Need help with this sequence
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How to prove $u_n=b^n/n!$ is a null sequence for $b>0$?
sequences-and-series
edited Nov 27 at 15:26
A. Pongrácz
5,218725
asked Nov 27 at 15:24
Sourish Bhattacharya
6
add a comment |
up vote
-3
down vote
favorite
How to prove $u_n=b^n/n!$ is a null sequence for $b>0$?
sequences-and-series
edited Nov 27 at 15:26
A. Pongrácz
5,218725
asked Nov 27 at 15:24
Sourish Bhattacharya
6
You are required to use MathJax
– Akash Roy
Nov 27 at 15:25
By the way, the downvotes are coming becaue you didn't show any of your own efforts. Usually I also downvote or vote to close for this reason, but I think it is acceptable in this particular case not to show substantial work: the solution hinges on one very technical idea, without which it is hard to see how to start. But I also understand why people downvote or close such questions.
– A. Pongrácz
Nov 27 at 15:36
At least if he had described the question well, the downvotes would not have come. The problem with the question lies in its presentation. However now it is okay.
– Akash Roy
Nov 27 at 15:40
Akash Roy: sure, but maybe we should also be more patient about that. The author of the post is a new contributor, and maybe (s)he is unexperienced in discussing mathematics, or could have difficulties speaking English.
– A. Pongrácz
Nov 27 at 15:44
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
How to prove $u_n=b^n/n!$ is a null sequence for $b>0$?
sequences-and-series
edited Nov 27 at 15:26
A. Pongrácz
5,218725
asked Nov 27 at 15:24
Sourish Bhattacharya
6
How to prove $u_n=b^n/n!$ is a null sequence for $b>0$?
sequences-and-series
sequences-and-series
edited Nov 27 at 15:26
A. Pongrácz
5,218725
asked Nov 27 at 15:24
Sourish Bhattacharya
6
edited Nov 27 at 15:26
A. Pongrácz
5,218725
asked Nov 27 at 15:24
Sourish Bhattacharya
6
edited Nov 27 at 15:26
A. Pongrácz
5,218725
edited Nov 27 at 15:26
A. Pongrácz
5,218725
edited Nov 27 at 15:26
A. Pongrácz
5,218725
5,218725
asked Nov 27 at 15:24
Sourish Bhattacharya
6
asked Nov 27 at 15:24
Sourish Bhattacharya
6
asked Nov 27 at 15:24
Sourish Bhattacharya
6
6
You are required to use MathJax
– Akash Roy
Nov 27 at 15:25
By the way, the downvotes are coming becaue you didn't show any of your own efforts. Usually I also downvote or vote to close for this reason, but I think it is acceptable in this particular case not to show substantial work: the solution hinges on one very technical idea, without which it is hard to see how to start. But I also understand why people downvote or close such questions.
– A. Pongrácz
Nov 27 at 15:36
At least if he had described the question well, the downvotes would not have come. The problem with the question lies in its presentation. However now it is okay.
– Akash Roy
Nov 27 at 15:40
Akash Roy: sure, but maybe we should also be more patient about that. The author of the post is a new contributor, and maybe (s)he is unexperienced in discussing mathematics, or could have difficulties speaking English.
– A. Pongrácz
Nov 27 at 15:44
add a comment |
You are required to use MathJax
– Akash Roy
Nov 27 at 15:25
By the way, the downvotes are coming becaue you didn't show any of your own efforts. Usually I also downvote or vote to close for this reason, but I think it is acceptable in this particular case not to show substantial work: the solution hinges on one very technical idea, without which it is hard to see how to start. But I also understand why people downvote or close such questions.
– A. Pongrácz
Nov 27 at 15:36
At least if he had described the question well, the downvotes would not have come. The problem with the question lies in its presentation. However now it is okay.
– Akash Roy
Nov 27 at 15:40
Akash Roy: sure, but maybe we should also be more patient about that. The author of the post is a new contributor, and maybe (s)he is unexperienced in discussing mathematics, or could have difficulties speaking English.
