Mixing
How to prove that the series $sum n^s e^{-n} (sge 0)$ converges?
$begingroup$
Prove that $ sum n^s cdot e^{-n} , s ge 0$ converge.
My attempt using ratio test:
$$ frac{A_{n+1}}{A_n} = frac{(n+1)^s}{e^{n+1}} cdot frac{e^n}{n^s} = frac{(n+1)^s}{ecdot n^s} $$
But how can I proceed from here?
calculus sequences-and-series limits
asked Feb 2 at 17:11
bm1125bm1125
69116
$endgroup$
add a comment |
$begingroup$
Prove that $ sum n^s cdot e^{-n} , s ge 0$ converge.
My attempt using ratio test:
$$ frac{A_{n+1}}{A_n} = frac{(n+1)^s}{e^{n+1}} cdot frac{e^n}{n^s} = frac{(n+1)^s}{ecdot n^s} $$
But how can I proceed from here?
calculus sequences-and-series limits
asked Feb 2 at 17:11
bm1125bm1125
69116
$endgroup$
2
$begingroup$
As a tip: you don't need $sgeq0$ for convergence. In fact, for any $sinBbb C$, the series will converge absolutely.
$endgroup$
– Clayton
Feb 2 at 17:27
add a comment |
$begingroup$
Prove that $ sum n^s cdot e^{-n} , s ge 0$ converge.
My attempt using ratio test:
$$ frac{A_{n+1}}{A_n} = frac{(n+1)^s}{e^{n+1}} cdot frac{e^n}{n^s} = frac{(n+1)^s}{ecdot n^s} $$
But how can I proceed from here?
calculus sequences-and-series limits
asked Feb 2 at 17:11
bm1125bm1125
69116
$endgroup$
Prove that $ sum n^s cdot e^{-n} , s ge 0$ converge.
My attempt using ratio test:
$$ frac{A_{n+1}}{A_n} = frac{(n+1)^s}{e^{n+1}} cdot frac{e^n}{n^s} = frac{(n+1)^s}{ecdot n^s} $$
But how can I proceed from here?
calculus sequences-and-series limits
calculus sequences-and-series limits
asked Feb 2 at 17:11
bm1125bm1125
69116
asked Feb 2 at 17:11
bm1125bm1125
69116
edited Feb 2 at 17:33
user587192
asked Feb 2 at 17:11
bm1125bm1125
69116
asked Feb 2 at 17:11
bm1125bm1125
69116
asked Feb 2 at 17:11
bm1125bm1125
69116
69116
2
$begingroup$
As a tip: you don't need $sgeq0$ for convergence. In fact, for any $sinBbb C$, the series will converge absolutely.
$endgroup$
– Clayton
Feb 2 at 17:27
add a comment |
2
$begingroup$
As a tip: you don't need $sgeq0$ for convergence. In fact, for any $sinBbb C$, the series will converge absolutely.
$endgroup$
– Clayton
Feb 2 at 17:27
2
2
$begingroup$
As a tip: you don't need $sgeq0$ for convergence. In fact, for any $sinBbb C$, the series will converge absolutely.
$endgroup$
– Clayton
Feb 2 at 17:27
$begingroup$
As a tip: you don't need $sgeq0$ for convergence. In fact, for any $sinBbb C$, the series will converge absolutely.
$endgroup$
– Clayton
Feb 2 at 17:27
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Rewrite $$frac{(n+1)^s}{e cdot n^s}$$ as $$left(frac{n+1}{n}right)^s cdot frac{1}{e}$$
If you need the next step, divide the top and bottom of the fraction by n to get
$$left(1+frac{1}{n}right)^s cdot frac{1}{e}$$
answered Feb 2 at 17:14
KeatingeKeatinge
23228
$endgroup$
$begingroup$
Thanks but actually Im still not sure how it is converge. I understand that the expression $ frac{n+1}{n}^s $ approaches $ e $ when $ n rightarrow infty $ but that doesn't prove that the original series converge. ?
$endgroup$
– bm1125
Feb 2 at 17:22
2
$begingroup$
@bm1125 The whole thing approaches 1/e < 1, not e, so it converges absolutely. The (n+1)/n part converges to just 1, because (n+1)/n = 1 + 1/n
$endgroup$
– Keatinge
Feb 2 at 17:24
$begingroup$
Thanks for clarification!!
