How to compute $(-1)^{n+1}n!(1-esum_{k=0}^nfrac{(-1)^k}{k!})$?
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I was doing some work on the integral $int_0^1 x^ne^x dx$ and I eventually came to this expression in terms of n $$int_0^1 x^ne^x dx=(-1)^{n+1}n!biggl(1-esum_{k=0}^nfrac{(-1)^k}{k!}biggr)$$ Now my question is this, how do I compute the limit of this expression as $n$ goes to $infty$? I recognize that $$lim_{ntoinfty}esum_{k=0}^{n}frac{(-1)^k}{k!}=1$$ per the series expression of $e^x$ at $-1$. This gives me an indeterminate form, but I can't apply L'Hopital's rule because I can't differentiate $n!$ (or $1/n!$ as the case would be after re-arranging), and I also can't think of a variable substitution that would help me here. Common sense and wolfram definitely imply that the limit is in fact $0$, but how can I show this?
calculus integration limits definite-integrals exponential-function
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up vote
2
down vote
favorite
I was doing some work on the integral $int_0^1 x^ne^x dx$ and I eventually came to this expression in terms of n $$int_0^1 x^ne^x dx=(-1)^{n+1}n!biggl(1-esum_{k=0}^nfrac{(-1)^k}{k!}biggr)$$ Now my question is this, how do I compute the limit of this expression as $n$ goes to $infty$? I recognize that $$lim_{ntoinfty}esum_{k=0}^{n}frac{(-1)^k}{k!}=1$$ per the series expression of $e^x$ at $-1$. This gives me an indeterminate form, but I can't apply L'Hopital's rule because I can't differentiate $n!$ (or $1/n!$ as the case would be after re-arranging), and I also can't think of a variable substitution that would help me here. Common sense and wolfram definitely imply that the limit is in fact $0$, but how can I show this?
calculus integration limits definite-integrals exponential-function
3
$n!=Gamma(n+1)$ is differentiable.
– J.G.
Nov 15 at 7:01
1
perhaps some properties of the gamma function/complex analysis would be useful
– rubikscube09
Nov 15 at 7:04
I'm not very familiar with the gamma function, is there any other way? And if not, how could I use the gamma function to do this? Also, would it be sufficient to show that the integrand approaches $0$ on the interval?
– JacksonFitzsimmons
Nov 15 at 7:06
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I was doing some work on the integral $int_0^1 x^ne^x dx$ and I eventually came to this expression in terms of n $$int_0^1 x^ne^x dx=(-1)^{n+1}n!biggl(1-esum_{k=0}^nfrac{(-1)^k}{k!}biggr)$$ Now my question is this, how do I compute the limit of this expression as $n$ goes to $infty$? I recognize that $$lim_{ntoinfty}esum_{k=0}^{n}frac{(-1)^k}{k!}=1$$ per the series expression of $e^x$ at $-1$. This gives me an indeterminate form, but I can't apply L'Hopital's rule because I can't differentiate $n!$ (or $1/n!$ as the case would be after re-arranging), and I also can't think of a variable substitution that would help me here. Common sense and wolfram definitely imply that the limit is in fact $0$, but how can I show this?
calculus integration limits definite-integrals exponential-function
I was doing some work on the integral $int_0^1 x^ne^x dx$ and I eventually came to this expression in terms of n $$int_0^1 x^ne^x dx=(-1)^{n+1}n!biggl(1-esum_{k=0}^nfrac{(-1)^k}{k!}biggr)$$ Now my question is this, how do I compute the limit of this expression as $n$ goes to $infty$? I recognize that $$lim_{ntoinfty}esum_{k=0}^{n}frac{(-1)^k}{k!}=1$$ per the series expression of $e^x$ at $-1$. This gives me an indeterminate form, but I can't apply L'Hopital's rule because I can't differentiate $n!$ (or $1/n!$ as the case would be after re-arranging), and I also can't think of a variable substitution that would help me here. Common sense and wolfram definitely imply that the limit is in fact $0$, but how can I show this?
calculus integration limits definite-integrals exponential-function
calculus integration limits definite-integrals exponential-function
edited Nov 15 at 12:11
user21820
38k441148
38k441148
asked Nov 15 at 6:51
JacksonFitzsimmons
551212
551212
3
$n!=Gamma(n+1)$ is differentiable.
