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Convergenge or divergence of series with e


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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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  • $begingroup$
    Given the range of answers below, can you show your work for your attempts?
    $endgroup$
    – Eric Towers
    1 hour ago


















0












$begingroup$


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







New contributor



Un Chico Más is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$












  • $begingroup$
    Given the range of answers below, can you show your work for your attempts?
    $endgroup$
    – Eric Towers
    1 hour ago














0












0








0


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







New contributor



Un Chico Más is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




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






share|cite|improve this question







New contributor



Un Chico Más is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










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asked 10 hours ago









Un Chico MásUn Chico Más

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  • $begingroup$
    Given the range of answers below, can you show your work for your attempts?
    $endgroup$
    – Eric Towers
    1 hour ago


















  • $begingroup$
    Given the range of answers below, can you show your work for your attempts?
    $endgroup$
    – Eric Towers
    1 hour ago
















$begingroup$
Given the range of answers below, can you show your work for your attempts?
$endgroup$
– Eric Towers
1 hour ago




$begingroup$
Given the range of answers below, can you show your work for your attempts?
$endgroup$
– Eric Towers
1 hour ago










5 Answers
5






active

oldest

votes


















6












$begingroup$

Since$$(forall ninmathbb N):sqrt[n]{e^{-n^2}}=e^{-n}to0,$$the series converges, by the root test.






share|cite|improve this answer











$endgroup$





















    5












    $begingroup$

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






    share|cite|improve this answer









    $endgroup$





















      3












      $begingroup$

      Also by ratio test:
      $$
      left|frac{a_{n+1}}{a_n}right| = frac{1}{e^{2n+1}} to 0 text{ as $n to infty$}
      $$



      Also by comparison test:
      $$
      e^{n^2} > n^2 implies frac{1}{e^{n^2}} < frac{1}{n^2}
      $$






      share|cite|improve this answer








      New contributor



      Ruben du Burck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      $endgroup$





















        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 :
          $frac{e^{-(n+1)^2}}{e^{-n^2}} = e^{-2n - 1} rightarrow 0$, which is a limit of absolute value $< 1$.






          share|cite|improve this answer









          $endgroup$
















            Your Answer








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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6












            $begingroup$

            Since$$(forall ninmathbb N):sqrt[n]{e^{-n^2}}=e^{-n}to0,$$the series converges, by the root test.






            share|cite|improve this answer











            $endgroup$


















              6












              $begingroup$

              Since$$(forall ninmathbb N):sqrt[n]{e^{-n^2}}=e^{-n}to0,$$the series converges, by the root test.






              share|cite|improve this answer











              $endgroup$
















                6












                6








                6





                $begingroup$

                Since$$(forall ninmathbb N):sqrt[n]{e^{-n^2}}=e^{-n}to0,$$the series converges, by the root test.






                share|cite|improve this answer











                $endgroup$



                Since$$(forall ninmathbb N):sqrt[n]{e^{-n^2}}=e^{-n}to0,$$the series converges, by the root test.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 10 hours ago

























                answered 10 hours ago









                José Carlos SantosJosé Carlos Santos

                195k24 gold badges153 silver badges270 bronze badges




                195k24 gold badges153 silver badges270 bronze badges

























                    5












                    $begingroup$

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






                    share|cite|improve this answer









                    $endgroup$


















                      5












                      $begingroup$

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






                      share|cite|improve this answer









                      $endgroup$
















                        5












                        5








                        5





                        $begingroup$

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






                        share|cite|improve this answer









                        $endgroup$



                        Since $e^{-n^2}leq e^{-n}$, 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 10 hours ago









                        saulspatzsaulspatz

                        20.4k4 gold badges16 silver badges36 bronze badges




                        20.4k4 gold badges16 silver badges36 bronze badges























                            3












                            $begingroup$

                            Also by ratio test:
                            $$
                            left|frac{a_{n+1}}{a_n}right| = frac{1}{e^{2n+1}} to 0 text{ as $n to infty$}
                            $$



                            Also by comparison test:
                            $$
                            e^{n^2} > n^2 implies frac{1}{e^{n^2}} < frac{1}{n^2}
                            $$






                            share|cite|improve this answer








                            New contributor



                            Ruben du Burck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.





                            $endgroup$


















                              3












                              $begingroup$

                              Also by ratio test:
                              $$
                              left|frac{a_{n+1}}{a_n}right| = frac{1}{e^{2n+1}} to 0 text{ as $n to infty$}
                              $$



                              Also by comparison test:
                              $$
                              e^{n^2} > n^2 implies frac{1}{e^{n^2}} < frac{1}{n^2}
                              $$






                              share|cite|improve this answer








                              New contributor



                              Ruben du Burck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.





                              $endgroup$
















                                3












                                3








                                3





                                $begingroup$

                                Also by ratio test:
                                $$
                                left|frac{a_{n+1}}{a_n}right| = frac{1}{e^{2n+1}} to 0 text{ as $n to infty$}
                                $$



                                Also by comparison test:
                                $$
                                e^{n^2} > n^2 implies frac{1}{e^{n^2}} < frac{1}{n^2}
                                $$






                                share|cite|improve this answer








                                New contributor



                                Ruben du Burck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.





                                $endgroup$



                                Also by ratio test:
                                $$
                                left|frac{a_{n+1}}{a_n}right| = frac{1}{e^{2n+1}} to 0 text{ as $n to infty$}
                                $$



                                Also by comparison test:
                                $$
                                e^{n^2} > n^2 implies frac{1}{e^{n^2}} < frac{1}{n^2}
                                $$







                                share|cite|improve this answer








                                New contributor



                                Ruben du Burck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.








                                share|cite|improve this answer



                                share|cite|improve this answer






                                New contributor



                                Ruben du Burck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.








                                answered 9 hours ago









                                Ruben du BurckRuben du Burck

                                5962 silver badges11 bronze badges




                                5962 silver badges11 bronze badges




                                New contributor



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                                Check out our Code of Conduct.

























                                    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 9 hours ago









                                        marty cohenmarty cohen

                                        77.7k5 gold badges49 silver badges133 bronze badges




                                        77.7k5 gold badges49 silver badges133 bronze badges























                                            1












                                            $begingroup$

                                            It converges also via the ratio test :
                                            $frac{e^{-(n+1)^2}}{e^{-n^2}} = e^{-2n - 1} rightarrow 0$, which is a limit of absolute value $< 1$.






                                            share|cite|improve this answer









                                            $endgroup$


















                                              1












                                              $begingroup$

                                              It converges also via the ratio test :
                                              $frac{e^{-(n+1)^2}}{e^{-n^2}} = e^{-2n - 1} rightarrow 0$, 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 :
                                                $frac{e^{-(n+1)^2}}{e^{-n^2}} = e^{-2n - 1} rightarrow 0$, which is a limit of absolute value $< 1$.






                                                share|cite|improve this answer









                                                $endgroup$



                                                It converges also via the ratio test :
                                                $frac{e^{-(n+1)^2}}{e^{-n^2}} = e^{-2n - 1} rightarrow 0$, which is a limit of absolute value $< 1$.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered 9 hours ago









                                                DLeMeurDLeMeur

                                                4599 bronze badges




                                                4599 bronze badges






















                                                    Un Chico Más is a new contributor. Be nice, and check out our Code of Conduct.










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