$endgroup$
– bm1125
Feb 2 at 17:30
add a comment |
$begingroup$
Hint The root test is more obvious here:
$$lim_n sqrt[n]{n^s cdot e^{-n}}= lim_n (sqrt[n]{n})^s e^{-1}= 1^s cdot e^{-1}=frac{1}{e}$$
answered Feb 2 at 17:40
N. S.N. S.
105k7115210
$endgroup$
add a comment |
$begingroup$
Since $n^{1/n} to 1$,
for fixed $s$,
$n^{s/n} to 1$
so
$n^{s/n} < 2$
for all large enough $n$.
Therefore
$begin{array}\
n^se^{-n}
&=(n^{s/n})^{n}e^{-n}\
&=(frac{n^{s/n}}{e})^n\
&<(frac{2}{e})^n
qquadtext{for all large enough } n\
end{array}
$
and the sum of these converges.
answered Feb 2 at 18:01
marty cohenmarty cohen
75.6k549130
$endgroup$
add a comment |
$begingroup$
We can also use the Cauchy's convergence test. This test tells us that a serie $sum_{k=1}^infty a_k$ is convergent if, and only if, for all $epsilon>0$ there is a $N_epsiloninmathbb{N}-{0}$ such that
$$
n> N_epsilon implies | a_{k+1}+a_{k+2}+ldots+a_{k+p}|<epsilon quad forall pinmathbb{N}
$$
Note that
begin{align}
left|
(n+1)^s e^{-(n+1)}+cdots+(n+p)^s e^{-(n+p)}
right|
=&
sum_{ell=1}^{p}(n+ell)^s e^{-(n+ell)}
\
leq&
sum_{ell=1}^{p}(n+p)^s e^{-(n+ell)}
\
leq&
sum_{ell=1}^{p}(n+p)^s e^{-n}
\
=&
p(n+p)^s e^{-n}
\
=&
frac{n^s}{e^n}
cdot
underbrace{
p(1+p/n)^s
}_{mathrm{limited}}
end{align}
Choose $ N_{p,s}inmathbb{N}$ big enough so that $n> N_{p,s}$ implies $(1+p/n)^s<2$. Then $n>N_{p,s}$ implies
begin{align}
left|
(n+1)^s e^{-(n+1)}+cdots+(n+p)^s e^{-(n+p)}
right|
leq &
frac{2pn^s}{e^n}
end{align}
For an arbitrary $ epsilon> 0 $ we can choose a large enough $ N_{p,s,epsilon} $ such that $n>N_{p,s,epsilon}$ implies $$frac{2pn^s}{e^n}<epsilon.$$ Choose $N_{epsilon}=max{N_{p,s,epsilon},N_{s,p}}$. Therefore, the convergence of the series follows .
answered Feb 2 at 18:40
MathOverviewMathOverview
8,95743164
$endgroup$
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
$begingroup$
Rewrite $$frac{(n+1)^s}{e cdot n^s}$$ as $$left(frac{n+1}{n}right)^s cdot frac{1}{e}$$
If you need the next step, divide the top and bottom of the fraction by n to get
$$left(1+frac{1}{n}right)^s cdot frac{1}{e}$$
answered Feb 2 at 17:14
KeatingeKeatinge
23228
$endgroup$
$begingroup$
Thanks but actually Im still not sure how it is converge. I understand that the expression $ frac{n+1}{n}^s $ approaches $ e $ when $ n rightarrow infty $ but that doesn't prove that the original series converge. ?
$endgroup$
– bm1125
Feb 2 at 17:22
2
$begingroup$
@bm1125 The whole thing approaches 1/e < 1, not e, so it converges absolutely. The (n+1)/n part converges to just 1, because (n+1)/n = 1 + 1/n
$endgroup$
– Keatinge
Feb 2 at 17:24
$begingroup$
Thanks for clarification!!
