Mixing
$sumlimits_{n=1}^infty frac{n^n}{n!}$ converges or diverges?
$begingroup$
$$sum_{n=1}^infty frac{n^n}{n!}$$
It is a convergent or divergent series? Explain how.
I am getting infinite value as $n$ tends to infinity. I have applied the ratio test and higher ratio test.
calculus sequences-and-series convergence
edited Sep 5 '17 at 3:10
J. M. is not a mathematician
61.1k5152290
asked Sep 5 '17 at 2:30
sammy asammy a
41
$endgroup$
|
show 1 more comment
$begingroup$
$$sum_{n=1}^infty frac{n^n}{n!}$$
It is a convergent or divergent series? Explain how.
I am getting infinite value as $n$ tends to infinity. I have applied the ratio test and higher ratio test.
calculus sequences-and-series convergence
edited Sep 5 '17 at 3:10
J. M. is not a mathematician
61.1k5152290
asked Sep 5 '17 at 2:30
sammy asammy a
41
$endgroup$
1
$begingroup$
The ratio test is overkill. This series doesn't even get past the divergence test.
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 2:37
$begingroup$
Using no tests whatsoever, you can tell that it will diverge just by plugging in values for $n>1$. The numerator will always be $n*n*cdots$ while the denominator will always be smaller: $n(n-1)(n-2)cdots$.
$endgroup$
– WaveX
Sep 5 '17 at 2:42
1
$begingroup$
@WaveX Technically, that is an application of the divergence test. :P
$endgroup$
– Xander Henderson
Sep 5 '17 at 2:56
$begingroup$
@Xander Henderson Yea I guess that's true, it has been well over a year since I've last looked at infinite series :)
$endgroup$
– WaveX
Sep 5 '17 at 3:03
1
$begingroup$
Phrasing? :) Any question of the form "Is it convergent or divergent?" can be answered immediately with a resounding "yes."
$endgroup$
– Clement C.
Sep 5 '17 at 3:09
|
show 1 more comment
$begingroup$
$$sum_{n=1}^infty frac{n^n}{n!}$$
It is a convergent or divergent series? Explain how.
I am getting infinite value as $n$ tends to infinity. I have applied the ratio test and higher ratio test.
calculus sequences-and-series convergence
edited Sep 5 '17 at 3:10
J. M. is not a mathematician
61.1k5152290
asked Sep 5 '17 at 2:30
sammy asammy a
41
$endgroup$
$$sum_{n=1}^infty frac{n^n}{n!}$$
It is a convergent or divergent series? Explain how.
I am getting infinite value as $n$ tends to infinity. I have applied the ratio test and higher ratio test.
calculus sequences-and-series convergence
calculus sequences-and-series convergence
edited Sep 5 '17 at 3:10
J. M. is not a mathematician
61.1k5152290
asked Sep 5 '17 at 2:30
sammy asammy a
41
edited Sep 5 '17 at 3:10
J. M. is not a mathematician
61.1k5152290
asked Sep 5 '17 at 2:30
sammy asammy a
41
edited Sep 5 '17 at 3:10
J. M. is not a mathematician
61.1k5152290
edited Sep 5 '17 at 3:10
J. M. is not a mathematician
61.1k5152290
edited Sep 5 '17 at 3:10
J. M. is not a mathematician
61.1k5152290
61.1k5152290
asked Sep 5 '17 at 2:30
sammy asammy a
41
asked Sep 5 '17 at 2:30
sammy asammy a
41
asked Sep 5 '17 at 2:30
sammy asammy a
41
41
1
$begingroup$
The ratio test is overkill. This series doesn't even get past the divergence test.
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 2:37
$begingroup$
Using no tests whatsoever, you can tell that it will diverge just by plugging in values for $n>1$. The numerator will always be $n*n*cdots$ while the denominator will always be smaller: $n(n-1)(n-2)cdots$.
$endgroup$
– WaveX
Sep 5 '17 at 2:42
1
$begingroup$
@WaveX Technically, that is an application of the divergence test. :P
$endgroup$
– Xander Henderson
Sep 5 '17 at 2:56
$begingroup$
@Xander Henderson Yea I guess that's true, it has been well over a year since I've last looked at infinite series :)
$endgroup$
– WaveX
Sep 5 '17 at 3:03
1
$begingroup$
Phrasing? :) Any question of the form "Is it convergent or divergent?" can be answered immediately with a resounding "yes."
$endgroup$
– Clement C.
Sep 5 '17 at 3:09
|
show 1 more comment
1
$begingroup$
The ratio test is overkill. This series doesn't even get past the divergence test.
