$lim_{n to infty} (-1)^n !ne+ (-1)^{(n+1)} n! = 0 : ?$
$lim_{n to infty} (-1)^n !ne+ (-1)^{(n+1)} n! = 0 : ?$
$!n$ is the subfactorial. Here's a plot made with wolfram:
Maybe the function goes to $0$ at infinity. How to prove this analytically? I wanted to have a better understanding of how this function behaves. If we changed $e$ to $2.7182$, which is close to the actual value of $e$ the graph changes to:
So it's only going to $0$ because of $e$?
real-analysis
asked Dec 11 '18 at 3:37
Pinteco
684313
add a comment |
$lim_{n to infty} (-1)^n !ne+ (-1)^{(n+1)} n! = 0 : ?$
$!n$ is the subfactorial. Here's a plot made with wolfram:
Maybe the function goes to $0$ at infinity. How to prove this analytically? I wanted to have a better understanding of how this function behaves. If we changed $e$ to $2.7182$, which is close to the actual value of $e$ the graph changes to:
So it's only going to $0$ because of $e$?
real-analysis
asked Dec 11 '18 at 3:37
Pinteco
684313
add a comment |
$lim_{n to infty} (-1)^n !ne+ (-1)^{(n+1)} n! = 0 : ?$
$!n$ is the subfactorial. Here's a plot made with wolfram:
Maybe the function goes to $0$ at infinity. How to prove this analytically? I wanted to have a better understanding of how this function behaves. If we changed $e$ to $2.7182$, which is close to the actual value of $e$ the graph changes to:
So it's only going to $0$ because of $e$?
real-analysis
asked Dec 11 '18 at 3:37
Pinteco
684313
$lim_{n to infty} (-1)^n !ne+ (-1)^{(n+1)} n! = 0 : ?$
$!n$ is the subfactorial. Here's a plot made with wolfram:
Maybe the function goes to $0$ at infinity. How to prove this analytically? I wanted to have a better understanding of how this function behaves. If we changed $e$ to $2.7182$, which is close to the actual value of $e$ the graph changes to:
So it's only going to $0$ because of $e$?
real-analysis
real-analysis
asked Dec 11 '18 at 3:37
Pinteco
684313
asked Dec 11 '18 at 3:37
Pinteco
684313
edited Dec 13 '18 at 22:14
asked Dec 11 '18 at 3:37
Pinteco
684313
asked Dec 11 '18 at 3:37
Pinteco
684313
asked Dec 11 '18 at 3:37
Pinteco
684313
684313
add a comment |
add a comment |
1 Answer
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Using that
$$!n=n!sum_{k=0}^n frac{(-1)^k}{k!},$$
we have
$$!n=frac{n!}{e}-n!sum_{k=n+1}^{infty} frac{(-1)^k}{k!}.$$
So
$$n!-e!n=en!sum_{k=n+1}^{infty} frac{(-1)^k}{k!}=esum_{k=n+1}^{infty} frac{n!(-1)^k}{k!}.$$
Think back to the proof that $e$ is irrational. Can you use similar logic to show that this last sum tends to $0$ as $n$ grows large? Can you use this to conclude that your original limit tends to $0$?
For your second question, think about what happens when we replace $e$ with some number very close to $e$. How does this differ from this limit?
answered Dec 11 '18 at 4:42
Carl Schildkraut
11.2k11441
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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Using that
$$!n=n!sum_{k=0}^n frac{(-1)^k}{k!},$$
we have
$$!n=frac{n!}{e}-n!sum_{k=n+1}^{infty} frac{(-1)^k}{k!}.$$
So
$$n!-e!n=en!sum_{k=n+1}^{infty} frac{(-1)^k}{k!}=esum_{k=n+1}^{infty} frac{n!(-1)^k}{k!}.$$
Think back to the proof that $e$ is irrational. Can you use similar logic to show that this last sum tends to $0$ as $n$ grows large? Can you use this to conclude that your original limit tends to $0$?
For your second question, think about what happens when we replace $e$ with some number very close to $e$. How does this differ from this limit?
answered Dec 11 '18 at 4:42
Carl Schildkraut
11.2k11441
add a comment |
Using that
$$!n=n!sum_{k=0}^n frac{(-1)^k}{k!},$$
we have
$$!n=frac{n!}{e}-n!sum_{k=n+1}^{infty} frac{(-1)^k}{k!}.$$
So
$$n!-e!n=en!sum_{k=n+1}^{infty} frac{(-1)^k}{k!}=esum_{k=n+1}^{infty} frac{n!(-1)^k}{k!}.$$
Think back to the proof that $e$ is irrational. Can you use similar logic to show that this last sum tends to $0$ as $n$ grows large? Can you use this to conclude that your original limit tends to $0$?
For your second question, think about what happens when we replace $e$ with some number very close to $e$. How does this differ from this limit?
answered Dec 11 '18 at 4:42
Carl Schildkraut
11.2k11441
add a comment |
Using that
$$!n=n!sum_{k=0}^n frac{(-1)^k}{k!},$$
we have
$$!n=frac{n!}{e}-n!sum_{k=n+1}^{infty} frac{(-1)^k}{k!}.$$
So
$$n!-e!n=en!sum_{k=n+1}^{infty} frac{(-1)^k}{k!}=esum_{k=n+1}^{infty} frac{n!(-1)^k}{k!}.$$
Think back to the proof that $e$ is irrational. Can you use similar logic to show that this last sum tends to $0$ as $n$ grows large? Can you use this to conclude that your original limit tends to $0$?
For your second question, think about what happens when we replace $e$ with some number very close to $e$. How does this differ from this limit?
answered Dec 11 '18 at 4:42
Carl Schildkraut
11.2k11441
Using that
$$!n=n!sum_{k=0}^n frac{(-1)^k}{k!},$$
we have
$$!n=frac{n!}{e}-n!sum_{k=n+1}^{infty} frac{(-1)^k}{k!}.$$
So
$$n!-e!n=en!sum_{k=n+1}^{infty} frac{(-1)^k}{k!}=esum_{k=n+1}^{infty} frac{n!(-1)^k}{k!}.$$
Think back to the proof that $e$ is irrational. Can you use similar logic to show that this last sum tends to $0$ as $n$ grows large? Can you use this to conclude that your original limit tends to $0$?
For your second question, think about what happens when we replace $e$ with some number very close to $e$. How does this differ from this limit?
answered Dec 11 '18 at 4:42
Carl Schildkraut
11.2k11441
answered Dec 11 '18 at 4:42
Carl Schildkraut
11.2k11441
answered Dec 11 '18 at 4:42
Carl Schildkraut
11.2k11441
answered Dec 11 '18 at 4:42
Carl Schildkraut
11.2k11441
11.2k11441
add a comment |
add a comment |
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