$endgroup$
– bm1125
Feb 2 at 17:30
add a comment |
$begingroup$
Rewrite $$frac{(n+1)^s}{e cdot n^s}$$ as $$left(frac{n+1}{n}right)^s cdot frac{1}{e}$$
If you need the next step, divide the top and bottom of the fraction by n to get
$$left(1+frac{1}{n}right)^s cdot frac{1}{e}$$
answered Feb 2 at 17:14
KeatingeKeatinge
23228
$endgroup$
$begingroup$
Thanks but actually Im still not sure how it is converge. I understand that the expression $ frac{n+1}{n}^s $ approaches $ e $ when $ n rightarrow infty $ but that doesn't prove that the original series converge. ?
$endgroup$
– bm1125
Feb 2 at 17:22
2
$begingroup$
@bm1125 The whole thing approaches 1/e < 1, not e, so it converges absolutely. The (n+1)/n part converges to just 1, because (n+1)/n = 1 + 1/n
$endgroup$
– Keatinge
Feb 2 at 17:24
$begingroup$
Thanks for clarification!!
$endgroup$
– bm1125
Feb 2 at 17:30
add a comment |
$begingroup$
Rewrite $$frac{(n+1)^s}{e cdot n^s}$$ as $$left(frac{n+1}{n}right)^s cdot frac{1}{e}$$
If you need the next step, divide the top and bottom of the fraction by n to get
$$left(1+frac{1}{n}right)^s cdot frac{1}{e}$$
answered Feb 2 at 17:14
KeatingeKeatinge
23228
$endgroup$
Rewrite $$frac{(n+1)^s}{e cdot n^s}$$ as $$left(frac{n+1}{n}right)^s cdot frac{1}{e}$$
If you need the next step, divide the top and bottom of the fraction by n to get
$$left(1+frac{1}{n}right)^s cdot frac{1}{e}$$
answered Feb 2 at 17:14
KeatingeKeatinge
23228
edited Feb 2 at 17:27
answered Feb 2 at 17:14
KeatingeKeatinge
23228
answered Feb 2 at 17:14
KeatingeKeatinge
23228
answered Feb 2 at 17:14
KeatingeKeatinge
23228
23228
$begingroup$
Thanks but actually Im still not sure how it is converge. I understand that the expression $ frac{n+1}{n}^s $ approaches $ e $ when $ n rightarrow infty $ but that doesn't prove that the original series converge. ?
$endgroup$
– bm1125
Feb 2 at 17:22
2
$begingroup$
@bm1125 The whole thing approaches 1/e < 1, not e, so it converges absolutely. The (n+1)/n part converges to just 1, because (n+1)/n = 1 + 1/n
$endgroup$
– Keatinge
Feb 2 at 17:24
$begingroup$
Thanks for clarification!!
$endgroup$
– bm1125
Feb 2 at 17:30
add a comment |
$begingroup$
Thanks but actually Im still not sure how it is converge. I understand that the expression $ frac{n+1}{n}^s $ approaches $ e $ when $ n rightarrow infty $ but that doesn't prove that the original series converge. ?
$endgroup$
– bm1125
Feb 2 at 17:22
2
$begingroup$
@bm1125 The whole thing approaches 1/e < 1, not e, so it converges absolutely. The (n+1)/n part converges to just 1, because (n+1)/n = 1 + 1/n
$endgroup$
– Keatinge
Feb 2 at 17:24
$begingroup$
Thanks for clarification!!
$endgroup$
– bm1125
Feb 2 at 17:30
$begingroup$
Thanks but actually Im still not sure how it is converge. I understand that the expression $ frac{n+1}{n}^s $ approaches $ e $ when $ n rightarrow infty $ but that doesn't prove that the original series converge. ?
$endgroup$
– bm1125
Feb 2 at 17:22
$begingroup$
Thanks but actually Im still not sure how it is converge. I understand that the expression $ frac{n+1}{n}^s $ approaches $ e $ when $ n rightarrow infty $ but that doesn't prove that the original series converge. ?
