Series with a logarithm












1












$begingroup$


I'm trying to solve this example (I have to tell if it's convergent or not)



$$sum_{n=2}^{infty}ln{left(frac{n^3+n+1}{n^3-n}right)}$$



I tried to do this by comparison:
$$sum_{n=2}^{infty}ln{left(frac{n^3+n+1}{n^3-n}right)} < frac{1}{n^2}$$
But I'm stuck at this point, I can't tell if it's true. Can you give me any tips? Thank you in advance.






sequences-and-series







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I'm trying to solve this example (I have to tell if it's convergent or not)



    $$sum_{n=2}^{infty}ln{left(frac{n^3+n+1}{n^3-n}right)}$$



    I tried to do this by comparison:
    $$sum_{n=2}^{infty}ln{left(frac{n^3+n+1}{n^3-n}right)} < frac{1}{n^2}$$
    But I'm stuck at this point, I can't tell if it's true. Can you give me any tips? Thank you in advance.






    sequences-and-series







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I'm trying to solve this example (I have to tell if it's convergent or not)



      $$sum_{n=2}^{infty}ln{left(frac{n^3+n+1}{n^3-n}right)}$$



      I tried to do this by comparison:
      $$sum_{n=2}^{infty}ln{left(frac{n^3+n+1}{n^3-n}right)} < frac{1}{n^2}$$
      But I'm stuck at this point, I can't tell if it's true. Can you give me any tips? Thank you in advance.






      sequences-and-series







      share|cite|improve this question











      $endgroup$




      I'm trying to solve this example (I have to tell if it's convergent or not)



      $$sum_{n=2}^{infty}ln{left(frac{n^3+n+1}{n^3-n}right)}$$



      I tried to do this by comparison:
      $$sum_{n=2}^{infty}ln{left(frac{n^3+n+1}{n^3-n}right)} < frac{1}{n^2}$$
      But I'm stuck at this point, I can't tell if it's true. Can you give me any tips? Thank you in advance.






      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 Dec 12 '18 at 21:54









      gimusi

      92.9k84594




      92.9k84594










      asked Dec 12 '18 at 21:46







      user609637





























          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          Yes it is true, indeed by $log(1+x)<x$ as $x>0$ we have that



          $$ln{left(frac{n^3+n+1}{n^3-n}right)}=ln{left(frac{n^3-n+2n+1}{n^3-n}right)}=lnleft(1+frac{2n+1}{n^3-n}right)<frac{2n+1}{n^3-n}sim frac{2}{n^2}$$



          then refer to limit comparison test.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much for your answer. Could you also tell me if Dirichlet's test can be used for proving divergence? Or it works only one way? For example if I multiply this logarithm by 5^n, can I use Dirichlet's test to prove the divergence?
            $endgroup$
            – user609637
            Dec 12 '18 at 22:29








          • 1




            $begingroup$
            @iforgotmypass Here we are using that both direct comparison test that is $$lnleft(1+frac{2n+1}{n^3-n}right)<frac{2n+1}{n^3-n} implies sum lnleft(1+frac{2n+1}{n^3-n}right)<sum frac{2n+1}{n^3-n}$$ and limit comparison test that is $$frac{frac{2n+1}{n^3-n}}{frac{2}{n^2}}to 1$$
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:35












          • $begingroup$
            I see, but it won't work when I will multiply it by (-5)^n, am I right? Then we have to prove that the partial sums aren't limited?
            $endgroup$
            – user609637
            Dec 12 '18 at 22:36








          • 1




            $begingroup$
            @iforgotmypass Yes of course we can also use that for a divergent case. What about for example $$sum_{n=2}^{infty}ln{left(frac{n^3+n^2+1}{n^3-n}right)}$$ Here we can use limit comaprison test.
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:37






          • 1




            $begingroup$
            @iforgotmypass For $$sum_{n=2}^{infty}(-1)^nln{left(frac{n^3+n+1}{n^3-n}right)}$$ we can use absolute convergence and for $$sum_{n=2}^{infty}(-5)^nln{left(frac{n^3+n+1}{n^3-n}right)}$$ just check that $$(-5)^nln{left(frac{n^3+n+1}{n^3-n}right)}not to 0$$
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:39












          Your Answer





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          1 Answer
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          1 Answer
          1






