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Convergenge or divergence of $ sum_n=1^infty e^-n^2$ [on hold]
Test for convergence/divergence of $sum_k=1^infty frack^2-1k^3+4$The Series( $sum_1^+ inftyfrac1n! + n$ convergence or divergence?Decide whether the series $sum_n=1^infty frac1+5^n1+6^n$ converges or divergesConvergence of the series $sum_n=1^inftycot^-1(n)$When does the series $sum_n=1^infty fraclnleft(1+frac1nright)n^a$ convergeWill this series with radical converge?Find the sum of the series $sum_n=1^inftyfrac12^n-1.$Does the series $sum_n=1^infty fracnsqrt[3]8n^5-1$ Converge?$sum_n=1^infty lnleft(fracnn+1right)$ convergence or divergenceDivergence of series with n factorial
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I can't find the convergence or divergence of the following series by using aspect ratio test or comparison test. The series is:
$$ sum_n=1^infty e^-n^2$$
Thanks.
real-analysis calculus sequences-and-series algebra-precalculus
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put on hold as off-topic by Martin R, Jam, metamorphy, mrtaurho, cmk Jun 28 at 0:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Jam, metamorphy, mrtaurho, cmk
add a comment |
$begingroup$
I can't find the convergence or divergence of the following series by using aspect ratio test or comparison test. The series is:
$$ sum_n=1^infty e^-n^2$$
Thanks.
real-analysis calculus sequences-and-series algebra-precalculus
$endgroup$
put on hold as off-topic by Martin R, Jam, metamorphy, mrtaurho, cmk Jun 28 at 0:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Jam, metamorphy, mrtaurho, cmk
1
$begingroup$
Given the range of answers below, can you show your work for your attempts?
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– Eric Towers
Jun 25 at 1:17
add a comment |
$begingroup$
I can't find the convergence or divergence of the following series by using aspect ratio test or comparison test. The series is:
$$ sum_n=1^infty e^-n^2$$
Thanks.
real-analysis calculus sequences-and-series algebra-precalculus
$endgroup$
I can't find the convergence or divergence of the following series by using aspect ratio test or comparison test. The series is:
$$ sum_n=1^infty e^-n^2$$
Thanks.
real-analysis calculus sequences-and-series algebra-precalculus
real-analysis calculus sequences-and-series algebra-precalculus
edited Jun 25 at 6:19
Asaf Karagila♦
313k34 gold badges448 silver badges782 bronze badges
313k34 gold badges448 silver badges782 bronze badges
asked Jun 24 at 16:33
Un Chico MásUn Chico Más
91 bronze badge
91 bronze badge
put on hold as off-topic by Martin R, Jam, metamorphy, mrtaurho, cmk Jun 28 at 0:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Jam, metamorphy, mrtaurho, cmk
put on hold as off-topic by Martin R, Jam, metamorphy, mrtaurho, cmk Jun 28 at 0:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Jam, metamorphy, mrtaurho, cmk
1
$begingroup$
Given the range of answers below, can you show your work for your attempts?
$endgroup$
– Eric Towers
Jun 25 at 1:17
add a comment |
1
$begingroup$
Given the range of answers below, can you show your work for your attempts?
$endgroup$
– Eric Towers
Jun 25 at 1:17
1
1
$begingroup$
Given the range of answers below, can you show your work for your attempts?
$endgroup$
– Eric Towers
Jun 25 at 1:17
$begingroup$
Given the range of answers below, can you show your work for your attempts?
$endgroup$
– Eric Towers
Jun 25 at 1:17
add a comment |
6 Answers
6
active
oldest
votes
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Since $e^-n^2leq e^-n$, the series converges by comparison with the geometric series $sum_n=1^infty e^-n$
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add a comment |
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Since$$(forall ninmathbb N):sqrt[n]e^-n^2=e^-nto0,$$the series converges, by the root test.