– A. Pongrácz
Nov 27 at 15:44
You are required to use MathJax
– Akash Roy
Nov 27 at 15:25
You are required to use MathJax
– Akash Roy
Nov 27 at 15:25
By the way, the downvotes are coming becaue you didn't show any of your own efforts. Usually I also downvote or vote to close for this reason, but I think it is acceptable in this particular case not to show substantial work: the solution hinges on one very technical idea, without which it is hard to see how to start. But I also understand why people downvote or close such questions.
– A. Pongrácz
Nov 27 at 15:36
By the way, the downvotes are coming becaue you didn't show any of your own efforts. Usually I also downvote or vote to close for this reason, but I think it is acceptable in this particular case not to show substantial work: the solution hinges on one very technical idea, without which it is hard to see how to start. But I also understand why people downvote or close such questions.
– A. Pongrácz
Nov 27 at 15:36
At least if he had described the question well, the downvotes would not have come. The problem with the question lies in its presentation. However now it is okay.
– Akash Roy
Nov 27 at 15:40
At least if he had described the question well, the downvotes would not have come. The problem with the question lies in its presentation. However now it is okay.
– Akash Roy
Nov 27 at 15:40
Akash Roy: sure, but maybe we should also be more patient about that. The author of the post is a new contributor, and maybe (s)he is unexperienced in discussing mathematics, or could have difficulties speaking English.
– A. Pongrácz
Nov 27 at 15:44
Akash Roy: sure, but maybe we should also be more patient about that. The author of the post is a new contributor, and maybe (s)he is unexperienced in discussing mathematics, or could have difficulties speaking English.
– A. Pongrácz
Nov 27 at 15:44
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
Hint: Note that
$$1+b+frac{b^2}{2!}+frac{b^3}{3!}+frac{b^4}{4!}+… = e^b$$
The series converges, so what does $frac{b^n}{n!}$ tend to as $n to infty$?
answered Nov 27 at 15:39
KM101
3,945417
add a comment |
up vote
1
down vote
Easiest way to do it is by noticing that
$$ sum frac{b^n}{n!} $$
converges as it can be done by the ratio test or just by the fact that it is $e^b$ and hence
$$ lim frac{b^n}{n!} =0 $$
answered Nov 27 at 15:32
Neymar
385113
add a comment |
up vote
1
down vote
Hint: let $k>b$ be any positive integer at least 10.
Then by trivial estimations, you can easily obtain for $ngeq k^2+k$ that $k^n/n!leq frac{k}{1}cdot frac{k}{2}cdot dots frac{k}{k}cdot dots frac{k}{k^2}cdot frac{k}{k^2+1}cdot dots frac{k}{k^2+k-1}cdot frac{k}{k^2+k} leq frac{k}{k^2+k}= frac{1}{k+1}$, which is a null sequence.
answered Nov 27 at 15:33
A. Pongrácz
5,218725
why do you do $k>b$?
– Neymar
Nov 27 at 15:38
I want the first $k$ fractions in the product be different from the last $k+1$ ($k$ at least 10 surely guarantees this). It is also easier to refer to these fractions if the base is an integer.
– A. Pongrácz
Nov 27 at 15:40
I really dont understand this method at all. Its very hard. also why you make $n=k^2+k$?
– Neymar
Nov 27 at 15:57
It is much simpler that the other suggestions in the sense that it is completely elementary. All the fractions $k/k^2, ldots, k/(k^2+k-1)$ are at most $1/k$, so they cancel out the first $k$ fractions. In the middle, every fraction is at most $1$. Hence, the product is at most $k/(k^2+k)$. You need $k^2$ to make the estimations completly trivial.
– A. Pongrácz
Nov 27 at 16:44
but my question is how you come up with $n=k^2+k$. Why are you allowed to do this?
