Calculating convergence of a sum
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1
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I have the sum:
$$S=sum_{n=1}^inftyfrac{n!}{n^n}$$
and I am using D'Alembert's test for convergence which states for some sum:
$$sum_{n=a}^infty u_n,,(aneqpminfty)$$
that it is convergent if:
$$lim_{ntoinfty}left|frac{u_{n+1}}{u_n}right|<1$$
so to begin with I know that:
$$L=lim_{ntoinfty}frac{frac{(n+1)!}{(n+1)^{n+1}}}{frac{n!}{n^n}}=lim_{ntoinfty}frac{n^n}{(n+1)^n}$$
Up to this point is fine but I am unsure if what I have done next is correct:
$$ln(L)=lim_{ntoinfty}nlnleft(frac{n}{n+1}right)=lim_{ntoinfty}frac{lnleft(1-frac{1}{n+1}right)}{frac{1}{n}}$$
now I used the substitution $u=frac{1}{n+1}$ and with rearrangement I believe I can obtain:
$$ln(L)=lim_{uto0}frac{(1-u)ln(1-u)}{u}$$
now since when $uto0$ both the top and bottom also tend to $0$ I can use L'Hopitals rule:
$$ln(L)=lim_{uto0}frac{-1-ln(1-u)}{1}=lim_{uto0}-bigl(1+ln(1-u)bigr)=-1$$
which now gives:
$$ln(L)=-1therefore L=e^{-1}approx0.368<1$$
and so there is convergence
In the question it states that I will have to use:
$$lim_{ntoinfty}left(1+frac1nright)^n=e$$
convergence summation
asked Dec 5 at 0:49
Henry Lee
1,707218
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up vote
1
down vote
favorite
I have the sum:
$$S=sum_{n=1}^inftyfrac{n!}{n^n}$$
and I am using D'Alembert's test for convergence which states for some sum:
$$sum_{n=a}^infty u_n,,(aneqpminfty)$$
that it is convergent if:
$$lim_{ntoinfty}left|frac{u_{n+1}}{u_n}right|<1$$
so to begin with I know that:
$$L=lim_{ntoinfty}frac{frac{(n+1)!}{(n+1)^{n+1}}}{frac{n!}{n^n}}=lim_{ntoinfty}frac{n^n}{(n+1)^n}$$
Up to this point is fine but I am unsure if what I have done next is correct:
$$ln(L)=lim_{ntoinfty}nlnleft(frac{n}{n+1}right)=lim_{ntoinfty}frac{lnleft(1-frac{1}{n+1}right)}{frac{1}{n}}$$
now I used the substitution $u=frac{1}{n+1}$ and with rearrangement I believe I can obtain:
$$ln(L)=lim_{uto0}frac{(1-u)ln(1-u)}{u}$$
now since when $uto0$ both the top and bottom also tend to $0$ I can use L'Hopitals rule:
$$ln(L)=lim_{uto0}frac{-1-ln(1-u)}{1}=lim_{uto0}-bigl(1+ln(1-u)bigr)=-1$$
which now gives:
$$ln(L)=-1therefore L=e^{-1}approx0.368<1$$
and so there is convergence
In the question it states that I will have to use:
$$lim_{ntoinfty}left(1+frac1nright)^n=e$$
convergence summation
asked Dec 5 at 0:49
Henry Lee
1,707218
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have the sum:
$$S=sum_{n=1}^inftyfrac{n!}{n^n}$$
and I am using D'Alembert's test for convergence which states for some sum:
$$sum_{n=a}^infty u_n,,(aneqpminfty)$$
that it is convergent if:
$$lim_{ntoinfty}left|frac{u_{n+1}}{u_n}right|<1$$
so to begin with I know that:
$$L=lim_{ntoinfty}frac{frac{(n+1)!}{(n+1)^{n+1}}}{frac{n!}{n^n}}=lim_{ntoinfty}frac{n^n}{(n+1)^n}$$
Up to this point is fine but I am unsure if what I have done next is correct:
$$ln(L)=lim_{ntoinfty}nlnleft(frac{n}{n+1}right)=lim_{ntoinfty}frac{lnleft(1-frac{1}{n+1}right)}{frac{1}{n}}$$
now I used the substitution $u=frac{1}{n+1}$ and with rearrangement I believe I can obtain:
$$ln(L)=lim_{uto0}frac{(1-u)ln(1-u)}{u}$$
now since when $uto0$ both the top and bottom also tend to $0$ I can use L'Hopitals rule:
$$ln(L)=lim_{uto0}frac{-1-ln(1-u)}{1}=lim_{uto0}-bigl(1+ln(1-u)bigr)=-1$$
which now gives:
$$ln(L)=-1therefore L=e^{-1}approx0.368<1$$
and so there is convergence
In the question it states that I will have to use:
$$lim_{ntoinfty}left(1+frac1nright)^n=e$$
convergence summation
asked Dec 5 at 0:49
Henry Lee
1,707218
I have the sum:
