6 at 11:22
Ani
61
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up vote
1
down vote
favorite
1
Intuitively I feel like the limit is going to 0. But while proving if I take log then it is coming infinity. Please solve the problem.
calculus limits
asked Aug 6 at 11:22
Ani
61
Why would you think the limit should be $0$?
– Suzet
Aug 6 at 11:25
@Suzet Wolfram alpha ahah!
– Davide Morgante
Aug 6 at 11:27
Because $m^2$ grows much faster than $m$ and hence $(1-frac1m+1)^m^2-m-1$ will go to $0$ much faster than $m$ goes to $infty$. Hence the whole limit should go to $0$.
– Ani
Aug 6 at 11:28
What about writing the expression as the exp of a log, factorizing and using a limited development of log in 0 ?
– Pjonin
Aug 6 at 11:28
Oh right, of course, I feel a little rusty now
– Suzet
Aug 6 at 11:29
 |Â
show 1 more comment
up vote
1
down vote
favorite
1
up vote
1
down vote
favorite
1
1
Intuitively I feel like the limit is going to 0. But while proving if I take log then it is coming infinity. Please solve the problem.
calculus limits
asked Aug 6 at 11:22
Ani
61
Intuitively I feel like the limit is going to 0. But while proving if I take log then it is coming infinity. Please solve the problem.
calculus limits
asked Aug 6 at 11:22
Ani
61
asked Aug 6 at 11:22
Ani
61
asked Aug 6 at 11:22
Ani
61
asked Aug 6 at 11:22
Ani
61
61
Why would you think the limit should be $0$?
– Suzet
Aug 6 at 11:25
@Suzet Wolfram alpha ahah!
– Davide Morgante
Aug 6 at 11:27
Because $m^2$ grows much faster than $m$ and hence $(1-frac1m+1)^m^2-m-1$ will go to $0$ much faster than $m$ goes to $infty$. Hence the whole limit should go to $0$.
– Ani
Aug 6 at 11:28
What about writing the expression as the exp of a log, factorizing and using a limited development of log in 0 ?
– Pjonin
Aug 6 at 11:28
Oh right, of course, I feel a little rusty now
– Suzet
Aug 6 at 11:29
 |Â
show 1 more comment
Why would you think the limit should be $0$?
– Suzet
Aug 6 at 11:25
@Suzet Wolfram alpha ahah!
– Davide Morgante
Aug 6 at 11:27
Because $m^2$ grows much faster than $m$ and hence $(1-frac1m+1)^m^2-m-1$ will go to $0$ much faster than $m$ goes to $infty$. Hence the whole limit should go to $0$.
– Ani
Aug 6 at 11:28
What about writing the expression as the exp of a log, factorizing and using a limited development of log in 0 ?
– Pjonin
Aug 6 at 11:28
Oh right, of course, I feel a little rusty now
– Suzet
Aug 6 at 11:29
Why would you think the limit should be $0$?
– Suzet
Aug 6 at 11:25
Why would you think the limit should be $0$?
– Suzet
Aug 6 at 11:25
@Suzet Wolfram alpha ahah!
– Davide Morgante
Aug 6 at 11:27
@Suzet Wolfram alpha ahah!
– Davide Morgante
Aug 6 at 11:27
Because $m^2$ grows much faster than $m$ and hence $(1-frac1m+1)^m^2-m-1$ will go to $0$ much faster than $m$ goes to $infty$. Hence the whole limit should go to $0$.
– Ani
Aug 6 at 11:28
Because $m^2$ grows much faster than $m$ and hence $(1-frac1m+1)^m^2-m-1$ will go to $0$ much faster than $m$ goes to $infty$. Hence the whole limit should go to $0$.
– Ani
Aug 6 at 11:28
What about writing the expression as the exp of a log, factorizing and using a limited development of log in 0 ?
– Pjonin
Aug 6 at 11:28
What about writing the expression as the exp of a log, factorizing and using a limited development of log in 0 ?
