which of the following series converges?












1














A. $sum_limits{n=1}^infty frac{1}{n}$



B. $sum_limits{n=0}^infty (1.5)^n$



C. $sum_limits{n=0}^infty frac{4n-1}{5n+1}$



D. $sum_limits{n=1}^infty (e)^{-2n}$



So my thinking so far is that everything diverges except D, but I don't know how to specifically show it. The reason I believe so is because as n increases $(1/e^{-2})^n$ will get extremely close to zero meaning that soon you will basically be adding 0 so it will converge. However I don't know if that is correct, or if it is the correct way to think about it. Is there a convergence test that will help me in this case?






sequences-and-series







share|cite|improve this question
























  • The argument is not sufficient because the behavior is the same in A, which diverges.
    – Yves Daoust
    Nov 23 '16 at 21:13






  • 1




    Do you know any convergence tests (e.g. integral test, comparison test, ratio test, root test, etc.)?
    – Dave
    Nov 23 '16 at 21:14










  • I just looked up the ration test, and i feel like it should help me show that D converges
    – L. Johnson
    Nov 23 '16 at 21:17
















1














A. $sum_limits{n=1}^infty frac{1}{n}$



B. $sum_limits{n=0}^infty (1.5)^n$



C. $sum_limits{n=0}^infty frac{4n-1}{5n+1}$



D. $sum_limits{n=1}^infty (e)^{-2n}$



So my thinking so far is that everything diverges except D, but I don't know how to specifically show it. The reason I believe so is because as n increases $(1/e^{-2})^n$ will get extremely close to zero meaning that soon you will basically be adding 0 so it will converge. However I don't know if that is correct, or if it is the correct way to think about it. Is there a convergence test that will help me in this case?






sequences-and-series







share|cite|improve this question
























  • The argument is not sufficient because the behavior is the same in A, which diverges.
    – Yves Daoust
    Nov 23 '16 at 21:13






  • 1




    Do you know any convergence tests (e.g. integral test, comparison test, ratio test, root test, etc.)?
    – Dave
    Nov 23 '16 at 21:14










  • I just looked up the ration test, and i feel like it should help me show that D converges
    – L. Johnson
    Nov 23 '16 at 21:17














1












1








1


1





A. $sum_limits{n=1}^infty frac{1}{n}$



B. $sum_limits{n=0}^infty (1.5)^n$



C. $sum_limits{n=0}^infty frac{4n-1}{5n+1}$



D. $sum_limits{n=1}^infty (e)^{-2n}$



So my thinking so far is that everything diverges except D, but I don't know how to specifically show it. The reason I believe so is because as n increases $(1/e^{-2})^n$ will get extremely close to zero meaning that soon you will basically be adding 0 so it will converge. However I don't know if that is correct, or if it is the correct way to think about it. Is there a convergence test that will help me in this case?






sequences-and-series







share|cite|improve this question















A. $sum_limits{n=1}^infty frac{1}{n}$



B. $sum_limits{n=0}^infty (1.5)^n$



C. $sum_limits{n=0}^infty frac{4n-1}{5n+1}$



D. $sum_limits{n=1}^infty (e)^{-2n}$



So my thinking so far is that everything diverges except D, but I don't know how to specifically show it. The reason I believe so is because as n increases $(1/e^{-2})^n$ will get extremely close to zero meaning that soon you will basically be adding 0 so it will converge. However I don't know if that is correct, or if it is the correct way to think about it. Is there a convergence test that will help me in this case?






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 '16 at 21:43









EnlightenedFunky

7211822




7211822










asked Nov 23 '16 at 21:10









L. Johnson

924




924












  • The argument is not sufficient because the behavior is the same in A, which diverges.
    – Yves Daoust
    Nov 23 '16 at 21:13






  • 1




    Do you know any convergence tests (e.g. integral test, comparison test, ratio test, root test, etc.)?
    – Dave
    Nov 23 '16 at 21:14










  • I just looked up the ration test, and i feel like it should help me show that D converges
    – L. Johnson
    Nov 23 '16 at 21:17


















  • The argument is not sufficient because the behavior is the same in A, which diverges.
    – Yves Daoust
    Nov 23 '16 at 21:13






  • 1




    Do you know any convergence tests (e.g. integral test, comparison test, ratio test, root test, etc.)?
    – Dave
    Nov 23 '16 at 21:14










  • I just looked up the ration test, and i feel like it should help me show that D converges
    – L. Johnson
    Nov 23 '16 at 21:17
















The argument is not sufficient because the behavior is the same in A, which diverges.
– Yves Daoust
Nov 23 '16 at 21:13




The argument is not sufficient because the behavior is the same in A, which diverges.
– Yves Daoust
Nov 23 '16 at 21:13




1




1




Do you know any convergence tests (e.g. integral test, comparison test, ratio test, root test, etc.)?
– Dave
Nov 23 '16 at 21:14




Do you know any convergence tests (e.g. integral test, comparison test, ratio test, root test, etc.)?
– Dave
Nov 23 '16 at 21:14












I just looked up the ration test, and i feel like it should help me show that D converges
– L. Johnson
Nov 23 '16 at 21:17




I just looked up the ration test, and i feel like it should help me show that D converges
– L. Johnson
Nov 23 '16 at 21:17










