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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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-1

1

$begingroup$

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

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edited Jun 25 at 6:19

Asaf Karagila♦

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asked Jun 24 at 16:33

Un Chico MásUn Chico Más

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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

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  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

1

$begingroup$

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

share|cite|improve this question

edited Jun 25 at 6:19

Asaf Karagila♦

313k3434 gold badges448448 silver badges782782 bronze badges

asked Jun 24 at 16:33

Un Chico MásUn Chico Más

911 bronze badge

$endgroup$

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

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  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 | 

$begingroup$

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

share|cite|improve this question

edited Jun 25 at 6:19

Asaf Karagila♦

313k3434 gold badges448448 silver badges782782 bronze badges

asked Jun 24 at 16:33

Un Chico MásUn Chico Más

911 bronze badge

$endgroup$

share|cite|improve this question

edited Jun 25 at 6:19

Asaf Karagila♦

313k3434 gold badges448448 silver badges782782 bronze badges

asked Jun 24 at 16:33

Un Chico MásUn Chico Más

911 bronze badge

share|cite|improve this question

edited Jun 25 at 6:19

Asaf Karagila♦

313k3434 gold badges448448 silver badges782782 bronze badges

asked Jun 24 at 16:33

Un Chico MásUn Chico Más

911 bronze badge

edited Jun 25 at 6:19

Asaf Karagila♦

313k3434 gold badges448448 silver badges782782 bronze badges

edited Jun 25 at 6:19

Asaf Karagila♦

313k3434 gold badges448448 silver badges782782 bronze badges

asked Jun 24 at 16:33

Un Chico MásUn Chico Más

911 bronze badge

asked Jun 24 at 16:33

Un Chico MásUn Chico Más

911 bronze badge

6 Answers
6

active

oldest

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9

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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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answered Jun 24 at 16:38

saulspatzsaulspatz

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add a comment | 

8

$begingroup$

Since$$(forall ninmathbb N):sqrt[n]e^-n^2=e^-nto0,$$the series converges, by the root test.

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edited Jun 24 at 16:38

answered Jun 24 at 16:35

José Carlos SantosJosé Carlos Santos

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add a comment | 

4

$begingroup$

Also by ratio test:
$$
left|fraca_n+1a_nright| = frac1e^2n+1 to 0 text as $n to infty$
$$

Also by comparison test:
$$
e^n^2 > n^2 implies frac1e^n^2 < frac1n^2
$$

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answered Jun 24 at 16:43

Ruben du BurckRuben du Burck

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add a comment | 

1

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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.

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answered Jun 24 at 16:40

marty cohenmarty cohen

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add a comment | 

1

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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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answered Jun 24 at 16:43

DLeMeurDLeMeur

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add a comment | 

0

$begingroup$

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 $N to infty$, $$sum_n=1^infty e^-n^2 < frac1e-1.$$

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answered Jun 25 at 6:50

heropupheropup

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add a comment | 

9

$begingroup$

Since $e^-n^2leq e^-n$, the series converges by comparison with the geometric series $sum_n=1^infty e^-n$

share|cite|improve this answer

answered Jun 24 at 16:38

saulspatzsaulspatz

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$endgroup$

add a comment | 

9

$begingroup$

Since $e^-n^2leq e^-n$, the series converges by comparison with the geometric series $sum_n=1^infty e^-n$

share|cite|improve this answer

answered Jun 24 at 16:38

saulspatzsaulspatz

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$endgroup$

add a comment | 

$begingroup$

Since $e^-n^2leq e^-n$, the series converges by comparison with the geometric series $sum_n=1^infty e^-n$

share|cite|improve this answer

answered Jun 24 at 16:38

saulspatzsaulspatz

20.7k44 gold badges1616 silver badges3737 bronze badges

$endgroup$

share|cite|improve this answer

answered Jun 24 at 16:38

saulspatzsaulspatz

20.7k44 gold badges1616 silver badges3737 bronze badges

answered Jun 24 at 16:38

saulspatzsaulspatz

20.7k44 gold badges1616 silver badges3737 bronze badges

answered Jun 24 at 16:38

saulspatzsaulspatz

20.7k44 gold badges1616 silver badges3737 bronze badges

8

$begingroup$

Since$$(forall ninmathbb N):sqrt[n]e^-n^2=e^-nto0,$$the series converges, by the root test.

share|cite|improve this answer

edited Jun 24 at 16:38

answered Jun 24 at 16:35

José Carlos SantosJosé Carlos Santos

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$endgroup$

add a comment | 

8

$begingroup$

Since$$(forall ninmathbb N):sqrt[n]e^-n^2=e^-nto0,$$the series converges, by the root test.

share|cite|improve this answer

edited Jun 24 at 16:38

answered Jun 24 at 16:35

José Carlos SantosJosé Carlos Santos

196k2424 gold badges154154 silver badges272272 bronze badges

$endgroup$

add a comment | 

$begingroup$

Since$$(forall ninmathbb N):sqrt[n]e^-n^2=e^-nto0,$$the series converges, by the root test.

