Mixing
Convergence of series with trigonometric function and factorial
0
$begingroup$
I'm stuck on this.. Given $sum_{n=1}^{infty}frac{n!}{n^n}cdot sin(n^2)$ I have to determine if it's convergent or not.
I can see that $sin(n^2)$ is bounded and so are the partial sums of it. But $frac{n!}{n^n}$ doesnt converge to $0$ to use Dirichlet's test
Also my intuition is that it's not convergence but I can't find another divergent sequence to use the comparison test
Any hints?
sequences-and-series convergence
asked Jan 28 at 18:57
VakiPitsiVakiPitsi
1998
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add a comment |
0
$begingroup$
I'm stuck on this.. Given $sum_{n=1}^{infty}frac{n!}{n^n}cdot sin(n^2)$ I have to determine if it's convergent or not.
I can see that $sin(n^2)$ is bounded and so are the partial sums of it. But $frac{n!}{n^n}$ doesnt converge to $0$ to use Dirichlet's test
Also my intuition is that it's not convergence but I can't find another divergent sequence to use the comparison test
Any hints?
sequences-and-series convergence
asked Jan 28 at 18:57
VakiPitsiVakiPitsi
1998
$endgroup$
1
$begingroup$
$n!/n^n$ does tend to $0$, and hence your serie converges absolutely, and hence converges
$endgroup$
– Thinking
Jan 28 at 19:03
1
$begingroup$
By the way, partial sums of $sin n^2$ are not bounded.
$endgroup$
– RRL
Jan 28 at 19:07
add a comment |
0
0
0
$begingroup$
I'm stuck on this.. Given $sum_{n=1}^{infty}frac{n!}{n^n}cdot sin(n^2)$ I have to determine if it's convergent or not.
I can see that $sin(n^2)$ is bounded and so are the partial sums of it. But $frac{n!}{n^n}$ doesnt converge to $0$ to use Dirichlet's test
Also my intuition is that it's not convergence but I can't find another divergent sequence to use the comparison test
Any hints?
sequences-and-series convergence
asked Jan 28 at 18:57
VakiPitsiVakiPitsi
1998
$endgroup$
I'm stuck on this.. Given $sum_{n=1}^{infty}frac{n!}{n^n}cdot sin(n^2)$ I have to determine if it's convergent or not.
I can see that $sin(n^2)$ is bounded and so are the partial sums of it. But $frac{n!}{n^n}$ doesnt converge to $0$ to use Dirichlet's test
Also my intuition is that it's not convergence but I can't find another divergent sequence to use the comparison test
Any hints?
sequences-and-series convergence
sequences-and-series convergence
asked Jan 28 at 18:57
VakiPitsiVakiPitsi
1998
asked Jan 28 at 18:57
VakiPitsiVakiPitsi
1998
asked Jan 28 at 18:57
VakiPitsiVakiPitsi
1998
asked Jan 28 at 18:57
VakiPitsiVakiPitsi
1998
asked Jan 28 at 18:57
VakiPitsiVakiPitsi
1998
1998
1
$begingroup$
$n!/n^n$ does tend to $0$, and hence your serie converges absolutely, and hence converges
$endgroup$
– Thinking
Jan 28 at 19:03
1
$begingroup$
By the way, partial sums of $sin n^2$ are not bounded.
$endgroup$
– RRL
Jan 28 at 19:07
add a comment |
1
$begingroup$
$n!/n^n$ does tend to $0$, and hence your serie converges absolutely, and hence converges
$endgroup$
– Thinking
Jan 28 at 19:03
1
$begingroup$
By the way, partial sums of $sin n^2$ are not bounded.
$endgroup$
– RRL
Jan 28 at 19:07
1
1
$begingroup$
$n!/n^n$ does tend to $0$, and hence your serie converges absolutely, and hence converges
$endgroup$
– Thinking
Jan 28 at 19:03
$begingroup$
$n!/n^n$ does tend to $0$, and hence your serie converges absolutely, and hence converges
$endgroup$
– Thinking
Jan 28 at 19:03
1
1
$begingroup$
By the way, partial sums of $sin n^2$ are not bounded.
$endgroup$
– RRL
Jan 28 at 19:07
$begingroup$
By the way, partial sums of $sin n^2$ are not bounded.