– J.G.
Nov 15 at 7:01
1
perhaps some properties of the gamma function/complex analysis would be useful
– rubikscube09
Nov 15 at 7:04
I'm not very familiar with the gamma function, is there any other way? And if not, how could I use the gamma function to do this? Also, would it be sufficient to show that the integrand approaches $0$ on the interval?
– JacksonFitzsimmons
Nov 15 at 7:06
add a comment |
3
$n!=Gamma(n+1)$ is differentiable.
– J.G.
Nov 15 at 7:01
1
perhaps some properties of the gamma function/complex analysis would be useful
– rubikscube09
Nov 15 at 7:04
I'm not very familiar with the gamma function, is there any other way? And if not, how could I use the gamma function to do this? Also, would it be sufficient to show that the integrand approaches $0$ on the interval?
– JacksonFitzsimmons
Nov 15 at 7:06
3
3
$n!=Gamma(n+1)$ is differentiable.
– J.G.
Nov 15 at 7:01
$n!=Gamma(n+1)$ is differentiable.
– J.G.
Nov 15 at 7:01
1
1
perhaps some properties of the gamma function/complex analysis would be useful
– rubikscube09
Nov 15 at 7:04
perhaps some properties of the gamma function/complex analysis would be useful
– rubikscube09
Nov 15 at 7:04
I'm not very familiar with the gamma function, is there any other way? And if not, how could I use the gamma function to do this? Also, would it be sufficient to show that the integrand approaches $0$ on the interval?
– JacksonFitzsimmons
Nov 15 at 7:06
I'm not very familiar with the gamma function, is there any other way? And if not, how could I use the gamma function to do this? Also, would it be sufficient to show that the integrand approaches $0$ on the interval?
– JacksonFitzsimmons
Nov 15 at 7:06
add a comment |
3 Answers
3
active
oldest
votes
up vote
5
down vote
accepted
When $xin [0,1]$, $e^xle e$, by first mean value theorem for definite integrals,
begin{align*}
I_n&lemax_{xin[0,1]}{e^x}int_0^1x^{n}~mathrm dx\
&=frac{e}{n+1}to0.
end{align*}
Since $I_n>0$, the limit is $0$ by squeeze theorem.
Thanks, I never remember the squeeze theorem
– JacksonFitzsimmons
Nov 15 at 7:08
add a comment |
up vote
4
down vote
Not as slick as Tianlalu's method, but for $yin (0,,1)$ we have
$$
lim_{ntoinfty}int_0^y x^n e^x dx le
lim_{ntoinfty}y^nint_0^y e^x dx=0.
$$
add a comment |
up vote
2
down vote
I don't know what you are 'allowed' to use but you could use the lagrange remainer for the exponential function:
$$exp(-1)=sum_{k=0}^nfrac{(-1)^k}{k!}+frac{e^xi}{(n+1)!}(-1)^{n+1}$$
for some $-1<xi<0$. If you estimate $e^xi $ and replace the sum you get convergence to $0$.
New contributor
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
When $xin [0,1]$, $e^xle e$, by first mean value theorem for definite integrals,
begin{align*}
I_n&lemax_{xin[0,1]}{e^x}int_0^1x^{n}~mathrm dx\
&=frac{e}{n+1}to0.
end{align*}
Since $I_n>0$, the limit is $0$ by squeeze theorem.
Thanks, I never remember the squeeze theorem
– JacksonFitzsimmons
Nov 15 at 7:08
add a comment |
up vote
5
down vote
accepted
When $xin [0,1]$, $e^xle e$, by first mean value theorem for definite integrals,
begin{align*}
I_n&lemax_{xin[0,1]}{e^x}int_0^1x^{n}~mathrm dx\
&=frac{e}{n+1}to0.
end{align*}
Since $I_n>0$, the limit is $0$ by squeeze theorem.
Thanks, I never remember the squeeze theorem
– JacksonFitzsimmons
Nov 15 at 7:08
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
When $xin [0,1]$, $e^xle e$, by first mean value theorem for definite integrals,
begin{align*}
I_n&lemax_{xin[0,1]}{e^x}int_0^1x^{n}~mathrm dx\
&=frac{e}{n+1}to0.
end{align*}
Since $I_n>0$, the limit is $0$ by squeeze theorem.