– Neymar
Nov 27 at 18:06
|
show 1 more 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
up vote
2
down vote
Hint: Note that
$$1+b+frac{b^2}{2!}+frac{b^3}{3!}+frac{b^4}{4!}+… = e^b$$
The series converges, so what does $frac{b^n}{n!}$ tend to as $n to infty$?
answered Nov 27 at 15:39
KM101
3,945417
add a comment |
up vote
2
down vote
Hint: Note that
$$1+b+frac{b^2}{2!}+frac{b^3}{3!}+frac{b^4}{4!}+… = e^b$$
The series converges, so what does $frac{b^n}{n!}$ tend to as $n to infty$?
answered Nov 27 at 15:39
KM101
3,945417
add a comment |
up vote
2
down vote
up vote
2
down vote
Hint: Note that
$$1+b+frac{b^2}{2!}+frac{b^3}{3!}+frac{b^4}{4!}+… = e^b$$
The series converges, so what does $frac{b^n}{n!}$ tend to as $n to infty$?
answered Nov 27 at 15:39
KM101
3,945417
Hint: Note that
$$1+b+frac{b^2}{2!}+frac{b^3}{3!}+frac{b^4}{4!}+… = e^b$$
The series converges, so what does $frac{b^n}{n!}$ tend to as $n to infty$?
answered Nov 27 at 15:39
KM101
3,945417
answered Nov 27 at 15:39
KM101
3,945417
answered Nov 27 at 15:39
KM101
3,945417
answered Nov 27 at 15:39
KM101
3,945417
3,945417
add a comment |
add a comment |
up vote
1
down vote
Easiest way to do it is by noticing that
$$ sum frac{b^n}{n!} $$
converges as it can be done by the ratio test or just by the fact that it is $e^b$ and hence
$$ lim frac{b^n}{n!} =0 $$
answered Nov 27 at 15:32
Neymar
385113
add a comment |
up vote
1
down vote
Easiest way to do it is by noticing that
$$ sum frac{b^n}{n!} $$
converges as it can be done by the ratio test or just by the fact that it is $e^b$ and hence
$$ lim frac{b^n}{n!} =0 $$
answered Nov 27 at 15:32
Neymar
385113
add a comment |
up vote
1
down vote
up vote
1
down vote
Easiest way to do it is by noticing that
$$ sum frac{b^n}{n!} $$
converges as it can be done by the ratio test or just by the fact that it is $e^b$ and hence
$$ lim frac{b^n}{n!} =0 $$
answered Nov 27 at 15:32
Neymar
385113
Easiest way to do it is by noticing that
$$ sum frac{b^n}{n!} $$
converges as it can be done by the ratio test or just by the fact that it is $e^b$ and hence
$$ lim frac{b^n}{n!} =0 $$
answered Nov 27 at 15:32
Neymar
385113
answered Nov 27 at 15:32
Neymar
385113
answered Nov 27 at 15:32
Neymar
385113
answered Nov 27 at 15:32
Neymar
385113
385113
add a comment |
add a comment |
up vote
1
down vote
Hint: let $k>b$ be any positive integer at least 10.
Then by trivial estimations, you can easily obtain for $ngeq k^2+k$ that $k^n/n!leq frac{k}{1}cdot frac{k}{2}cdot dots frac{k}{k}cdot dots frac{k}{k^2}cdot frac{k}{k^2+1}cdot dots frac{k}{k^2+k-1}cdot frac{k}{k^2+k} leq frac{k}{k^2+k}= frac{1}{k+1}$, which is a null sequence.
answered Nov 27 at 15:33
A. Pongrácz
5,218725
why do you do $k>b$?
– Neymar
Nov 27 at 15:38
I want the first $k$ fractions in the product be different from the last $k+1$ ($k$ at least 10 surely guarantees this). It is also easier to refer to these fractions if the base is an integer.
– A. Pongrácz
Nov 27 at 15:40
I really dont understand this method at all. Its very hard. also why you make $n=k^2+k$?
– Neymar
Nov 27 at 15:57
It is much simpler that the other suggestions in the sense that it is completely elementary. All the fractions $k/k^2, ldots, k/(k^2+k-1)$ are at most $1/k$, so they cancel out the first $k$ fractions. In the middle, every fraction is at most $1$. Hence, the product is at most $k/(k^2+k)$. You need $k^2$ to make the estimations completly trivial.