$endgroup$
– RRL
Jan 28 at 19:07
add a comment |
2 Answers
2
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oldest
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1
$begingroup$
Hint
$$frac{n!}{n^n}=frac{1}{n}cdotfrac{2}{n}cdot...cdotfrac{n}{n}leq frac{1}{n}cdotfrac{2}{n}cdot 1 cdot 1 cdot... cdot 1=frac{2}{n^2}$$
answered Jan 28 at 19:06
N. S.N. S.
105k7114210
$endgroup$
$begingroup$
So if $frac{n!}{n^n} leq frac{2}{n^2}$ that means that $frac{n!}{n^n} leq frac{1}{n}$ and $frac{n!}{n^n} rightarrow 0$. So I can use Dirichlet's test? Is that right?
$endgroup$
– VakiPitsi
Jan 28 at 19:19
$begingroup$
No nevermind Dirichlet's test cause as RRL mention the partial sum is not bounded But we can say that since $left | sin(x^2) right | leq1Rightarrow left |frac{n!}{n^n}sin(x^2) right |leq frac{1}{n^2}$ and thus the series converges right?
$endgroup$
– VakiPitsi
Jan 28 at 19:25
1
$begingroup$
@VakiPitsi Exactly. You should maybe emphasize that you are proving absolute convergence, hence convergence.
$endgroup$
– N. S.
Jan 28 at 19:47
1
$begingroup$
Can the downvoter please explain what is wrong with the hint.
$endgroup$
– N. S.
Jan 28 at 19:48
$begingroup$
I wasn't the downvoter, but did you mean $frac 2 {n} . 1$ rather than $frac 1 {n} . 1$ ?
$endgroup$
– J. W. Tanner
Jan 28 at 20:00
|
show 1 more comment
0
$begingroup$
Hint: if $k$ is the integer part of $n/2$, then $$frac{n!}{n^n} leq frac{k!}{n^k} leq (k/n)^k leq 2^{-k}.$$
answered Jan 28 at 19:01
MindlackMindlack
4,910211
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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
1
$begingroup$
Hint
$$frac{n!}{n^n}=frac{1}{n}cdotfrac{2}{n}cdot...cdotfrac{n}{n}leq frac{1}{n}cdotfrac{2}{n}cdot 1 cdot 1 cdot... cdot 1=frac{2}{n^2}$$
answered Jan 28 at 19:06
N. S.N. S.
105k7114210
$endgroup$
$begingroup$
So if $frac{n!}{n^n} leq frac{2}{n^2}$ that means that $frac{n!}{n^n} leq frac{1}{n}$ and $frac{n!}{n^n} rightarrow 0$. So I can use Dirichlet's test? Is that right?
$endgroup$
– VakiPitsi
Jan 28 at 19:19
$begingroup$
No nevermind Dirichlet's test cause as RRL mention the partial sum is not bounded But we can say that since $left | sin(x^2) right | leq1Rightarrow left |frac{n!}{n^n}sin(x^2) right |leq frac{1}{n^2}$ and thus the series converges right?
$endgroup$
– VakiPitsi
Jan 28 at 19:25
1
$begingroup$
@VakiPitsi Exactly. You should maybe emphasize that you are proving absolute convergence, hence convergence.
$endgroup$
– N. S.
Jan 28 at 19:47
1
$begingroup$
Can the downvoter please explain what is wrong with the hint.
$endgroup$
– N. S.
Jan 28 at 19:48
$begingroup$
I wasn't the downvoter, but did you mean $frac 2 {n} . 1$ rather than $frac 1 {n} . 1$ ?
$endgroup$
– J. W. Tanner
Jan 28 at 20:00
|
show 1 more comment
1
$begingroup$
Hint
$$frac{n!}{n^n}=frac{1}{n}cdotfrac{2}{n}cdot...cdotfrac{n}{n}leq frac{1}{n}cdotfrac{2}{n}cdot 1 cdot 1 cdot... cdot 1=frac{2}{n^2}$$
answered Jan 28 at 19:06
N. S.N. S.
105k7114210
$endgroup$
$begingroup$
So if $frac{n!}{n^n} leq frac{2}{n^2}$ that means that $frac{n!}{n^n} leq frac{1}{n}$ and $frac{n!}{n^n} rightarrow 0$. So I can use Dirichlet's test? Is that right?