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add a comment |
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Also by ratio test:
$$
left|fraca_n+1a_nright| = frac1e^2n+1 to 0 text as $n to infty$
$$
Also by comparison test:
$$
e^n^2 > n^2 implies frac1e^n^2 < frac1n^2
$$
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add a comment |
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The series converges rather violently.
Since
$n^2 ge n$,
$e^-n^2 le e^-n$,
and the sum of that converges
so your series also converges.
$endgroup$
add a comment |
$begingroup$
It converges also via the ratio test :
$frace^-(n+1)^2e^-n^2 = e^-2n - 1 rightarrow 0$, which is a limit of absolute value $< 1$.
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add a comment |
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Another way: Since $e^-n > 0$ for all $n$, $$sum_n=1^N e^-n^2 = sum_n=1^N prod_k=1^n e^-n < sum_n=1^N prod_k=1^n e^-1 = sum_n=1^N e^-n = frac1 - e^-Ne - 1.$$ Then as $N to infty$, $$sum_n=1^infty e^-n^2 < frac1e-1.$$
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add a comment |
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $e^-n^2leq e^-n$, the series converges by comparison with the geometric series $sum_n=1^infty e^-n$
$endgroup$
add a comment |
$begingroup$
Since $e^-n^2leq e^-n$, the series converges by comparison with the geometric series $sum_n=1^infty e^-n$
$endgroup$
add a comment |
$begingroup$
Since $e^-n^2leq e^-n$, the series converges by comparison with the geometric series $sum_n=1^infty e^-n$
$endgroup$
Since $e^-n^2leq e^-n$, the series converges by comparison with the geometric series $sum_n=1^infty e^-n$
answered Jun 24 at 16:38
saulspatzsaulspatz
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20.7k4 gold badges16 silver badges37 bronze badges
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$begingroup$
Since$$(forall ninmathbb N):sqrt[n]e^-n^2=e^-nto0,$$the series converges, by the root test.
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add a comment |
$begingroup$
Since$$(forall ninmathbb N):sqrt[n]e^-n^2=e^-nto0,$$the series converges, by the root test.
$endgroup$
add a comment |
$begingroup$
Since$$(forall ninmathbb N):sqrt[n]e^-n^2=e^-nto0,$$the series converges, by the root test.
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Since$$(forall ninmathbb N):sqrt[n]e^-n^2=e^-nto0,$$the series converges, by the root test.
edited Jun 24 at 16:38
answered Jun 24 at 16:35
José Carlos SantosJosé Carlos Santos
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196k24 gold badges154 silver badges272 bronze badges
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add a comment |
$begingroup$
Also by ratio test:
$$
left|fraca_n+1a_nright| = frac1e^2n+1 to 0 text as $n to infty$
$$
Also by comparison test:
$$
e^n^2 > n^2 implies frac1e^n^2 < frac1n^2
$$
$endgroup$
add a comment |
$begingroup$
Also by ratio test:
$$
left|fraca_n+1a_nright| = frac1e^2n+1 to 0 text as $n to infty$
$$
Also by comparison test:
$$
e^n^2 > n^2 implies frac1e^n^2 < frac1n^2
$$
$endgroup$
add a comment |
$begingroup$
Also by ratio test:
$$
left|fraca_n+1a_nright| = frac1e^2n+1 to 0 text as $n to infty$
$$
Also by comparison test:
$$
e^n^2 > n^2 implies frac1e^n^2 < frac1n^2
$$
$endgroup$
Also by ratio test:
$$
left|fraca_n+1a_nright| = frac1e^2n+1 to 0 text as $n to infty$
$$
Also by comparison test:
$$
e^n^2 > n^2 implies frac1e^n^2 < frac1n^2
$$
answered Jun 24 at 16:43
Ruben du BurckRuben du Burck
9332 silver badges14 bronze badges
9332 silver badges14 bronze badges
add a comment |
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$begingroup$
The series converges rather violently.