$endgroup$
– bm1125
Feb 2 at 17:22
2
2
$begingroup$
@bm1125 The whole thing approaches 1/e < 1, not e, so it converges absolutely. The (n+1)/n part converges to just 1, because (n+1)/n = 1 + 1/n
$endgroup$
– Keatinge
Feb 2 at 17:24
$begingroup$
@bm1125 The whole thing approaches 1/e < 1, not e, so it converges absolutely. The (n+1)/n part converges to just 1, because (n+1)/n = 1 + 1/n
$endgroup$
– Keatinge
Feb 2 at 17:24
$begingroup$
Thanks for clarification!!
$endgroup$
– bm1125
Feb 2 at 17:30
$begingroup$
Thanks for clarification!!
$endgroup$
– bm1125
Feb 2 at 17:30
add a comment |
$begingroup$
Hint The root test is more obvious here:
$$lim_n sqrt[n]{n^s cdot e^{-n}}= lim_n (sqrt[n]{n})^s e^{-1}= 1^s cdot e^{-1}=frac{1}{e}$$
answered Feb 2 at 17:40
N. S.N. S.
105k7115210
$endgroup$
add a comment |
$begingroup$
Hint The root test is more obvious here:
$$lim_n sqrt[n]{n^s cdot e^{-n}}= lim_n (sqrt[n]{n})^s e^{-1}= 1^s cdot e^{-1}=frac{1}{e}$$
answered Feb 2 at 17:40
N. S.N. S.
105k7115210
$endgroup$
add a comment |
$begingroup$
Hint The root test is more obvious here:
$$lim_n sqrt[n]{n^s cdot e^{-n}}= lim_n (sqrt[n]{n})^s e^{-1}= 1^s cdot e^{-1}=frac{1}{e}$$
answered Feb 2 at 17:40
N. S.N. S.
105k7115210
$endgroup$
Hint The root test is more obvious here:
$$lim_n sqrt[n]{n^s cdot e^{-n}}= lim_n (sqrt[n]{n})^s e^{-1}= 1^s cdot e^{-1}=frac{1}{e}$$
answered Feb 2 at 17:40
N. S.N. S.
105k7115210
answered Feb 2 at 17:40
N. S.N. S.
105k7115210
answered Feb 2 at 17:40
N. S.N. S.
105k7115210
answered Feb 2 at 17:40
N. S.N. S.
105k7115210
105k7115210
add a comment |
add a comment |
$begingroup$
Since $n^{1/n} to 1$,
for fixed $s$,
$n^{s/n} to 1$
so
$n^{s/n} < 2$
for all large enough $n$.
Therefore
$begin{array}\
n^se^{-n}
&=(n^{s/n})^{n}e^{-n}\
&=(frac{n^{s/n}}{e})^n\
&<(frac{2}{e})^n
qquadtext{for all large enough } n\
end{array}
$
and the sum of these converges.
answered Feb 2 at 18:01
marty cohenmarty cohen
75.6k549130
$endgroup$
add a comment |
$begingroup$
Since $n^{1/n} to 1$,
for fixed $s$,
$n^{s/n} to 1$
so
$n^{s/n} < 2$
for all large enough $n$.
Therefore
$begin{array}\
n^se^{-n}
&=(n^{s/n})^{n}e^{-n}\
&=(frac{n^{s/n}}{e})^n\
&<(frac{2}{e})^n
qquadtext{for all large enough } n\
end{array}
$
and the sum of these converges.
answered Feb 2 at 18:01
marty cohenmarty cohen
75.6k549130
$endgroup$
add a comment |
$begingroup$
Since $n^{1/n} to 1$,
for fixed $s$,
$n^{s/n} to 1$
so
$n^{s/n} < 2$
for all large enough $n$.
Therefore
$begin{array}\
n^se^{-n}
&=(n^{s/n})^{n}e^{-n}\
&=(frac{n^{s/n}}{e})^n\
&<(frac{2}{e})^n
qquadtext{for all large enough } n\
end{array}
$
and the sum of these converges.
answered Feb 2 at 18:01
marty cohenmarty cohen
75.6k549130
$endgroup$
Since $n^{1/n} to 1$,
for fixed $s$,
$n^{s/n} to 1$
so
$n^{s/n} < 2$
for all large enough $n$.