$endgroup$
– VakiPitsi
Jan 28 at 19:19
$begingroup$
No nevermind Dirichlet's test cause as RRL mention the partial sum is not bounded But we can say that since $left | sin(x^2) right | leq1Rightarrow left |frac{n!}{n^n}sin(x^2) right |leq frac{1}{n^2}$ and thus the series converges right?
$endgroup$
– VakiPitsi
Jan 28 at 19:25
1
$begingroup$
@VakiPitsi Exactly. You should maybe emphasize that you are proving absolute convergence, hence convergence.
$endgroup$
– N. S.
Jan 28 at 19:47
1
$begingroup$
Can the downvoter please explain what is wrong with the hint.
$endgroup$
– N. S.
Jan 28 at 19:48
$begingroup$
I wasn't the downvoter, but did you mean $frac 2 {n} . 1$ rather than $frac 1 {n} . 1$ ?
$endgroup$
– J. W. Tanner
Jan 28 at 20:00
|
show 1 more comment
1
1
1
$begingroup$
Hint
$$frac{n!}{n^n}=frac{1}{n}cdotfrac{2}{n}cdot...cdotfrac{n}{n}leq frac{1}{n}cdotfrac{2}{n}cdot 1 cdot 1 cdot... cdot 1=frac{2}{n^2}$$
answered Jan 28 at 19:06
N. S.N. S.
105k7114210
$endgroup$
Hint
$$frac{n!}{n^n}=frac{1}{n}cdotfrac{2}{n}cdot...cdotfrac{n}{n}leq frac{1}{n}cdotfrac{2}{n}cdot 1 cdot 1 cdot... cdot 1=frac{2}{n^2}$$
answered Jan 28 at 19:06
N. S.N. S.
105k7114210
edited Jan 28 at 21:07
answered Jan 28 at 19:06
N. S.N. S.
105k7114210
answered Jan 28 at 19:06
N. S.N. S.
105k7114210
answered Jan 28 at 19:06
N. S.N. S.
105k7114210
105k7114210
$begingroup$
So if $frac{n!}{n^n} leq frac{2}{n^2}$ that means that $frac{n!}{n^n} leq frac{1}{n}$ and $frac{n!}{n^n} rightarrow 0$. So I can use Dirichlet's test? Is that right?
$endgroup$
– VakiPitsi
Jan 28 at 19:19
$begingroup$
No nevermind Dirichlet's test cause as RRL mention the partial sum is not bounded But we can say that since $left | sin(x^2) right | leq1Rightarrow left |frac{n!}{n^n}sin(x^2) right |leq frac{1}{n^2}$ and thus the series converges right?
$endgroup$
– VakiPitsi
Jan 28 at 19:25
1
$begingroup$
@VakiPitsi Exactly. You should maybe emphasize that you are proving absolute convergence, hence convergence.
$endgroup$
– N. S.
Jan 28 at 19:47
1
$begingroup$
Can the downvoter please explain what is wrong with the hint.
$endgroup$
– N. S.
Jan 28 at 19:48
$begingroup$
I wasn't the downvoter, but did you mean $frac 2 {n} . 1$ rather than $frac 1 {n} . 1$ ?
$endgroup$
– J. W. Tanner
Jan 28 at 20:00
|
show 1 more comment
$begingroup$
So if $frac{n!}{n^n} leq frac{2}{n^2}$ that means that $frac{n!}{n^n} leq frac{1}{n}$ and $frac{n!}{n^n} rightarrow 0$. So I can use Dirichlet's test? Is that right?
$endgroup$
– VakiPitsi
Jan 28 at 19:19
$begingroup$
No nevermind Dirichlet's test cause as RRL mention the partial sum is not bounded But we can say that since $left | sin(x^2) right | leq1Rightarrow left |frac{n!}{n^n}sin(x^2) right |leq frac{1}{n^2}$ and thus the series converges right?
$endgroup$
– VakiPitsi
Jan 28 at 19:25
1
$begingroup$
@VakiPitsi Exactly. You should maybe emphasize that you are proving absolute convergence, hence convergence.
$endgroup$
– N. S.
Jan 28 at 19:47
1
$begingroup$
Can the downvoter please explain what is wrong with the hint.
$endgroup$
– N. S.
Jan 28 at 19:48
$begingroup$
I wasn't the downvoter, but did you mean $frac 2 {n} . 1$ rather than $frac 1 {n} . 1$ ?