– Pjonin
Aug 6 at 11:28
Oh right, of course, I feel a little rusty now
– Suzet
Aug 6 at 11:29
Oh right, of course, I feel a little rusty now
– Suzet
Aug 6 at 11:29
 |Â
show 1 more comment
1 Answer
1
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votes
up vote
2
down vote
By root test we have
$$sqrt[m]a_m=sqrt[m]mleft(1-frac1m+1right)^fracm^2-m-1mto 1cdot frac1e=frac1e<1$$
indeed
$$sqrt[m]m=e^fraclog mmto e^0=1$$
and
$$left(1-frac1m+1right)^fracm^2-m-1m=left[left(1-frac1m+1right)^m+1right]^fracm^2-m-1m(m+1)to frac1e$$
therefore
$$a_m=mleft(1-frac1m+1right)^m^2-m-1to 0$$
answered Aug 6 at 11:57
gimusi
65.4k73684
Thanks a lot. Never crossed my mind. Was thinking in the direction of finding the limit rigorously.
– Ani
Aug 6 at 12:01
@Ani Root test in this case is a very effective way.
– gimusi
Aug 6 at 12:02
But isn't root test is for the test of convergence of a series?
– Ani
Aug 7 at 3:59
@Ani Even if not always clearly pointed out it is of course valid also for sequences. Moreover note that if €$(a_n)^1/nto L <1$ the series $sum a_n$ converges but then $a_n to 0$.
– gimusi
Aug 7 at 5:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
By root test we have
$$sqrt[m]a_m=sqrt[m]mleft(1-frac1m+1right)^fracm^2-m-1mto 1cdot frac1e=frac1e<1$$
indeed
$$sqrt[m]m=e^fraclog mmto e^0=1$$
and
$$left(1-frac1m+1right)^fracm^2-m-1m=left[left(1-frac1m+1right)^m+1right]^fracm^2-m-1m(m+1)to frac1e$$
therefore
$$a_m=mleft(1-frac1m+1right)^m^2-m-1to 0$$
answered Aug 6 at 11:57
gimusi
65.4k73684
Thanks a lot. Never crossed my mind. Was thinking in the direction of finding the limit rigorously.
– Ani
Aug 6 at 12:01
@Ani Root test in this case is a very effective way.
– gimusi
Aug 6 at 12:02
But isn't root test is for the test of convergence of a series?
– Ani
Aug 7 at 3:59
@Ani Even if not always clearly pointed out it is of course valid also for sequences. Moreover note that if €$(a_n)^1/nto L <1$ the series $sum a_n$ converges but then $a_n to 0$.
– gimusi
Aug 7 at 5:03
add a comment |Â
up vote
2
down vote
By root test we have
$$sqrt[m]a_m=sqrt[m]mleft(1-frac1m+1right)^fracm^2-m-1mto 1cdot frac1e=frac1e<1$$
indeed
$$sqrt[m]m=e^fraclog mmto e^0=1$$
and
$$left(1-frac1m+1right)^fracm^2-m-1m=left[left(1-frac1m+1right)^m+1right]^fracm^2-m-1m(m+1)to frac1e$$
therefore
$$a_m=mleft(1-frac1m+1right)^m^2-m-1to 0$$
answered Aug 6 at 11:57
gimusi
65.4k73684
Thanks a lot. Never crossed my mind. Was thinking in the direction of finding the limit rigorously.
– Ani
Aug 6 at 12:01
@Ani Root test in this case is a very effective way.
– gimusi
Aug 6 at 12:02
But isn't root test is for the test of convergence of a series?
– Ani
Aug 7 at 3:59
@Ani Even if not always clearly pointed out it is of course valid also for sequences. Moreover note that if €$(a_n)^1/nto L <1$ the series $sum a_n$ converges but then $a_n to 0$.
