Does $sum_{n=1}^{infty} frac{3+(-1)^n}{n}$ converge or diverge?












6












$begingroup$


I'm having trouble figuring out if the following series converges or diverges.



$$sum_{n=1}^{infty} frac{3+(-1)^n}{n}$$



Here's my thinking:



$$frac{2}{n} leq frac{3+(-1)^n}{n}$$



Since $sum_{n=1}^{infty} frac{2}{n}$ diverges, then so does $sum_{n=1}^{infty} frac{3+(-1)^n}{n}$



Is that correct?






calculus sequences-and-series divergent-series







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  • 9




    $begingroup$
    Yes. correct. continue.
    $endgroup$
    – hamam_Abdallah
    Dec 10 '18 at 20:47
















6












$begingroup$


I'm having trouble figuring out if the following series converges or diverges.



$$sum_{n=1}^{infty} frac{3+(-1)^n}{n}$$



Here's my thinking:



$$frac{2}{n} leq frac{3+(-1)^n}{n}$$



Since $sum_{n=1}^{infty} frac{2}{n}$ diverges, then so does $sum_{n=1}^{infty} frac{3+(-1)^n}{n}$



Is that correct?






calculus sequences-and-series divergent-series







share|cite|improve this question











$endgroup$








  • 9




    $begingroup$
    Yes. correct. continue.
    $endgroup$
    – hamam_Abdallah
    Dec 10 '18 at 20:47














6












6








6





$begingroup$


I'm having trouble figuring out if the following series converges or diverges.



$$sum_{n=1}^{infty} frac{3+(-1)^n}{n}$$



Here's my thinking:



$$frac{2}{n} leq frac{3+(-1)^n}{n}$$



Since $sum_{n=1}^{infty} frac{2}{n}$ diverges, then so does $sum_{n=1}^{infty} frac{3+(-1)^n}{n}$



Is that correct?






calculus sequences-and-series divergent-series







share|cite|improve this question











$endgroup$




I'm having trouble figuring out if the following series converges or diverges.



$$sum_{n=1}^{infty} frac{3+(-1)^n}{n}$$



Here's my thinking:



$$frac{2}{n} leq frac{3+(-1)^n}{n}$$



Since $sum_{n=1}^{infty} frac{2}{n}$ diverges, then so does $sum_{n=1}^{infty} frac{3+(-1)^n}{n}$



Is that correct?






calculus sequences-and-series divergent-series




calculus sequences-and-series divergent-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 22 '18 at 8:37









choco_addicted

8,08261947




8,08261947










asked Dec 10 '18 at 20:46









James MitchellJames Mitchell

25627




25627








  • 9




    $begingroup$
    Yes. correct. continue.
    $endgroup$
    – hamam_Abdallah
    Dec 10 '18 at 20:47














  • 9




    $begingroup$
    Yes. correct. continue.
    $endgroup$
    – hamam_Abdallah
    Dec 10 '18 at 20:47








9




9




$begingroup$
Yes. correct. continue.
$endgroup$
– hamam_Abdallah
Dec 10 '18 at 20:47




$begingroup$
Yes. correct. continue.
$endgroup$
– hamam_Abdallah
Dec 10 '18 at 20:47










2 Answers
2






active

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0












$begingroup$

Note that
$$frac{2}{n}leqfrac{3+(-1)^n}{n},$$
so by the comparison criteria, your series diverges.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Yes your prove is perfectly fine, indeed note, as an alternative



    $$sum_{n=1}^{N} frac{3+(-1)^n}{n}=sum_{n=1}^{N} frac{3}{n}+sum_{n=1}^{N} frac{(-1)^n}{n}$$



    and taking the limit $Nto infty$ the first series on the RHS diverges whereas the second one converges (by Leibniz).