2 Answers
2






active

oldest

votes


















1














Hint. One may recall that each series that converges has its general term tending to $0$, here
$$
lim_{n to infty}(1.5)^n=? qquad lim_{n to infty}frac{4n-1}{5n+1}=?
$$ One may recall that the geometric series $displaystyle sum_{n=0}^infty x^n$ is convergent if and only if $|x|<1$. Concerning $A$, one may recall the nature of the harmonic series.






share|cite|improve this answer





























    0















    • A diverges as harmonic


    • B divergent as geometric$(1.5>1)$


    • C the general term doesn't go to zero


    • D converges as geometric$(e^{-2}<1)$.







    share|cite|improve this answer



















    • 2




      Is this a "hint" or a complete solution?
      – JMoravitz
      Nov 23 '16 at 21:16










    • To me it looks like a solution. And compact enough. Now the OP needs to interpret and understand...
      – imranfat
      Nov 23 '16 at 21:29











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    2 Answers
    2






    active

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    2 Answers
    2






    active

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    active

    oldest

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    active

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    1














    Hint. One may recall that each series that converges has its general term tending to $0$, here
    $$
    lim_{n to infty}(1.5)^n=? qquad lim_{n to infty}frac{4n-1}{5n+1}=?
    $$ One may recall that the geometric series $displaystyle sum_{n=0}^infty x^n$ is convergent if and only if $|x|<1$. Concerning $A$, one may recall the nature of the harmonic series.






    share|cite|improve this answer


























      1














      Hint. One may recall that each series that converges has its general term tending to $0$, here
      $$
      lim_{n to infty}(1.5)^n=? qquad lim_{n to infty}frac{4n-1}{5n+1}=?
      $$ One may recall that the geometric series $displaystyle sum_{n=0}^infty x^n$ is convergent if and only if $|x|<1$. Concerning $A$, one may recall the nature of the harmonic series.






      share|cite|improve this answer
























        1












        1








        1






        Hint. One may recall that each series that converges has its general term tending to $0$, here
        $$
        lim_{n to infty}(1.5)^n=? qquad lim_{n to infty}frac{4n-1}{5n+1}=?
        $$ One may recall that the geometric series $displaystyle sum_{n=0}^infty x^n$ is convergent if and only if $|x|<1$. Concerning $A$, one may recall the nature of the harmonic series.






        share|cite|improve this answer












        Hint. One may recall that each series that converges has its general term tending to $0$, here
        $$
        lim_{n to infty}(1.5)^n=? qquad lim_{n to infty}frac{4n-1}{5n+1}=?
        $$ One may recall that the geometric series $displaystyle sum_{n=0}^infty x^n$ is convergent if and only if $|x|<1$. Concerning $A$, one may recall the nature of the harmonic series.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 23 '16 at 21:14









        Olivier Oloa

        107k17175293




        107k17175293























            0















            • A diverges as harmonic


            • B divergent as geometric$(1.5>1)$


            • C the general term doesn't go to zero


            • D converges as geometric$(e^{-2}<1)$.







            share|cite|improve this answer



















            • 2




              Is this a "hint" or a complete solution?
              – JMoravitz
              Nov 23 '16 at 21:16










            • To me it looks like a solution. And compact enough. Now the OP needs to interpret and understand...
              – imranfat
              Nov 23 '16 at 21:29
















            0















            • A diverges as harmonic


            • B divergent as geometric$(1.5>1)$


            • C the general term doesn't go to zero


            • D converges as geometric$(e^{-2}<1)$.







            share|cite|improve this answer



















            • 2




              Is this a "hint" or a complete solution?
              – JMoravitz
              Nov 23 '16 at 21:16










            • To me it looks like a solution. And compact enough. Now the OP needs to interpret and understand...
              – imranfat
              Nov 23 '16 at 21:29














            0












            0








            0







            • A diverges as harmonic


            • B divergent as geometric$(1.5>1)$


            • C the general term doesn't go to zero


            • D converges as geometric$(e^{-2}<1)$.







            share|cite|improve this answer















            • A diverges as harmonic


            • B divergent as geometric$(1.5>1)$


            • C the general term doesn't go to zero


            • D converges as geometric$(e^{-2}<1)$.








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 23 '16 at 21:30

























            answered Nov 23 '16 at 21:14









            hamam_Abdallah

            37.8k21634




            37.8k21634








            • 2




              Is this a "hint" or a complete solution?
              – JMoravitz
              Nov 23 '16 at 21:16










            • To me it looks like a solution. And compact enough. Now the OP needs to interpret and understand...
              – imranfat
              Nov 23 '16 at 21:29














            • 2




              Is this a "hint" or a complete solution?
              – JMoravitz
              Nov 23 '16 at 21:16










            • To me it looks like a solution. And compact enough. Now the OP needs to interpret and understand...
              – imranfat
              Nov 23 '16 at 21:29








            2




            2




            Is this a "hint" or a complete solution?
            – JMoravitz
            Nov 23 '16 at 21:16




            Is this a "hint" or a complete solution?
            – JMoravitz
            Nov 23 '16 at 21:16












            To me it looks like a solution. And compact enough. Now the OP needs to interpret and understand...
            – imranfat
            Nov 23 '16 at 21:29




            To me it looks like a solution. And compact enough. Now the OP needs to interpret and understand...
            – imranfat
            Nov 23 '16 at 21:29


















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