Therefore
$begin{array}\
n^se^{-n}
&=(n^{s/n})^{n}e^{-n}\
&=(frac{n^{s/n}}{e})^n\
&<(frac{2}{e})^n
qquadtext{for all large enough } n\
end{array}
$
and the sum of these converges.
answered Feb 2 at 18:01
marty cohenmarty cohen
75.6k549130
answered Feb 2 at 18:01
marty cohenmarty cohen
75.6k549130
answered Feb 2 at 18:01
marty cohenmarty cohen
75.6k549130
answered Feb 2 at 18:01
marty cohenmarty cohen
75.6k549130
75.6k549130
add a comment |
add a comment |
$begingroup$
We can also use the Cauchy's convergence test. This test tells us that a serie $sum_{k=1}^infty a_k$ is convergent if, and only if, for all $epsilon>0$ there is a $N_epsiloninmathbb{N}-{0}$ such that
$$
n> N_epsilon implies | a_{k+1}+a_{k+2}+ldots+a_{k+p}|<epsilon quad forall pinmathbb{N}
$$
Note that
begin{align}
left|
(n+1)^s e^{-(n+1)}+cdots+(n+p)^s e^{-(n+p)}
right|
=&
sum_{ell=1}^{p}(n+ell)^s e^{-(n+ell)}
\
leq&
sum_{ell=1}^{p}(n+p)^s e^{-(n+ell)}
\
leq&
sum_{ell=1}^{p}(n+p)^s e^{-n}
\
=&
p(n+p)^s e^{-n}
\
=&
frac{n^s}{e^n}
cdot
underbrace{
p(1+p/n)^s
}_{mathrm{limited}}
end{align}
Choose $ N_{p,s}inmathbb{N}$ big enough so that $n> N_{p,s}$ implies $(1+p/n)^s<2$. Then $n>N_{p,s}$ implies
begin{align}
left|
(n+1)^s e^{-(n+1)}+cdots+(n+p)^s e^{-(n+p)}
right|
leq &
frac{2pn^s}{e^n}
end{align}
For an arbitrary $ epsilon> 0 $ we can choose a large enough $ N_{p,s,epsilon} $ such that $n>N_{p,s,epsilon}$ implies $$frac{2pn^s}{e^n}<epsilon.$$ Choose $N_{epsilon}=max{N_{p,s,epsilon},N_{s,p}}$. Therefore, the convergence of the series follows .
answered Feb 2 at 18:40
MathOverviewMathOverview
8,95743164
$endgroup$
add a comment |
$begingroup$
We can also use the Cauchy's convergence test. This test tells us that a serie $sum_{k=1}^infty a_k$ is convergent if, and only if, for all $epsilon>0$ there is a $N_epsiloninmathbb{N}-{0}$ such that
$$
n> N_epsilon implies | a_{k+1}+a_{k+2}+ldots+a_{k+p}|<epsilon quad forall pinmathbb{N}
$$
Note that
begin{align}
left|
(n+1)^s e^{-(n+1)}+cdots+(n+p)^s e^{-(n+p)}
right|
=&
sum_{ell=1}^{p}(n+ell)^s e^{-(n+ell)}
\
leq&
sum_{ell=1}^{p}(n+p)^s e^{-(n+ell)}
\
leq&
sum_{ell=1}^{p}(n+p)^s e^{-n}
\
=&
p(n+p)^s e^{-n}
\
=&
frac{n^s}{e^n}
cdot
underbrace{
p(1+p/n)^s
}_{mathrm{limited}}
end{align}
Choose $ N_{p,s}inmathbb{N}$ big enough so that $n> N_{p,s}$ implies $(1+p/n)^s<2$. Then $n>N_{p,s}$ implies
begin{align}
left|
(n+1)^s e^{-(n+1)}+cdots+(n+p)^s e^{-(n+p)}
right|
leq &
frac{2pn^s}{e^n}
end{align}
For an arbitrary $ epsilon> 0 $ we can choose a large enough $ N_{p,s,epsilon} $ such that $n>N_{p,s,epsilon}$ implies $$frac{2pn^s}{e^n}<epsilon.$$ Choose $N_{epsilon}=max{N_{p,s,epsilon},N_{s,p}}$. Therefore, the convergence of the series follows .