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 2:37
$begingroup$
Using no tests whatsoever, you can tell that it will diverge just by plugging in values for $n>1$. The numerator will always be $n*n*cdots$ while the denominator will always be smaller: $n(n-1)(n-2)cdots$.
$endgroup$
– WaveX
Sep 5 '17 at 2:42
1
$begingroup$
@WaveX Technically, that is an application of the divergence test. :P
$endgroup$
– Xander Henderson
Sep 5 '17 at 2:56
$begingroup$
@Xander Henderson Yea I guess that's true, it has been well over a year since I've last looked at infinite series :)
$endgroup$
– WaveX
Sep 5 '17 at 3:03
1
$begingroup$
Phrasing? :) Any question of the form "Is it convergent or divergent?" can be answered immediately with a resounding "yes."
$endgroup$
– Clement C.
Sep 5 '17 at 3:09
1
1
$begingroup$
The ratio test is overkill. This series doesn't even get past the divergence test.
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 2:37
$begingroup$
The ratio test is overkill. This series doesn't even get past the divergence test.
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 2:37
$begingroup$
Using no tests whatsoever, you can tell that it will diverge just by plugging in values for $n>1$. The numerator will always be $n*n*cdots$ while the denominator will always be smaller: $n(n-1)(n-2)cdots$.
$endgroup$
– WaveX
Sep 5 '17 at 2:42
$begingroup$
Using no tests whatsoever, you can tell that it will diverge just by plugging in values for $n>1$. The numerator will always be $n*n*cdots$ while the denominator will always be smaller: $n(n-1)(n-2)cdots$.
$endgroup$
– WaveX
Sep 5 '17 at 2:42
1
1
$begingroup$
@WaveX Technically, that is an application of the divergence test. :P
$endgroup$
– Xander Henderson
Sep 5 '17 at 2:56
$begingroup$
@WaveX Technically, that is an application of the divergence test. :P
$endgroup$
– Xander Henderson
Sep 5 '17 at 2:56
$begingroup$
@Xander Henderson Yea I guess that's true, it has been well over a year since I've last looked at infinite series :)
$endgroup$
– WaveX
Sep 5 '17 at 3:03
$begingroup$
@Xander Henderson Yea I guess that's true, it has been well over a year since I've last looked at infinite series :)
$endgroup$
– WaveX
Sep 5 '17 at 3:03
1
1
$begingroup$
Phrasing? :) Any question of the form "Is it convergent or divergent?" can be answered immediately with a resounding "yes."
$endgroup$
– Clement C.
Sep 5 '17 at 3:09
$begingroup$
Phrasing? :) Any question of the form "Is it convergent or divergent?" can be answered immediately with a resounding "yes."
$endgroup$
– Clement C.
Sep 5 '17 at 3:09
|
show 1 more comment
4 Answers
4
active
oldest
votes
$begingroup$
Hint: Note that
$$frac{a_{n+1}}{a_n}=frac{frac{(n+1)^{n+1}}{(n+1)!}}{frac{n^n}{n!}}=frac{(n+1)^{n+1}n!}{n^n(n+1)!}=frac{(n+1)^{n+1}}{n^n(n+1)}=left(frac{n+1}nright)^n;.$$
answered Sep 5 '17 at 2:35
WenWen
1,633417
$endgroup$
add a comment |
$begingroup$
The numerator grows faster than the denominator. For any value of $n>1$, we have $$n^n>n!$$ because they are both products of $n$ factors, but on the LHS, each factor equals $n$, and on the RHS, every factor but one is smaller than $n$. Therefore, for $n>1$, we have $a_n=frac{n^n}{n!}>1$. Since $a_nnotto 0$, the series cannot converge, by the divergence test.
In fact, the terms in this sum grow without bound, but even without that, we have shown that the series diverges.
answered Sep 5 '17 at 2:35
G Tony JacobsG Tony Jacobs
25.9k43686
$endgroup$
2
$begingroup$
It is enough to argue that $n^n>n!$ implies that $frac{n^n}{n!}>1$ cannot converge to zero.
$endgroup$
– lhf
Sep 5 '17 at 2:38
$begingroup$
True. Are you saying I should edit my answer?
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 2:38
$begingroup$
No, unless you feel it makes the answer clearer.
$endgroup$
– lhf
Sep 5 '17 at 2:39
$begingroup$
Yes, the terms are greater than $1$ for $n>1,$ but this does not imply they grow without bound. (They do grow without bound, but you did not show it, and you don't need it.)
$endgroup$
– zhw.
Sep 5 '17 at 3:07
$begingroup$
Of course it doesn't imply that they grow without bound, but nobody said it does.
