Mixing
How can you proof that $lim_n to infty n!=infty$?
Clash Royale CLAN TAG#URR8PPP
up vote
-2
down vote
favorite
When solving a limit of a succession last class involving $n!$, the professor said we could proof that $lim_n to infty n!$ is $infty$, but he left it as some sort of homework the actual process of the proof.
So far I tried an induction approach, since the first terms of the succession (I'm using the recursive defintion of $a_n=na_n-1$) are greater than the previous, I stated that $$n!>(n-1)!$$ and for $n+1$ then $$(n+1)!=(n+1)n!$$ And since $n+1>0$ $$(n+1)n!>n!$$ The thing is that I'm not sure about how should I finish the proof, is there some sort of $delta-epsilon$ process involved?
Thanks for reading
limits formal-proofs
asked Aug 1 at 11:55
Jesús Isea
225
add a comment |Â
up vote
-2
down vote
favorite
When solving a limit of a succession last class involving $n!$, the professor said we could proof that $lim_n to infty n!$ is $infty$, but he left it as some sort of homework the actual process of the proof.
So far I tried an induction approach, since the first terms of the succession (I'm using the recursive defintion of $a_n=na_n-1$) are greater than the previous, I stated that $$n!>(n-1)!$$ and for $n+1$ then $$(n+1)!=(n+1)n!$$ And since $n+1>0$ $$(n+1)n!>n!$$ The thing is that I'm not sure about how should I finish the proof, is there some sort of $delta-epsilon$ process involved?
Thanks for reading
limits formal-proofs
asked Aug 1 at 11:55
Jesús Isea
225
Suggestion: write down the definition of $lim_{a_n|=infty$ and take it from there.
– uniquesolution
Aug 1 at 11:57
1
We only need to observe that $n!ge n$.
– gimusi
Aug 1 at 11:59
1
Just a small remark: $delta-epsilon$ is usually used in continuity proofs, there is no '$delta$' in the definition of a limit of a sequence
– user190080
Aug 1 at 12:57
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
When solving a limit of a succession last class involving $n!$, the professor said we could proof that $lim_n to infty n!$ is $infty$, but he left it as some sort of homework the actual process of the proof.
So far I tried an induction approach, since the first terms of the succession (I'm using the recursive defintion of $a_n=na_n-1$) are greater than the previous, I stated that $$n!>(n-1)!$$ and for $n+1$ then $$(n+1)!=(n+1)n!$$ And since $n+1>0$ $$(n+1)n!>n!$$ The thing is that I'm not sure about how should I finish the proof, is there some sort of $delta-epsilon$ process involved?
Thanks for reading
limits formal-proofs
asked Aug 1 at 11:55
Jesús Isea
225
When solving a limit of a succession last class involving $n!$, the professor said we could proof that $lim_n to infty n!$ is $infty$, but he left it as some sort of homework the actual process of the proof.
So far I tried an induction approach, since the first terms of the succession (I'm using the recursive defintion of $a_n=na_n-1$) are greater than the previous, I stated that $$n!>(n-1)!$$ and for $n+1$ then $$(n+1)!=(n+1)n!$$ And since $n+1>0$ $$(n+1)n!>n!$$ The thing is that I'm not sure about how should I finish the proof, is there some sort of $delta-epsilon$ process involved?
Thanks for reading
limits formal-proofs
asked Aug 1 at 11:55
Jesús Isea
225
asked Aug 1 at 11:55
Jesús Isea
225
asked Aug 1 at 11:55
Jesús Isea
225
asked Aug 1 at 11:55
Jesús Isea
225
225
Suggestion: write down the definition of $lim_{a_n|=infty$ and take it from there.
– uniquesolution
Aug 1 at 11:57
1
We only need to observe that $n!ge n$.
– gimusi
Aug 1 at 11:59
1
Just a small remark: $delta-epsilon$ is usually used in continuity proofs, there is no '$delta$' in the definition of a limit of a sequence
– user190080
Aug 1 at 12:57
add a comment |Â
Suggestion: write down the definition of $lim_{a_n|=infty$ and take it from there.
– uniquesolution
Aug 1 at 11:57
1
We only need to observe that $n!ge n$.
– gimusi
Aug 1 at 11:59
1
Just a small remark: $delta-epsilon$ is usually used in continuity proofs, there is no '$delta$' in the definition of a limit of a sequence
– user190080
Aug 1 at 12:57
Suggestion: write down the definition of $lim_{a_n|=infty$ and take it from there.
– uniquesolution
Aug 1 at 11:57
Suggestion: write down the definition of $lim_{a_n|=infty$ and take it from there.
– uniquesolution
Aug 1 at 11:57
1
1
We only need to observe that $n!ge n$.