– gimusi
Aug 7 at 5:03
add a comment |Â
up vote
2
down vote
up vote
2
down vote
By root test we have
$$sqrt[m]a_m=sqrt[m]mleft(1-frac1m+1right)^fracm^2-m-1mto 1cdot frac1e=frac1e<1$$
indeed
$$sqrt[m]m=e^fraclog mmto e^0=1$$
and
$$left(1-frac1m+1right)^fracm^2-m-1m=left[left(1-frac1m+1right)^m+1right]^fracm^2-m-1m(m+1)to frac1e$$
therefore
$$a_m=mleft(1-frac1m+1right)^m^2-m-1to 0$$
answered Aug 6 at 11:57
gimusi
65.4k73684
By root test we have
$$sqrt[m]a_m=sqrt[m]mleft(1-frac1m+1right)^fracm^2-m-1mto 1cdot frac1e=frac1e<1$$
indeed
$$sqrt[m]m=e^fraclog mmto e^0=1$$
and
$$left(1-frac1m+1right)^fracm^2-m-1m=left[left(1-frac1m+1right)^m+1right]^fracm^2-m-1m(m+1)to frac1e$$
therefore
$$a_m=mleft(1-frac1m+1right)^m^2-m-1to 0$$
answered Aug 6 at 11:57
gimusi
65.4k73684
answered Aug 6 at 11:57
gimusi
65.4k73684
answered Aug 6 at 11:57
gimusi
65.4k73684
answered Aug 6 at 11:57
gimusi
65.4k73684
65.4k73684
Thanks a lot. Never crossed my mind. Was thinking in the direction of finding the limit rigorously.
– Ani
Aug 6 at 12:01
@Ani Root test in this case is a very effective way.
– gimusi
Aug 6 at 12:02
But isn't root test is for the test of convergence of a series?
– Ani
Aug 7 at 3:59
@Ani Even if not always clearly pointed out it is of course valid also for sequences. Moreover note that if €$(a_n)^1/nto L <1$ the series $sum a_n$ converges but then $a_n to 0$.
– gimusi
Aug 7 at 5:03
add a comment |Â
Thanks a lot. Never crossed my mind. Was thinking in the direction of finding the limit rigorously.
– Ani
Aug 6 at 12:01
@Ani Root test in this case is a very effective way.
– gimusi
Aug 6 at 12:02
But isn't root test is for the test of convergence of a series?
– Ani
Aug 7 at 3:59
@Ani Even if not always clearly pointed out it is of course valid also for sequences. Moreover note that if €$(a_n)^1/nto L <1$ the series $sum a_n$ converges but then $a_n to 0$.
– gimusi
Aug 7 at 5:03
Thanks a lot. Never crossed my mind. Was thinking in the direction of finding the limit rigorously.
– Ani
Aug 6 at 12:01
Thanks a lot. Never crossed my mind. Was thinking in the direction of finding the limit rigorously.
– Ani
Aug 6 at 12:01
@Ani Root test in this case is a very effective way.
– gimusi
Aug 6 at 12:02
@Ani Root test in this case is a very effective way.
– gimusi
Aug 6 at 12:02
But isn't root test is for the test of convergence of a series?
– Ani
Aug 7 at 3:59
But isn't root test is for the test of convergence of a series?
– Ani
Aug 7 at 3:59
@Ani Even if not always clearly pointed out it is of course valid also for sequences. Moreover note that if €$(a_n)^1/nto L <1$ the series $sum a_n$ converges but then $a_n to 0$.
– gimusi
Aug 7 at 5:03
@Ani Even if not always clearly pointed out it is of course valid also for sequences. Moreover note that if €$(a_n)^1/nto L <1$ the series $sum a_n$ converges but then $a_n to 0$.
– gimusi
Aug 7 at 5:03
add a comment |Â
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Why would you think the limit should be $0$?
– Suzet
Aug 6 at 11:25
@Suzet Wolfram alpha ahah!
– Davide Morgante
Aug 6 at 11:27
Because $m^2$ grows much faster than $m$ and hence $(1-frac1m+1)^m^2-m-1$ will go to $0$ much faster than $m$ goes to $infty$. Hence the whole limit should go to $0$.
– Ani
Aug 6 at 11:28
What about writing the expression as the exp of a log, factorizing and using a limited development of log in 0 ?
– Pjonin
Aug 6 at 11:28
Oh right, of course, I feel a little rusty now
– Suzet
Aug 6 at 11:29