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 3:53
add a comment |
$begingroup$
This series is diverge since $lim_inftydfrac{n^n}{n!}=infty$.
answered Sep 5 '17 at 2:43
NosratiNosrati
26.5k62354
$endgroup$
add a comment |
$begingroup$
To expand on Wen's answer,
$$rho=lim_{nrightarrowinfty}bigg(frac{n+1}{n}bigg)^n = lim_{nrightarrowinfty}bigg(1 + frac{1}{n}bigg)^n = e $$
and since $e > 1$, the series diverges by the ratio test.
answered Sep 5 '17 at 2:48
R.EvetR.Evet
908
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Note that
$$frac{a_{n+1}}{a_n}=frac{frac{(n+1)^{n+1}}{(n+1)!}}{frac{n^n}{n!}}=frac{(n+1)^{n+1}n!}{n^n(n+1)!}=frac{(n+1)^{n+1}}{n^n(n+1)}=left(frac{n+1}nright)^n;.$$
answered Sep 5 '17 at 2:35
WenWen
1,633417
$endgroup$
add a comment |
$begingroup$
Hint: Note that
$$frac{a_{n+1}}{a_n}=frac{frac{(n+1)^{n+1}}{(n+1)!}}{frac{n^n}{n!}}=frac{(n+1)^{n+1}n!}{n^n(n+1)!}=frac{(n+1)^{n+1}}{n^n(n+1)}=left(frac{n+1}nright)^n;.$$
answered Sep 5 '17 at 2:35
WenWen
1,633417
$endgroup$
add a comment |
$begingroup$
Hint: Note that
$$frac{a_{n+1}}{a_n}=frac{frac{(n+1)^{n+1}}{(n+1)!}}{frac{n^n}{n!}}=frac{(n+1)^{n+1}n!}{n^n(n+1)!}=frac{(n+1)^{n+1}}{n^n(n+1)}=left(frac{n+1}nright)^n;.$$
answered Sep 5 '17 at 2:35
WenWen
1,633417
$endgroup$
Hint: Note that
$$frac{a_{n+1}}{a_n}=frac{frac{(n+1)^{n+1}}{(n+1)!}}{frac{n^n}{n!}}=frac{(n+1)^{n+1}n!}{n^n(n+1)!}=frac{(n+1)^{n+1}}{n^n(n+1)}=left(frac{n+1}nright)^n;.$$
answered Sep 5 '17 at 2:35
WenWen
1,633417
answered Sep 5 '17 at 2:35
WenWen
1,633417
answered Sep 5 '17 at 2:35
WenWen
1,633417
answered Sep 5 '17 at 2:35
WenWen
1,633417
1,633417
add a comment |
add a comment |
$begingroup$
The numerator grows faster than the denominator. For any value of $n>1$, we have $$n^n>n!$$ because they are both products of $n$ factors, but on the LHS, each factor equals $n$, and on the RHS, every factor but one is smaller than $n$. Therefore, for $n>1$, we have $a_n=frac{n^n}{n!}>1$. Since $a_nnotto 0$, the series cannot converge, by the divergence test.
In fact, the terms in this sum grow without bound, but even without that, we have shown that the series diverges.
answered Sep 5 '17 at 2:35
G Tony JacobsG Tony Jacobs
25.9k43686
$endgroup$
2
$begingroup$
It is enough to argue that $n^n>n!$ implies that $frac{n^n}{n!}>1$ cannot converge to zero.
$endgroup$
– lhf
Sep 5 '17 at 2:38
$begingroup$
True. Are you saying I should edit my answer?
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 2:38
$begingroup$
No, unless you feel it makes the answer clearer.
$endgroup$
– lhf
Sep 5 '17 at 2:39
$begingroup$
Yes, the terms are greater than $1$ for $n>1,$ but this does not imply they grow without bound. (They do grow without bound, but you did not show it, and you don't need it.)
$endgroup$
– zhw.
Sep 5 '17 at 3:07
$begingroup$
Of course it doesn't imply that they grow without bound, but nobody said it does.
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 3:53
add a comment |
$begingroup$
The numerator grows faster than the denominator. For any value of $n>1$, we have $$n^n>n!$$ because they are both products of $n$ factors, but on the LHS, each factor equals $n$, and on the RHS, every factor but one is smaller than $n$. Therefore, for $n>1$, we have $a_n=frac{n^n}{n!}>1$. Since $a_nnotto 0$, the series cannot converge, by the divergence test.
In fact, the terms in this sum grow without bound, but even without that, we have shown that the series diverges.
answered Sep 5 '17 at 2:35
G Tony JacobsG Tony Jacobs
25.9k43686
$endgroup$
2
$begingroup$
It is enough to argue that $n^n>n!$ implies that $frac{n^n}{n!}>1$ cannot converge to zero.