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      Please don't casually split up sums like this without justification. (This one is OK but you have to show why.)
      $endgroup$
      – Ethan Bolker
      Dec 10 '18 at 21:02










    • $begingroup$
      @EthanBolker Yes you are right! I fix, Thanks
      $endgroup$
      – gimusi
      Dec 10 '18 at 21:03











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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

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    0












    $begingroup$

    Note that
    $$frac{2}{n}leqfrac{3+(-1)^n}{n},$$
    so by the comparison criteria, your series diverges.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Note that
      $$frac{2}{n}leqfrac{3+(-1)^n}{n},$$
      so by the comparison criteria, your series diverges.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Note that
        $$frac{2}{n}leqfrac{3+(-1)^n}{n},$$
        so by the comparison criteria, your series diverges.






        share|cite|improve this answer









        $endgroup$



        Note that
        $$frac{2}{n}leqfrac{3+(-1)^n}{n},$$
        so by the comparison criteria, your series diverges.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 20:58









        José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda

        802110




        802110























            0












            $begingroup$

            Yes your prove is perfectly fine, indeed note, as an alternative



            $$sum_{n=1}^{N} frac{3+(-1)^n}{n}=sum_{n=1}^{N} frac{3}{n}+sum_{n=1}^{N} frac{(-1)^n}{n}$$



            and taking the limit $Nto infty$ the first series on the RHS diverges whereas the second one converges (by Leibniz).






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              Please don't casually split up sums like this without justification. (This one is OK but you have to show why.)
              $endgroup$
              – Ethan Bolker
              Dec 10 '18 at 21:02










            • $begingroup$
              @EthanBolker Yes you are right! I fix, Thanks
              $endgroup$
              – gimusi
              Dec 10 '18 at 21:03
















            0












            $begingroup$

            Yes your prove is perfectly fine, indeed note, as an alternative



            $$sum_{n=1}^{N} frac{3+(-1)^n}{n}=sum_{n=1}^{N} frac{3}{n}+sum_{n=1}^{N} frac{(-1)^n}{n}$$



            and taking the limit $Nto infty$ the first series on the RHS diverges whereas the second one converges (by Leibniz).






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              Please don't casually split up sums like this without justification. (This one is OK but you have to show why.)
              $endgroup$
              – Ethan Bolker
              Dec 10 '18 at 21:02










            • $begingroup$
              @EthanBolker Yes you are right! I fix, Thanks
              $endgroup$
              – gimusi
              Dec 10 '18 at 21:03














            0












            0








            0





            $begingroup$

            Yes your prove is perfectly fine, indeed note, as an alternative



            $$sum_{n=1}^{N} frac{3+(-1)^n}{n}=sum_{n=1}^{N} frac{3}{n}+sum_{n=1}^{N} frac{(-1)^n}{n}$$



            and taking the limit $Nto infty$ the first series on the RHS diverges whereas the second one converges (by Leibniz).






            share|cite|improve this answer











            $endgroup$



            Yes your prove is perfectly fine, indeed note, as an alternative



            $$sum_{n=1}^{N} frac{3+(-1)^n}{n}=sum_{n=1}^{N} frac{3}{n}+sum_{n=1}^{N} frac{(-1)^n}{n}$$



            and taking the limit $Nto infty$ the first series on the RHS diverges whereas the second one converges (by Leibniz).







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 10 '18 at 21:04

























            answered Dec 10 '18 at 20:57









            gimusigimusi

            92.8k84494




            92.8k84494








            • 2




              $begingroup$
              Please don't casually split up sums like this without justification. (This one is OK but you have to show why.)
              $endgroup$
              – Ethan Bolker
              Dec 10 '18 at 21:02










            • $begingroup$
              @EthanBolker Yes you are right! I fix, Thanks
              $endgroup$
              – gimusi
              Dec 10 '18 at 21:03














            • 2




              $begingroup$
              Please don't casually split up sums like this without justification. (This one is OK but you have to show why.)
              $endgroup$
              – Ethan Bolker
              Dec 10 '18 at 21:02










            • $begingroup$
              @EthanBolker Yes you are right! I fix, Thanks
              $endgroup$
              – gimusi
              Dec 10 '18 at 21:03








            2




            2




            $begingroup$
            Please don't casually split up sums like this without justification. (This one is OK but you have to show why.)
            $endgroup$
            – Ethan Bolker
            Dec 10 '18 at 21:02




            $begingroup$
            Please don't casually split up sums like this without justification. (This one is OK but you have to show why.)
            $endgroup$
            – Ethan Bolker
            Dec 10 '18 at 21:02












            $begingroup$
            @EthanBolker Yes you are right! I fix, Thanks
            $endgroup$
            – gimusi
            Dec 10 '18 at 21:03




            $begingroup$
            @EthanBolker Yes you are right! I fix, Thanks
            $endgroup$
            – gimusi
            Dec 10 '18 at 21:03


















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