– gimusi
Aug 1 at 11:59
We only need to observe that $n!ge n$.
– gimusi
Aug 1 at 11:59
1
1
Just a small remark: $delta-epsilon$ is usually used in continuity proofs, there is no '$delta$' in the definition of a limit of a sequence
– user190080
Aug 1 at 12:57
Just a small remark: $delta-epsilon$ is usually used in continuity proofs, there is no '$delta$' in the definition of a limit of a sequence
– user190080
Aug 1 at 12:57
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
Note that $n$ is a non-negative integer (because you're talking about $n!$, which is only defined for non-negative integers), so you're not using $delta$-$epsilon$, but $N$-$epsilon$ (we're working with sequences here, as opposed to functions $mathbbR to mathbbR$). Furthermore, because the limit is not finite, you shouldn't be using $epsilon$.
To clarify, we want to show that $$lim_n to inftyn! = +inftytext.$$
To show this, the definition is as follows:
$lim_n to inftyn! = +infty$ means:
For every $alpha in mathbbR$ there is a positive integer $N$ such that for all $n > N$, $n! > alpha$.
See also https://en.wikipedia.org/wiki/Limit_of_a_sequence#Infinite_limits.
To show this, we may approach as follows.
Let $alpha in mathbbR$ be arbitrary.
Suppose $alpha < 1$ and let $N = 1$. Then for each $n geq 1$,
$$n! = n(n-1)cdots1geq underbrace1 cdot 1 cdots 1_n text times = 1^n=1> alphatext.$$
Now suppose $alpha geq 1$.
If $alpha geq 1$, take $N = lceilalpharceil+1$, so that for each $n geq N$,
$$n! = n(n-1)cdots1geq ngeq N=lceilalpharceil+1geqalpha+1>alpha$$
as desired.
answered Aug 1 at 12:43
Clarinetist
10.3k32767
add a comment |Â
up vote
3
down vote
It suffices simply to note that
$$n!=n(n-1)ldots 2cdot 1ge n to infty$$
answered Aug 1 at 11:56
gimusi
64k73480
add a comment |Â
up vote
2
down vote
The behaviour of infinite products of positive reals can be reduced to an infinite-series problem. You want to show $ln n!=sum_k=1^nln ktoinfty$ as $ntoinfty$; this follows from $ln kge ln 2>0$ for $kge 2$. Although this technique is worth knowing, in this case it's overkill, but a kind of overkill that makes obvious a slicker approach: prove by induction that $n!ge 2^n-1$ for $nge 1$. (In particular, to go from $n=k$ to $n=k+1$ multiply by $k+1ge 2$.)
answered Aug 1 at 12:00
J.G.
12.8k11423
add a comment |Â
up vote
0
down vote
Hint:
$$n! geq n$$ for all $ninmathbb N$.
answered Aug 1 at 12:01
5xum
81.8k382146
Already given above-mentioned
– Devendra Singh Rana
Aug 1 at 13:03
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note that $n$ is a non-negative integer (because you're talking about $n!$, which is only defined for non-negative integers), so you're not using $delta$-$epsilon$, but $N$-$epsilon$ (we're working with sequences here, as opposed to functions $mathbbR to mathbbR$). Furthermore, because the limit is not finite, you shouldn't be using $epsilon$.
To clarify, we want to show that $$lim_n to inftyn! = +inftytext.$$
To show this, the definition is as follows:
$lim_n to inftyn! = +infty$ means:
For every $alpha in mathbbR$ there is a positive integer $N$ such that for all $n > N$, $n! > alpha$.
See also https://en.wikipedia.org/wiki/Limit_of_a_sequence#Infinite_limits.
To show this, we may approach as follows.
Let $alpha in mathbbR$ be arbitrary.
Suppose $alpha < 1$ and let $N = 1$. Then for each $n geq 1$,
$$n! = n(n-1)cdots1geq underbrace1 cdot 1 cdots 1_n text times = 1^n=1> alphatext.$$
Now suppose $alpha geq 1$.
If $alpha geq 1$, take $N = lceilalpharceil+1$, so that for each $n geq N$,
$$n! = n(n-1)cdots1geq ngeq N=lceilalpharceil+1geqalpha+1>alpha$$
as desired.
answered Aug 1 at 12:43
Clarinetist
10.3k32767
add a comment |Â
up vote
1
down vote
accepted
Note that $n$ is a non-negative integer (because you're talking about $n!$, which is only defined for non-negative integers), so you're not using $delta$-$epsilon$, but $N$-$epsilon$ (we're working with sequences here, as opposed to functions $mathbbR to mathbbR$). Furthermore, because the limit is not finite, you shouldn't be using $epsilon$.
