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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












$begingroup$


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.










share|cite|improve this question











$endgroup$



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

















-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.










share|cite|improve this question











$endgroup$



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













-1












-1








-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.










share|cite|improve this question











$endgroup$




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






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edited Jun 25 at 6:19









Asaf Karagila

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313k34 gold badges448 silver badges782 bronze badges










asked Jun 24 at 16:33









Un Chico MásUn Chico Más

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







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












  • 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










6 Answers
6






active

oldest

votes


















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









$endgroup$




















    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











    $endgroup$




















      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









      $endgroup$




















        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









        $endgroup$




















          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









          $endgroup$




















            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









            $endgroup$





















              6 Answers
              6






              active

              oldest

              votes








              6 Answers
              6






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              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









              $endgroup$

















                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









                $endgroup$















                  9












                  9








                  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









                  $endgroup$



                  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












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jun 24 at 16:38









                  saulspatzsaulspatz

                  20.7k4 gold badges16 silver badges37 bronze badges




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                      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











                      $endgroup$

















                        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











                        $endgroup$















                          8












                          8








                          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











                          $endgroup$



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







                          share|cite|improve this answer














                          share|cite|improve this answer



                          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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                              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









                              $endgroup$

















                                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









                                $endgroup$















                                  4












                                  4








                                  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









                                  $endgroup$



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



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







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered Jun 24 at 16:43









                                  Ruben du BurckRuben du Burck

                                  9332 silver badges14 bronze badges




                                  9332 silver badges14 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









                                      $endgroup$

















                                        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









                                        $endgroup$















                                          1












                                          1








                                          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









                                          $endgroup$



                                          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.







                                          share|cite|improve this answer












                                          share|cite|improve this answer



                                          share|cite|improve this answer










                                          answered Jun 24 at 16:40









                                          marty cohenmarty cohen

                                          78k5 gold badges49 silver badges134 bronze badges




                                          78k5 gold badges49 silver badges134 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









                                              $endgroup$

















                                                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









                                                $endgroup$















                                                  1












                                                  1








                                                  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









                                                  $endgroup$



                                                  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 $< 1$.







                                                  share|cite|improve this answer












                                                  share|cite|improve this answer



                                                  share|cite|improve this answer










                                                  answered Jun 24 at 16:43









                                                  DLeMeurDLeMeur

                                                  4509 bronze badges




                                                  4509 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









                                                      $endgroup$

















                                                        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









                                                        $endgroup$















                                                          0












                                                          0








                                                          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









                                                          $endgroup$



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







                                                          share|cite|improve this answer












                                                          share|cite|improve this answer



                                                          share|cite|improve this answer










                                                          answered Jun 25 at 6:50









                                                          heropupheropup

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