Since
$n^2 ge n$,
$e^-n^2 le e^-n$,
and the sum of that converges
so your series also converges.
$endgroup$
add a comment |
$begingroup$
The series converges rather violently.
Since
$n^2 ge n$,
$e^-n^2 le e^-n$,
and the sum of that converges
so your series also converges.
$endgroup$
add a comment |
$begingroup$
The series converges rather violently.
Since
$n^2 ge n$,
$e^-n^2 le e^-n$,
and the sum of that converges
so your series also converges.
$endgroup$
The series converges rather violently.
Since
$n^2 ge n$,
$e^-n^2 le e^-n$,
and the sum of that converges
so your series also converges.
answered Jun 24 at 16:40
marty cohenmarty cohen
78k5 gold badges49 silver badges134 bronze badges
78k5 gold badges49 silver badges134 bronze badges
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$begingroup$
It converges also via the ratio test :
$frace^-(n+1)^2e^-n^2 = e^-2n - 1 rightarrow 0$, which is a limit of absolute value $< 1$.
$endgroup$
add a comment |
$begingroup$
It converges also via the ratio test :
$frace^-(n+1)^2e^-n^2 = e^-2n - 1 rightarrow 0$, which is a limit of absolute value $< 1$.
$endgroup$
add a comment |
$begingroup$
It converges also via the ratio test :
$frace^-(n+1)^2e^-n^2 = e^-2n - 1 rightarrow 0$, which is a limit of absolute value $< 1$.
$endgroup$
It converges also via the ratio test :
$frace^-(n+1)^2e^-n^2 = e^-2n - 1 rightarrow 0$, which is a limit of absolute value $< 1$.
answered Jun 24 at 16:43
DLeMeurDLeMeur
4509 bronze badges
4509 bronze badges
add a comment |
add a comment |
$begingroup$
Another way: Since $e^-n > 0$ for all $n$, $$sum_n=1^N e^-n^2 = sum_n=1^N prod_k=1^n e^-n < sum_n=1^N prod_k=1^n e^-1 = sum_n=1^N e^-n = frac1 - e^-Ne - 1.$$ Then as $N to infty$, $$sum_n=1^infty e^-n^2 < frac1e-1.$$
$endgroup$
add a comment |
$begingroup$
Another way: Since $e^-n > 0$ for all $n$, $$sum_n=1^N e^-n^2 = sum_n=1^N prod_k=1^n e^-n < sum_n=1^N prod_k=1^n e^-1 = sum_n=1^N e^-n = frac1 - e^-Ne - 1.$$ Then as $N to infty$, $$sum_n=1^infty e^-n^2 < frac1e-1.$$
$endgroup$
add a comment |
$begingroup$
Another way: Since $e^-n > 0$ for all $n$, $$sum_n=1^N e^-n^2 = sum_n=1^N prod_k=1^n e^-n < sum_n=1^N prod_k=1^n e^-1 = sum_n=1^N e^-n = frac1 - e^-Ne - 1.$$ Then as $N to infty$, $$sum_n=1^infty e^-n^2 < frac1e-1.$$
$endgroup$
Another way: Since $e^-n > 0$ for all $n$, $$sum_n=1^N e^-n^2 = sum_n=1^N prod_k=1^n e^-n < sum_n=1^N prod_k=1^n e^-1 = sum_n=1^N e^-n = frac1 - e^-Ne - 1.$$ Then as $N to infty$, $$sum_n=1^infty e^-n^2 < frac1e-1.$$
answered Jun 25 at 6:50
heropupheropup
68k8 gold badges66 silver badges107 bronze badges
68k8 gold badges66 silver badges107 bronze badges
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CxYcoXPgyZD4E 8U2BEuVG,Wk,O8utd6WUEk4qKWXyIvz0AQUCWBTx
1
$begingroup$
Given the range of answers below, can you show your work for your attempts?
$endgroup$
– Eric Towers
Jun 25 at 1:17