share|cite|improve this answer

edited Jun 24 at 16:38

answered Jun 24 at 16:35

José Carlos SantosJosé Carlos Santos

196k2424 gold badges154154 silver badges272272 bronze badges

$endgroup$

share|cite|improve this answer

edited Jun 24 at 16:38

answered Jun 24 at 16:35

José Carlos SantosJosé Carlos Santos

196k2424 gold badges154154 silver badges272272 bronze badges

answered Jun 24 at 16:35

José Carlos SantosJosé Carlos Santos

196k2424 gold badges154154 silver badges272272 bronze badges

answered Jun 24 at 16:35

José Carlos SantosJosé Carlos Santos

196k2424 gold badges154154 silver badges272272 bronze badges

4

$begingroup$

Also by ratio test:
$$
left|fraca_n+1a_nright| = frac1e^2n+1 to 0 text as $n to infty$
$$

Also by comparison test:
$$
e^n^2 > n^2 implies frac1e^n^2 < frac1n^2
$$

share|cite|improve this answer

answered Jun 24 at 16:43

Ruben du BurckRuben du Burck

93322 silver badges1414 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

Also by ratio test:
$$
left|fraca_n+1a_nright| = frac1e^2n+1 to 0 text as $n to infty$
$$

Also by comparison test:
$$
e^n^2 > n^2 implies frac1e^n^2 < frac1n^2
$$

share|cite|improve this answer

answered Jun 24 at 16:43

Ruben du BurckRuben du Burck

93322 silver badges1414 bronze badges

$endgroup$

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:
$$
e^n^2 > n^2 implies frac1e^n^2 < frac1n^2
$$

share|cite|improve this answer

answered Jun 24 at 16:43

Ruben du BurckRuben du Burck

93322 silver badges1414 bronze badges

$endgroup$

share|cite|improve this answer

answered Jun 24 at 16:43

Ruben du BurckRuben du Burck

93322 silver badges1414 bronze badges

answered Jun 24 at 16:43

Ruben du BurckRuben du Burck

93322 silver badges1414 bronze badges

answered Jun 24 at 16:43

Ruben du BurckRuben du Burck

93322 silver badges1414 bronze badges

1

$begingroup$

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.

share|cite|improve this answer

answered Jun 24 at 16:40

marty cohenmarty cohen

78k55 gold badges4949 silver badges134134 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered Jun 24 at 16:40

marty cohenmarty cohen

78k55 gold badges4949 silver badges134134 bronze badges

$endgroup$

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.

share|cite|improve this answer

answered Jun 24 at 16:40

marty cohenmarty cohen

78k55 gold badges4949 silver badges134134 bronze badges

$endgroup$

share|cite|improve this answer

answered Jun 24 at 16:40

marty cohenmarty cohen

78k55 gold badges4949 silver badges134134 bronze badges

answered Jun 24 at 16:40

marty cohenmarty cohen

78k55 gold badges4949 silver badges134134 bronze badges

answered Jun 24 at 16:40

marty cohenmarty cohen

78k55 gold badges4949 silver badges134134 bronze badges

1

$begingroup$

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 $< 1$.

share|cite|improve this answer

answered Jun 24 at 16:43

DLeMeurDLeMeur

45099 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

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 $< 1$.

share|cite|improve this answer

answered Jun 24 at 16:43

DLeMeurDLeMeur

45099 bronze badges

$endgroup$

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 $< 1$.

share|cite|improve this answer

answered Jun 24 at 16:43

DLeMeurDLeMeur

45099 bronze badges

$endgroup$

share|cite|improve this answer

answered Jun 24 at 16:43

DLeMeurDLeMeur

45099 bronze badges

answered Jun 24 at 16:43

DLeMeurDLeMeur

45099 bronze badges

answered Jun 24 at 16:43

DLeMeurDLeMeur

45099 bronze badges

0

$begingroup$

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 $N to infty$, $$sum_n=1^infty e^-n^2 < frac1e-1.$$

share|cite|improve this answer

answered Jun 25 at 6:50

heropupheropup

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$endgroup$

add a comment | 

0

$begingroup$

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 $N to infty$, $$sum_n=1^infty e^-n^2 < frac1e-1.$$

share|cite|improve this answer

answered Jun 25 at 6:50

heropupheropup

68k88 gold badges6666 silver badges107107 bronze badges

$endgroup$

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 $N to infty$, $$sum_n=1^infty e^-n^2 < frac1e-1.$$

share|cite|improve this answer

answered Jun 25 at 6:50

heropupheropup

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$endgroup$

share|cite|improve this answer

answered Jun 25 at 6:50

heropupheropup

68k88 gold badges6666 silver badges107107 bronze badges

answered Jun 25 at 6:50

heropupheropup

68k88 gold badges6666 silver badges107107 bronze badges

answered Jun 25 at 6:50

heropupheropup

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