To clarify, we want to show that $$lim_n to inftyn! = +inftytext.$$
To show this, the definition is as follows:
$lim_n to inftyn! = +infty$ means:
For every $alpha in mathbbR$ there is a positive integer $N$ such that for all $n > N$, $n! > alpha$.
See also https://en.wikipedia.org/wiki/Limit_of_a_sequence#Infinite_limits.
To show this, we may approach as follows.
Let $alpha in mathbbR$ be arbitrary.
Suppose $alpha < 1$ and let $N = 1$. Then for each $n geq 1$,
$$n! = n(n-1)cdots1geq underbrace1 cdot 1 cdots 1_n text times = 1^n=1> alphatext.$$
Now suppose $alpha geq 1$.
If $alpha geq 1$, take $N = lceilalpharceil+1$, so that for each $n geq N$,
$$n! = n(n-1)cdots1geq ngeq N=lceilalpharceil+1geqalpha+1>alpha$$
as desired.
answered Aug 1 at 12:43
Clarinetist
10.3k32767
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note that $n$ is a non-negative integer (because you're talking about $n!$, which is only defined for non-negative integers), so you're not using $delta$-$epsilon$, but $N$-$epsilon$ (we're working with sequences here, as opposed to functions $mathbbR to mathbbR$). Furthermore, because the limit is not finite, you shouldn't be using $epsilon$.
To clarify, we want to show that $$lim_n to inftyn! = +inftytext.$$
To show this, the definition is as follows:
$lim_n to inftyn! = +infty$ means:
For every $alpha in mathbbR$ there is a positive integer $N$ such that for all $n > N$, $n! > alpha$.
See also https://en.wikipedia.org/wiki/Limit_of_a_sequence#Infinite_limits.
To show this, we may approach as follows.
Let $alpha in mathbbR$ be arbitrary.
Suppose $alpha < 1$ and let $N = 1$. Then for each $n geq 1$,
$$n! = n(n-1)cdots1geq underbrace1 cdot 1 cdots 1_n text times = 1^n=1> alphatext.$$
Now suppose $alpha geq 1$.
If $alpha geq 1$, take $N = lceilalpharceil+1$, so that for each $n geq N$,
$$n! = n(n-1)cdots1geq ngeq N=lceilalpharceil+1geqalpha+1>alpha$$
as desired.
answered Aug 1 at 12:43
Clarinetist
10.3k32767
Note that $n$ is a non-negative integer (because you're talking about $n!$, which is only defined for non-negative integers), so you're not using $delta$-$epsilon$, but $N$-$epsilon$ (we're working with sequences here, as opposed to functions $mathbbR to mathbbR$). Furthermore, because the limit is not finite, you shouldn't be using $epsilon$.
To clarify, we want to show that $$lim_n to inftyn! = +inftytext.$$
To show this, the definition is as follows:
$lim_n to inftyn! = +infty$ means:
For every $alpha in mathbbR$ there is a positive integer $N$ such that for all $n > N$, $n! > alpha$.
See also https://en.wikipedia.org/wiki/Limit_of_a_sequence#Infinite_limits.
To show this, we may approach as follows.
Let $alpha in mathbbR$ be arbitrary.
Suppose $alpha < 1$ and let $N = 1$. Then for each $n geq 1$,
$$n! = n(n-1)cdots1geq underbrace1 cdot 1 cdots 1_n text times = 1^n=1> alphatext.$$
Now suppose $alpha geq 1$.
If $alpha geq 1$, take $N = lceilalpharceil+1$, so that for each $n geq N$,
$$n! = n(n-1)cdots1geq ngeq N=lceilalpharceil+1geqalpha+1>alpha$$
as desired.
answered Aug 1 at 12:43
Clarinetist
10.3k32767
edited Aug 1 at 12:49
answered Aug 1 at 12:43
Clarinetist
10.3k32767
answered Aug 1 at 12:43
Clarinetist
10.3k32767
answered Aug 1 at 12:43
Clarinetist
10.3k32767
10.3k32767
add a comment |Â
add a comment |Â
up vote
3
down vote
It suffices simply to note that
$$n!=n(n-1)ldots 2cdot 1ge n to infty$$
answered Aug 1 at 11:56
gimusi
64k73480
add a comment |Â
up vote
3
down vote
It suffices simply to note that
$$n!=n(n-1)ldots 2cdot 1ge n to infty$$
answered Aug 1 at 11:56
gimusi
64k73480
add a comment |Â
up vote
3
down vote
up vote
3
down vote
It suffices simply to note that
$$n!=n(n-1)ldots 2cdot 1ge n to infty$$
answered Aug 1 at 11:56
gimusi
64k73480
It suffices simply to note that
$$n!=n(n-1)ldots 2cdot 1ge n to infty$$
answered Aug 1 at 11:56
gimusi
64k73480
answered Aug 1 at 11:56
gimusi
64k73480
answered Aug 1 at 11:56
gimusi
64k73480
answered Aug 1 at 11:56
gimusi
64k73480
64k73480
add a comment |Â
add a comment |Â
up vote
2
down vote
The behaviour of infinite products of positive reals can be reduced to an infinite-series problem. You want to show $ln n!=sum_k=1^nln ktoinfty$ as $ntoinfty$; this follows from $ln kge ln 2>0$ for $kge 2$. Although this technique is worth knowing, in this case it's overkill, but a kind of overkill that makes obvious a slicker approach: prove by induction that $n!ge 2^n-1$ for $nge 1$. (In particular, to go from $n=k$ to $n=k+1$ multiply by $k+1ge 2$.)