answered Feb 2 at 18:40
MathOverviewMathOverview
8,95743164
$endgroup$
add a comment |
$begingroup$
We can also use the Cauchy's convergence test. This test tells us that a serie $sum_{k=1}^infty a_k$ is convergent if, and only if, for all $epsilon>0$ there is a $N_epsiloninmathbb{N}-{0}$ such that
$$
n> N_epsilon implies | a_{k+1}+a_{k+2}+ldots+a_{k+p}|<epsilon quad forall pinmathbb{N}
$$
Note that
begin{align}
left|
(n+1)^s e^{-(n+1)}+cdots+(n+p)^s e^{-(n+p)}
right|
=&
sum_{ell=1}^{p}(n+ell)^s e^{-(n+ell)}
\
leq&
sum_{ell=1}^{p}(n+p)^s e^{-(n+ell)}
\
leq&
sum_{ell=1}^{p}(n+p)^s e^{-n}
\
=&
p(n+p)^s e^{-n}
\
=&
frac{n^s}{e^n}
cdot
underbrace{
p(1+p/n)^s
}_{mathrm{limited}}
end{align}
Choose $ N_{p,s}inmathbb{N}$ big enough so that $n> N_{p,s}$ implies $(1+p/n)^s<2$. Then $n>N_{p,s}$ implies
begin{align}
left|
(n+1)^s e^{-(n+1)}+cdots+(n+p)^s e^{-(n+p)}
right|
leq &
frac{2pn^s}{e^n}
end{align}
For an arbitrary $ epsilon> 0 $ we can choose a large enough $ N_{p,s,epsilon} $ such that $n>N_{p,s,epsilon}$ implies $$frac{2pn^s}{e^n}<epsilon.$$ Choose $N_{epsilon}=max{N_{p,s,epsilon},N_{s,p}}$. Therefore, the convergence of the series follows .
answered Feb 2 at 18:40
MathOverviewMathOverview
8,95743164
$endgroup$
We can also use the Cauchy's convergence test. This test tells us that a serie $sum_{k=1}^infty a_k$ is convergent if, and only if, for all $epsilon>0$ there is a $N_epsiloninmathbb{N}-{0}$ such that
$$
n> N_epsilon implies | a_{k+1}+a_{k+2}+ldots+a_{k+p}|<epsilon quad forall pinmathbb{N}
$$
Note that
begin{align}
left|
(n+1)^s e^{-(n+1)}+cdots+(n+p)^s e^{-(n+p)}
right|
=&
sum_{ell=1}^{p}(n+ell)^s e^{-(n+ell)}
\
leq&
sum_{ell=1}^{p}(n+p)^s e^{-(n+ell)}
\
leq&
sum_{ell=1}^{p}(n+p)^s e^{-n}
\
=&
p(n+p)^s e^{-n}
\
=&
frac{n^s}{e^n}
cdot
underbrace{
p(1+p/n)^s
}_{mathrm{limited}}
end{align}
Choose $ N_{p,s}inmathbb{N}$ big enough so that $n> N_{p,s}$ implies $(1+p/n)^s<2$. Then $n>N_{p,s}$ implies
begin{align}
left|
(n+1)^s e^{-(n+1)}+cdots+(n+p)^s e^{-(n+p)}
right|
leq &
frac{2pn^s}{e^n}
end{align}
For an arbitrary $ epsilon> 0 $ we can choose a large enough $ N_{p,s,epsilon} $ such that $n>N_{p,s,epsilon}$ implies $$frac{2pn^s}{e^n}<epsilon.$$ Choose $N_{epsilon}=max{N_{p,s,epsilon},N_{s,p}}$. Therefore, the convergence of the series follows .
answered Feb 2 at 18:40
MathOverviewMathOverview
8,95743164
edited Feb 2 at 18:47
answered Feb 2 at 18:40
MathOverviewMathOverview
8,95743164
answered Feb 2 at 18:40
MathOverviewMathOverview
8,95743164
answered Feb 2 at 18:40
MathOverviewMathOverview
8,95743164
8,95743164
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$begingroup$
As a tip: you don't need $sgeq0$ for convergence. In fact, for any $sinBbb C$, the series will converge absolutely.
$endgroup$
– Clayton
Feb 2 at 17:27