$endgroup$
– lhf
Sep 5 '17 at 2:38
$begingroup$
True. Are you saying I should edit my answer?
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 2:38
$begingroup$
No, unless you feel it makes the answer clearer.
$endgroup$
– lhf
Sep 5 '17 at 2:39
$begingroup$
Yes, the terms are greater than $1$ for $n>1,$ but this does not imply they grow without bound. (They do grow without bound, but you did not show it, and you don't need it.)
$endgroup$
– zhw.
Sep 5 '17 at 3:07
$begingroup$
Of course it doesn't imply that they grow without bound, but nobody said it does.
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 3:53
add a comment |
$begingroup$
The numerator grows faster than the denominator. For any value of $n>1$, we have $$n^n>n!$$ because they are both products of $n$ factors, but on the LHS, each factor equals $n$, and on the RHS, every factor but one is smaller than $n$. Therefore, for $n>1$, we have $a_n=frac{n^n}{n!}>1$. Since $a_nnotto 0$, the series cannot converge, by the divergence test.
In fact, the terms in this sum grow without bound, but even without that, we have shown that the series diverges.
answered Sep 5 '17 at 2:35
G Tony JacobsG Tony Jacobs
25.9k43686
$endgroup$
The numerator grows faster than the denominator. For any value of $n>1$, we have $$n^n>n!$$ because they are both products of $n$ factors, but on the LHS, each factor equals $n$, and on the RHS, every factor but one is smaller than $n$. Therefore, for $n>1$, we have $a_n=frac{n^n}{n!}>1$. Since $a_nnotto 0$, the series cannot converge, by the divergence test.
In fact, the terms in this sum grow without bound, but even without that, we have shown that the series diverges.
answered Sep 5 '17 at 2:35
G Tony JacobsG Tony Jacobs
25.9k43686
edited Sep 5 '17 at 3:55
answered Sep 5 '17 at 2:35
G Tony JacobsG Tony Jacobs
25.9k43686
answered Sep 5 '17 at 2:35
G Tony JacobsG Tony Jacobs
25.9k43686
answered Sep 5 '17 at 2:35
G Tony JacobsG Tony Jacobs
25.9k43686
25.9k43686
2
$begingroup$
It is enough to argue that $n^n>n!$ implies that $frac{n^n}{n!}>1$ cannot converge to zero.
$endgroup$
– lhf
Sep 5 '17 at 2:38
$begingroup$
True. Are you saying I should edit my answer?
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 2:38
$begingroup$
No, unless you feel it makes the answer clearer.
$endgroup$
– lhf
Sep 5 '17 at 2:39
$begingroup$
Yes, the terms are greater than $1$ for $n>1,$ but this does not imply they grow without bound. (They do grow without bound, but you did not show it, and you don't need it.)
$endgroup$
– zhw.
Sep 5 '17 at 3:07
$begingroup$
Of course it doesn't imply that they grow without bound, but nobody said it does.
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 3:53
add a comment |
2
$begingroup$
It is enough to argue that $n^n>n!$ implies that $frac{n^n}{n!}>1$ cannot converge to zero.
$endgroup$
– lhf
Sep 5 '17 at 2:38
$begingroup$
True. Are you saying I should edit my answer?
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 2:38
$begingroup$
No, unless you feel it makes the answer clearer.
$endgroup$
– lhf
Sep 5 '17 at 2:39
$begingroup$
Yes, the terms are greater than $1$ for $n>1,$ but this does not imply they grow without bound. (They do grow without bound, but you did not show it, and you don't need it.)
$endgroup$
– zhw.
Sep 5 '17 at 3:07
$begingroup$
Of course it doesn't imply that they grow without bound, but nobody said it does.
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 3:53
2
2
$begingroup$
It is enough to argue that $n^n>n!$ implies that $frac{n^n}{n!}>1$ cannot converge to zero.
$endgroup$
– lhf
Sep 5 '17 at 2:38
$begingroup$
It is enough to argue that $n^n>n!$ implies that $frac{n^n}{n!}>1$ cannot converge to zero.
$endgroup$
– lhf
Sep 5 '17 at 2:38
$begingroup$
True. Are you saying I should edit my answer?
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 2:38
$begingroup$
True. Are you saying I should edit my answer?
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 2:38
$begingroup$
No, unless you feel it makes the answer clearer.
$endgroup$
– lhf
Sep 5 '17 at 2:39
$begingroup$
No, unless you feel it makes the answer clearer.