answered Aug 1 at 12:00
J.G.
12.8k11423
add a comment |Â
up vote
2
down vote
The behaviour of infinite products of positive reals can be reduced to an infinite-series problem. You want to show $ln n!=sum_k=1^nln ktoinfty$ as $ntoinfty$; this follows from $ln kge ln 2>0$ for $kge 2$. Although this technique is worth knowing, in this case it's overkill, but a kind of overkill that makes obvious a slicker approach: prove by induction that $n!ge 2^n-1$ for $nge 1$. (In particular, to go from $n=k$ to $n=k+1$ multiply by $k+1ge 2$.)
answered Aug 1 at 12:00
J.G.
12.8k11423
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The behaviour of infinite products of positive reals can be reduced to an infinite-series problem. You want to show $ln n!=sum_k=1^nln ktoinfty$ as $ntoinfty$; this follows from $ln kge ln 2>0$ for $kge 2$. Although this technique is worth knowing, in this case it's overkill, but a kind of overkill that makes obvious a slicker approach: prove by induction that $n!ge 2^n-1$ for $nge 1$. (In particular, to go from $n=k$ to $n=k+1$ multiply by $k+1ge 2$.)
answered Aug 1 at 12:00
J.G.
12.8k11423
The behaviour of infinite products of positive reals can be reduced to an infinite-series problem. You want to show $ln n!=sum_k=1^nln ktoinfty$ as $ntoinfty$; this follows from $ln kge ln 2>0$ for $kge 2$. Although this technique is worth knowing, in this case it's overkill, but a kind of overkill that makes obvious a slicker approach: prove by induction that $n!ge 2^n-1$ for $nge 1$. (In particular, to go from $n=k$ to $n=k+1$ multiply by $k+1ge 2$.)
answered Aug 1 at 12:00
J.G.
12.8k11423
answered Aug 1 at 12:00
J.G.
12.8k11423
answered Aug 1 at 12:00
J.G.
12.8k11423
answered Aug 1 at 12:00
J.G.
12.8k11423
12.8k11423
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint:
$$n! geq n$$ for all $ninmathbb N$.
answered Aug 1 at 12:01
5xum
81.8k382146
Already given above-mentioned
– Devendra Singh Rana
Aug 1 at 13:03
add a comment |Â
up vote
0
down vote
Hint:
$$n! geq n$$ for all $ninmathbb N$.
answered Aug 1 at 12:01
5xum
81.8k382146
Already given above-mentioned
– Devendra Singh Rana
Aug 1 at 13:03
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
$$n! geq n$$ for all $ninmathbb N$.
answered Aug 1 at 12:01
5xum
81.8k382146
Hint:
$$n! geq n$$ for all $ninmathbb N$.
answered Aug 1 at 12:01
5xum
81.8k382146
edited Aug 1 at 12:12
answered Aug 1 at 12:01
5xum
81.8k382146
answered Aug 1 at 12:01
5xum
81.8k382146
answered Aug 1 at 12:01
5xum
81.8k382146
81.8k382146
Already given above-mentioned
– Devendra Singh Rana
Aug 1 at 13:03
add a comment |Â
Already given above-mentioned
– Devendra Singh Rana
Aug 1 at 13:03
Already given above-mentioned
– Devendra Singh Rana
Aug 1 at 13:03
Already given above-mentioned
– Devendra Singh Rana
Aug 1 at 13:03
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2869000%2fhow-can-you-proof-that-lim-n-to-infty-n-infty%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Suggestion: write down the definition of $lim_{a_n|=infty$ and take it from there.
– uniquesolution
Aug 1 at 11:57
1
We only need to observe that $n!ge n$.
– gimusi
Aug 1 at 11:59
1
Just a small remark: $delta-epsilon$ is usually used in continuity proofs, there is no '$delta$' in the definition of a limit of a sequence
– user190080
Aug 1 at 12:57