When $xin [0,1]$, $e^xle e$, by first mean value theorem for definite integrals,
begin{align*}
I_n&lemax_{xin[0,1]}{e^x}int_0^1x^{n}~mathrm dx\
&=frac{e}{n+1}to0.
end{align*}
Since $I_n>0$, the limit is $0$ by squeeze theorem.
answered Nov 15 at 7:05
Tianlalu
2,451632
2,451632
Thanks, I never remember the squeeze theorem
– JacksonFitzsimmons
Nov 15 at 7:08
add a comment |
Thanks, I never remember the squeeze theorem
– JacksonFitzsimmons
Nov 15 at 7:08
Thanks, I never remember the squeeze theorem
– JacksonFitzsimmons
Nov 15 at 7:08
Thanks, I never remember the squeeze theorem
– JacksonFitzsimmons
Nov 15 at 7:08
add a comment |
up vote
4
down vote
Not as slick as Tianlalu's method, but for $yin (0,,1)$ we have
$$
lim_{ntoinfty}int_0^y x^n e^x dx le
lim_{ntoinfty}y^nint_0^y e^x dx=0.
$$
add a comment |
up vote
4
down vote
Not as slick as Tianlalu's method, but for $yin (0,,1)$ we have
$$
lim_{ntoinfty}int_0^y x^n e^x dx le
lim_{ntoinfty}y^nint_0^y e^x dx=0.
$$
add a comment |
up vote
4
down vote
up vote
4
down vote
Not as slick as Tianlalu's method, but for $yin (0,,1)$ we have
$$
lim_{ntoinfty}int_0^y x^n e^x dx le
lim_{ntoinfty}y^nint_0^y e^x dx=0.
$$
Not as slick as Tianlalu's method, but for $yin (0,,1)$ we have
$$
lim_{ntoinfty}int_0^y x^n e^x dx le
lim_{ntoinfty}y^nint_0^y e^x dx=0.
$$
edited Nov 15 at 7:31
Tianlalu
2,451632
2,451632
answered Nov 15 at 7:28
J.G.
18k11830
18k11830
add a comment |
add a comment |
up vote
2
down vote
I don't know what you are 'allowed' to use but you could use the lagrange remainer for the exponential function:
$$exp(-1)=sum_{k=0}^nfrac{(-1)^k}{k!}+frac{e^xi}{(n+1)!}(-1)^{n+1}$$
for some $-1<xi<0$. If you estimate $e^xi $ and replace the sum you get convergence to $0$.
New contributor
add a comment |
up vote
2
down vote
I don't know what you are 'allowed' to use but you could use the lagrange remainer for the exponential function:
$$exp(-1)=sum_{k=0}^nfrac{(-1)^k}{k!}+frac{e^xi}{(n+1)!}(-1)^{n+1}$$
for some $-1<xi<0$. If you estimate $e^xi $ and replace the sum you get convergence to $0$.
New contributor
add a comment |
up vote
2
down vote
up vote
2
down vote
I don't know what you are 'allowed' to use but you could use the lagrange remainer for the exponential function:
$$exp(-1)=sum_{k=0}^nfrac{(-1)^k}{k!}+frac{e^xi}{(n+1)!}(-1)^{n+1}$$
for some $-1<xi<0$. If you estimate $e^xi $ and replace the sum you get convergence to $0$.
New contributor
I don't know what you are 'allowed' to use but you could use the lagrange remainer for the exponential function:
$$exp(-1)=sum_{k=0}^nfrac{(-1)^k}{k!}+frac{e^xi}{(n+1)!}(-1)^{n+1}$$
for some $-1<xi<0$. If you estimate $e^xi $ and replace the sum you get convergence to $0$.
New contributor
New contributor
answered Nov 15 at 7:13
maxmilgram
3587
3587
New contributor
New contributor
add a comment |
add a comment |
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3
$n!=Gamma(n+1)$ is differentiable.
– J.G.
Nov 15 at 7:01
1
perhaps some properties of the gamma function/complex analysis would be useful
– rubikscube09
Nov 15 at 7:04
I'm not very familiar with the gamma function, is there any other way? And if not, how could I use the gamma function to do this? Also, would it be sufficient to show that the integrand approaches $0$ on the interval?
– JacksonFitzsimmons
Nov 15 at 7:06