– A. Pongrácz
Nov 27 at 16:44
but my question is how you come up with $n=k^2+k$. Why are you allowed to do this?
– Neymar
Nov 27 at 18:06
|
show 1 more comment
up vote
1
down vote
Hint: let $k>b$ be any positive integer at least 10.
Then by trivial estimations, you can easily obtain for $ngeq k^2+k$ that $k^n/n!leq frac{k}{1}cdot frac{k}{2}cdot dots frac{k}{k}cdot dots frac{k}{k^2}cdot frac{k}{k^2+1}cdot dots frac{k}{k^2+k-1}cdot frac{k}{k^2+k} leq frac{k}{k^2+k}= frac{1}{k+1}$, which is a null sequence.
answered Nov 27 at 15:33
A. Pongrácz
5,218725
why do you do $k>b$?
– Neymar
Nov 27 at 15:38
I want the first $k$ fractions in the product be different from the last $k+1$ ($k$ at least 10 surely guarantees this). It is also easier to refer to these fractions if the base is an integer.
– A. Pongrácz
Nov 27 at 15:40
I really dont understand this method at all. Its very hard. also why you make $n=k^2+k$?
– Neymar
Nov 27 at 15:57
It is much simpler that the other suggestions in the sense that it is completely elementary. All the fractions $k/k^2, ldots, k/(k^2+k-1)$ are at most $1/k$, so they cancel out the first $k$ fractions. In the middle, every fraction is at most $1$. Hence, the product is at most $k/(k^2+k)$. You need $k^2$ to make the estimations completly trivial.
– A. Pongrácz
Nov 27 at 16:44
but my question is how you come up with $n=k^2+k$. Why are you allowed to do this?
– Neymar
Nov 27 at 18:06
|
show 1 more comment
up vote
1
down vote
up vote
1
down vote
Hint: let $k>b$ be any positive integer at least 10.
Then by trivial estimations, you can easily obtain for $ngeq k^2+k$ that $k^n/n!leq frac{k}{1}cdot frac{k}{2}cdot dots frac{k}{k}cdot dots frac{k}{k^2}cdot frac{k}{k^2+1}cdot dots frac{k}{k^2+k-1}cdot frac{k}{k^2+k} leq frac{k}{k^2+k}= frac{1}{k+1}$, which is a null sequence.
answered Nov 27 at 15:33
A. Pongrácz
5,218725
Hint: let $k>b$ be any positive integer at least 10.
Then by trivial estimations, you can easily obtain for $ngeq k^2+k$ that $k^n/n!leq frac{k}{1}cdot frac{k}{2}cdot dots frac{k}{k}cdot dots frac{k}{k^2}cdot frac{k}{k^2+1}cdot dots frac{k}{k^2+k-1}cdot frac{k}{k^2+k} leq frac{k}{k^2+k}= frac{1}{k+1}$, which is a null sequence.
answered Nov 27 at 15:33
A. Pongrácz
5,218725
edited Nov 27 at 18:17
answered Nov 27 at 15:33
A. Pongrácz
5,218725
answered Nov 27 at 15:33
A. Pongrácz
5,218725
answered Nov 27 at 15:33
A. Pongrácz
5,218725
5,218725
why do you do $k>b$?
– Neymar
Nov 27 at 15:38
I want the first $k$ fractions in the product be different from the last $k+1$ ($k$ at least 10 surely guarantees this). It is also easier to refer to these fractions if the base is an integer.
– A. Pongrácz
Nov 27 at 15:40
I really dont understand this method at all. Its very hard. also why you make $n=k^2+k$?
– Neymar
Nov 27 at 15:57
It is much simpler that the other suggestions in the sense that it is completely elementary. All the fractions $k/k^2, ldots, k/(k^2+k-1)$ are at most $1/k$, so they cancel out the first $k$ fractions. In the middle, every fraction is at most $1$. Hence, the product is at most $k/(k^2+k)$. You need $k^2$ to make the estimations completly trivial.