$$S=sum_{n=1}^inftyfrac{n!}{n^n}$$
and I am using D'Alembert's test for convergence which states for some sum:
$$sum_{n=a}^infty u_n,,(aneqpminfty)$$
that it is convergent if:
$$lim_{ntoinfty}left|frac{u_{n+1}}{u_n}right|<1$$
so to begin with I know that:
$$L=lim_{ntoinfty}frac{frac{(n+1)!}{(n+1)^{n+1}}}{frac{n!}{n^n}}=lim_{ntoinfty}frac{n^n}{(n+1)^n}$$
Up to this point is fine but I am unsure if what I have done next is correct:
$$ln(L)=lim_{ntoinfty}nlnleft(frac{n}{n+1}right)=lim_{ntoinfty}frac{lnleft(1-frac{1}{n+1}right)}{frac{1}{n}}$$
now I used the substitution $u=frac{1}{n+1}$ and with rearrangement I believe I can obtain:
$$ln(L)=lim_{uto0}frac{(1-u)ln(1-u)}{u}$$
now since when $uto0$ both the top and bottom also tend to $0$ I can use L'Hopitals rule:
$$ln(L)=lim_{uto0}frac{-1-ln(1-u)}{1}=lim_{uto0}-bigl(1+ln(1-u)bigr)=-1$$
which now gives:
$$ln(L)=-1therefore L=e^{-1}approx0.368<1$$
and so there is convergence
In the question it states that I will have to use:
$$lim_{ntoinfty}left(1+frac1nright)^n=e$$
convergence summation
convergence summation
asked Dec 5 at 0:49
Henry Lee
1,707218
asked Dec 5 at 0:49
Henry Lee
1,707218
asked Dec 5 at 0:49
Henry Lee
1,707218
asked Dec 5 at 0:49
Henry Lee
1,707218
asked Dec 5 at 0:49
Henry Lee
1,707218
1,707218
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add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Yes, that looks right.
Note that
$$lim_{ntoinfty}frac{n^n}{(n+1)^n}=lim_{ntoinfty}frac{1}{(1+frac{1}{n})^n}=frac{1}{e}$$
answered Dec 5 at 0:51
N. S.
101k5108203
add a comment |
up vote
1
down vote
Yes that's correct, more simply form here
$$frac{n^n}{(n+1)^n}=frac1{left(1+frac1nright)^n} to frac1e$$
answered Dec 5 at 0:52
gimusi
92.7k94495
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes, that looks right.
Note that
$$lim_{ntoinfty}frac{n^n}{(n+1)^n}=lim_{ntoinfty}frac{1}{(1+frac{1}{n})^n}=frac{1}{e}$$
answered Dec 5 at 0:51
N. S.
101k5108203
add a comment |
up vote
2
down vote
accepted
Yes, that looks right.
Note that
$$lim_{ntoinfty}frac{n^n}{(n+1)^n}=lim_{ntoinfty}frac{1}{(1+frac{1}{n})^n}=frac{1}{e}$$
answered Dec 5 at 0:51
N. S.
101k5108203
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes, that looks right.
Note that
$$lim_{ntoinfty}frac{n^n}{(n+1)^n}=lim_{ntoinfty}frac{1}{(1+frac{1}{n})^n}=frac{1}{e}$$
answered Dec 5 at 0:51
N. S.
101k5108203
Yes, that looks right.
Note that
$$lim_{ntoinfty}frac{n^n}{(n+1)^n}=lim_{ntoinfty}frac{1}{(1+frac{1}{n})^n}=frac{1}{e}$$
answered Dec 5 at 0:51
N. S.
101k5108203
answered Dec 5 at 0:51
N. S.
101k5108203
answered Dec 5 at 0:51
N. S.
101k5108203
answered Dec 5 at 0:51
N. S.
101k5108203
101k5108203
add a comment |
add a comment |
up vote
1
down vote
Yes that's correct, more simply form here
$$frac{n^n}{(n+1)^n}=frac1{left(1+frac1nright)^n} to frac1e$$
answered Dec 5 at 0:52
gimusi
92.7k94495
add a comment |
up vote
1
down vote
Yes that's correct, more simply form here
$$frac{n^n}{(n+1)^n}=frac1{left(1+frac1nright)^n} to frac1e$$
answered Dec 5 at 0:52
gimusi
92.7k94495
add a comment |
up vote
1
down vote
up vote
1
down vote
Yes that's correct, more simply form here
$$frac{n^n}{(n+1)^n}=frac1{left(1+frac1nright)^n} to frac1e$$
answered Dec 5 at 0:52
gimusi
92.7k94495
Yes that's correct, more simply form here
$$frac{n^n}{(n+1)^n}=frac1{left(1+frac1nright)^n} to frac1e$$
answered Dec 5 at 0:52
gimusi
92.7k94495
answered Dec 5 at 0:52
gimusi
92.7k94495
answered Dec 5 at 0:52
gimusi
92.7k94495
answered Dec 5 at 0:52
gimusi
92.7k94495
92.7k94495
add a comment |
add a comment |
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