$endgroup$
– lhf
Sep 5 '17 at 2:39
$begingroup$
Yes, the terms are greater than $1$ for $n>1,$ but this does not imply they grow without bound. (They do grow without bound, but you did not show it, and you don't need it.)
$endgroup$
– zhw.
Sep 5 '17 at 3:07
$begingroup$
Yes, the terms are greater than $1$ for $n>1,$ but this does not imply they grow without bound. (They do grow without bound, but you did not show it, and you don't need it.)
$endgroup$
– zhw.
Sep 5 '17 at 3:07
$begingroup$
Of course it doesn't imply that they grow without bound, but nobody said it does.
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 3:53
$begingroup$
Of course it doesn't imply that they grow without bound, but nobody said it does.
$endgroup$
– G Tony Jacobs
Sep 5 '17 at 3:53
add a comment |
$begingroup$
This series is diverge since $lim_inftydfrac{n^n}{n!}=infty$.
answered Sep 5 '17 at 2:43
NosratiNosrati
26.5k62354
$endgroup$
add a comment |
$begingroup$
This series is diverge since $lim_inftydfrac{n^n}{n!}=infty$.
answered Sep 5 '17 at 2:43
NosratiNosrati
26.5k62354
$endgroup$
add a comment |
$begingroup$
This series is diverge since $lim_inftydfrac{n^n}{n!}=infty$.
answered Sep 5 '17 at 2:43
NosratiNosrati
26.5k62354
$endgroup$
This series is diverge since $lim_inftydfrac{n^n}{n!}=infty$.
answered Sep 5 '17 at 2:43
NosratiNosrati
26.5k62354
answered Sep 5 '17 at 2:43
NosratiNosrati
26.5k62354
answered Sep 5 '17 at 2:43
NosratiNosrati
26.5k62354
answered Sep 5 '17 at 2:43
NosratiNosrati
26.5k62354
26.5k62354
add a comment |
add a comment |
$begingroup$
To expand on Wen's answer,
$$rho=lim_{nrightarrowinfty}bigg(frac{n+1}{n}bigg)^n = lim_{nrightarrowinfty}bigg(1 + frac{1}{n}bigg)^n = e $$
and since $e > 1$, the series diverges by the ratio test.
answered Sep 5 '17 at 2:48
R.EvetR.Evet
908
$endgroup$
add a comment |
$begingroup$
To expand on Wen's answer,
$$rho=lim_{nrightarrowinfty}bigg(frac{n+1}{n}bigg)^n = lim_{nrightarrowinfty}bigg(1 + frac{1}{n}bigg)^n = e $$
and since $e > 1$, the series diverges by the ratio test.
answered Sep 5 '17 at 2:48
R.EvetR.Evet
908
$endgroup$
add a comment |
$begingroup$
To expand on Wen's answer,
$$rho=lim_{nrightarrowinfty}bigg(frac{n+1}{n}bigg)^n = lim_{nrightarrowinfty}bigg(1 + frac{1}{n}bigg)^n = e $$
and since $e > 1$, the series diverges by the ratio test.
answered Sep 5 '17 at 2:48
R.EvetR.Evet
908
$endgroup$
To expand on Wen's answer,
$$rho=lim_{nrightarrowinfty}bigg(frac{n+1}{n}bigg)^n = lim_{nrightarrowinfty}bigg(1 + frac{1}{n}bigg)^n = e $$
and since $e > 1$, the series diverges by the ratio test.
answered Sep 5 '17 at 2:48
R.EvetR.Evet
908
answered Sep 5 '17 at 2:48
R.EvetR.Evet
908
answered Sep 5 '17 at 2:48
R.EvetR.Evet
908
answered Sep 5 '17 at 2:48
R.EvetR.Evet
908
908
add a comment |
add a comment |
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1
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The ratio test is overkill. This series doesn't even get past the divergence test.
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– G Tony Jacobs
Sep 5 '17 at 2:37
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Using no tests whatsoever, you can tell that it will diverge just by plugging in values for $n>1$. The numerator will always be $n*n*cdots$ while the denominator will always be smaller: $n(n-1)(n-2)cdots$.
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– WaveX
Sep 5 '17 at 2:42
1
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@WaveX Technically, that is an application of the divergence test. :P
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– Xander Henderson
Sep 5 '17 at 2:56
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@Xander Henderson Yea I guess that's true, it has been well over a year since I've last looked at infinite series :)
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– WaveX
Sep 5 '17 at 3:03
1
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Phrasing? :) Any question of the form "Is it convergent or divergent?" can be answered immediately with a resounding "yes."
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– Clement C.
Sep 5 '17 at 3:09