$endgroup$
– J. W. Tanner
Jan 28 at 20:00
$begingroup$
So if $frac{n!}{n^n} leq frac{2}{n^2}$ that means that $frac{n!}{n^n} leq frac{1}{n}$ and $frac{n!}{n^n} rightarrow 0$. So I can use Dirichlet's test? Is that right?
$endgroup$
– VakiPitsi
Jan 28 at 19:19
$begingroup$
So if $frac{n!}{n^n} leq frac{2}{n^2}$ that means that $frac{n!}{n^n} leq frac{1}{n}$ and $frac{n!}{n^n} rightarrow 0$. So I can use Dirichlet's test? Is that right?
$endgroup$
– VakiPitsi
Jan 28 at 19:19
$begingroup$
No nevermind Dirichlet's test cause as RRL mention the partial sum is not bounded But we can say that since $left | sin(x^2) right | leq1Rightarrow left |frac{n!}{n^n}sin(x^2) right |leq frac{1}{n^2}$ and thus the series converges right?
$endgroup$
– VakiPitsi
Jan 28 at 19:25
$begingroup$
No nevermind Dirichlet's test cause as RRL mention the partial sum is not bounded But we can say that since $left | sin(x^2) right | leq1Rightarrow left |frac{n!}{n^n}sin(x^2) right |leq frac{1}{n^2}$ and thus the series converges right?
$endgroup$
– VakiPitsi
Jan 28 at 19:25
1
1
$begingroup$
@VakiPitsi Exactly. You should maybe emphasize that you are proving absolute convergence, hence convergence.
$endgroup$
– N. S.
Jan 28 at 19:47
$begingroup$
@VakiPitsi Exactly. You should maybe emphasize that you are proving absolute convergence, hence convergence.
$endgroup$
– N. S.
Jan 28 at 19:47
1
1
$begingroup$
Can the downvoter please explain what is wrong with the hint.
$endgroup$
– N. S.
Jan 28 at 19:48
$begingroup$
Can the downvoter please explain what is wrong with the hint.
$endgroup$
– N. S.
Jan 28 at 19:48
$begingroup$
I wasn't the downvoter, but did you mean $frac 2 {n} . 1$ rather than $frac 1 {n} . 1$ ?
$endgroup$
– J. W. Tanner
Jan 28 at 20:00
$begingroup$
I wasn't the downvoter, but did you mean $frac 2 {n} . 1$ rather than $frac 1 {n} . 1$ ?
$endgroup$
– J. W. Tanner
Jan 28 at 20:00
|
show 1 more comment
0
$begingroup$
Hint: if $k$ is the integer part of $n/2$, then $$frac{n!}{n^n} leq frac{k!}{n^k} leq (k/n)^k leq 2^{-k}.$$
answered Jan 28 at 19:01
MindlackMindlack
4,910211
$endgroup$
add a comment |
0
$begingroup$
Hint: if $k$ is the integer part of $n/2$, then $$frac{n!}{n^n} leq frac{k!}{n^k} leq (k/n)^k leq 2^{-k}.$$
answered Jan 28 at 19:01
MindlackMindlack
4,910211
$endgroup$
add a comment |
0
0
0
$begingroup$
Hint: if $k$ is the integer part of $n/2$, then $$frac{n!}{n^n} leq frac{k!}{n^k} leq (k/n)^k leq 2^{-k}.$$
answered Jan 28 at 19:01
MindlackMindlack
4,910211
$endgroup$
Hint: if $k$ is the integer part of $n/2$, then $$frac{n!}{n^n} leq frac{k!}{n^k} leq (k/n)^k leq 2^{-k}.$$
answered Jan 28 at 19:01
MindlackMindlack
4,910211
answered Jan 28 at 19:01
MindlackMindlack
4,910211
answered Jan 28 at 19:01
MindlackMindlack
4,910211
answered Jan 28 at 19:01
MindlackMindlack
4,910211
4,910211
add a comment |
add a comment |
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$begingroup$
$n!/n^n$ does tend to $0$, and hence your serie converges absolutely, and hence converges
$endgroup$
– Thinking
Jan 28 at 19:03
1
$begingroup$
By the way, partial sums of $sin n^2$ are not bounded.
$endgroup$
– RRL
Jan 28 at 19:07