– A. Pongrácz
Nov 27 at 16:44
but my question is how you come up with $n=k^2+k$. Why are you allowed to do this?
– Neymar
Nov 27 at 18:06
|
show 1 more comment
why do you do $k>b$?
– Neymar
Nov 27 at 15:38
I want the first $k$ fractions in the product be different from the last $k+1$ ($k$ at least 10 surely guarantees this). It is also easier to refer to these fractions if the base is an integer.
– A. Pongrácz
Nov 27 at 15:40
I really dont understand this method at all. Its very hard. also why you make $n=k^2+k$?
– Neymar
Nov 27 at 15:57
It is much simpler that the other suggestions in the sense that it is completely elementary. All the fractions $k/k^2, ldots, k/(k^2+k-1)$ are at most $1/k$, so they cancel out the first $k$ fractions. In the middle, every fraction is at most $1$. Hence, the product is at most $k/(k^2+k)$. You need $k^2$ to make the estimations completly trivial.
– A. Pongrácz
Nov 27 at 16:44
but my question is how you come up with $n=k^2+k$. Why are you allowed to do this?
– Neymar
Nov 27 at 18:06
why do you do $k>b$?
– Neymar
Nov 27 at 15:38
why do you do $k>b$?
– Neymar
Nov 27 at 15:38
I want the first $k$ fractions in the product be different from the last $k+1$ ($k$ at least 10 surely guarantees this). It is also easier to refer to these fractions if the base is an integer.
– A. Pongrácz
Nov 27 at 15:40
I want the first $k$ fractions in the product be different from the last $k+1$ ($k$ at least 10 surely guarantees this). It is also easier to refer to these fractions if the base is an integer.
– A. Pongrácz
Nov 27 at 15:40
I really dont understand this method at all. Its very hard. also why you make $n=k^2+k$?
– Neymar
Nov 27 at 15:57
I really dont understand this method at all. Its very hard. also why you make $n=k^2+k$?
– Neymar
Nov 27 at 15:57
It is much simpler that the other suggestions in the sense that it is completely elementary. All the fractions $k/k^2, ldots, k/(k^2+k-1)$ are at most $1/k$, so they cancel out the first $k$ fractions. In the middle, every fraction is at most $1$. Hence, the product is at most $k/(k^2+k)$. You need $k^2$ to make the estimations completly trivial.
– A. Pongrácz
Nov 27 at 16:44
It is much simpler that the other suggestions in the sense that it is completely elementary. All the fractions $k/k^2, ldots, k/(k^2+k-1)$ are at most $1/k$, so they cancel out the first $k$ fractions. In the middle, every fraction is at most $1$. Hence, the product is at most $k/(k^2+k)$. You need $k^2$ to make the estimations completly trivial.
– A. Pongrácz
Nov 27 at 16:44
but my question is how you come up with $n=k^2+k$. Why are you allowed to do this?
– Neymar
Nov 27 at 18:06
but my question is how you come up with $n=k^2+k$. Why are you allowed to do this?
– Neymar
Nov 27 at 18:06
|
show 1 more comment
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– Akash Roy
Nov 27 at 15:25
By the way, the downvotes are coming becaue you didn't show any of your own efforts. Usually I also downvote or vote to close for this reason, but I think it is acceptable in this particular case not to show substantial work: the solution hinges on one very technical idea, without which it is hard to see how to start. But I also understand why people downvote or close such questions.
– A. Pongrácz
Nov 27 at 15:36
At least if he had described the question well, the downvotes would not have come. The problem with the question lies in its presentation. However now it is okay.
– Akash Roy
Nov 27 at 15:40
Akash Roy: sure, but maybe we should also be more patient about that. The author of the post is a new contributor, and maybe (s)he is unexperienced in discussing mathematics, or could have difficulties speaking English.
– A. Pongrácz
Nov 27 at 15:44