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          4












          $begingroup$

          Yes it is true, indeed by $log(1+x)<x$ as $x>0$ we have that



          $$ln{left(frac{n^3+n+1}{n^3-n}right)}=ln{left(frac{n^3-n+2n+1}{n^3-n}right)}=lnleft(1+frac{2n+1}{n^3-n}right)<frac{2n+1}{n^3-n}sim frac{2}{n^2}$$



          then refer to limit comparison test.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much for your answer. Could you also tell me if Dirichlet's test can be used for proving divergence? Or it works only one way? For example if I multiply this logarithm by 5^n, can I use Dirichlet's test to prove the divergence?
            $endgroup$
            – user609637
            Dec 12 '18 at 22:29








          • 1




            $begingroup$
            @iforgotmypass Here we are using that both direct comparison test that is $$lnleft(1+frac{2n+1}{n^3-n}right)<frac{2n+1}{n^3-n} implies sum lnleft(1+frac{2n+1}{n^3-n}right)<sum frac{2n+1}{n^3-n}$$ and limit comparison test that is $$frac{frac{2n+1}{n^3-n}}{frac{2}{n^2}}to 1$$
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:35












          • $begingroup$
            I see, but it won't work when I will multiply it by (-5)^n, am I right? Then we have to prove that the partial sums aren't limited?
            $endgroup$
            – user609637
            Dec 12 '18 at 22:36








          • 1




            $begingroup$
            @iforgotmypass Yes of course we can also use that for a divergent case. What about for example $$sum_{n=2}^{infty}ln{left(frac{n^3+n^2+1}{n^3-n}right)}$$ Here we can use limit comaprison test.
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:37






          • 1




            $begingroup$
            @iforgotmypass For $$sum_{n=2}^{infty}(-1)^nln{left(frac{n^3+n+1}{n^3-n}right)}$$ we can use absolute convergence and for $$sum_{n=2}^{infty}(-5)^nln{left(frac{n^3+n+1}{n^3-n}right)}$$ just check that $$(-5)^nln{left(frac{n^3+n+1}{n^3-n}right)}not to 0$$
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:39
















          4












          $begingroup$

          Yes it is true, indeed by $log(1+x)<x$ as $x>0$ we have that



          $$ln{left(frac{n^3+n+1}{n^3-n}right)}=ln{left(frac{n^3-n+2n+1}{n^3-n}right)}=lnleft(1+frac{2n+1}{n^3-n}right)<frac{2n+1}{n^3-n}sim frac{2}{n^2}$$



          then refer to limit comparison test.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much for your answer. Could you also tell me if Dirichlet's test can be used for proving divergence? Or it works only one way? For example if I multiply this logarithm by 5^n, can I use Dirichlet's test to prove the divergence?
            $endgroup$
            – user609637
            Dec 12 '18 at 22:29








          • 1




            $begingroup$
            @iforgotmypass Here we are using that both direct comparison test that is $$lnleft(1+frac{2n+1}{n^3-n}right)<frac{2n+1}{n^3-n} implies sum lnleft(1+frac{2n+1}{n^3-n}right)<sum frac{2n+1}{n^3-n}$$ and limit comparison test that is $$frac{frac{2n+1}{n^3-n}}{frac{2}{n^2}}to 1$$
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:35












          • $begingroup$
            I see, but it won't work when I will multiply it by (-5)^n, am I right? Then we have to prove that the partial sums aren't limited?
            $endgroup$
            – user609637
            Dec 12 '18 at 22:36








          • 1




            $begingroup$
            @iforgotmypass Yes of course we can also use that for a divergent case. What about for example $$sum_{n=2}^{infty}ln{left(frac{n^3+n^2+1}{n^3-n}right)}$$ Here we can use limit comaprison test.
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:37






          • 1




            $begingroup$
            @iforgotmypass For $$sum_{n=2}^{infty}(-1)^nln{left(frac{n^3+n+1}{n^3-n}right)}$$ we can use absolute convergence and for $$sum_{n=2}^{infty}(-5)^nln{left(frac{n^3+n+1}{n^3-n}right)}$$ just check that $$(-5)^nln{left(frac{n^3+n+1}{n^3-n}right)}not to 0$$
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:39














          4












          4








          4





          $begingroup$

          Yes it is true, indeed by $log(1+x)<x$ as $x>0$ we have that



          $$ln{left(frac{n^3+n+1}{n^3-n}right)}=ln{left(frac{n^3-n+2n+1}{n^3-n}right)}=lnleft(1+frac{2n+1}{n^3-n}right)<frac{2n+1}{n^3-n}sim frac{2}{n^2}$$



          then refer to limit comparison test.






          share|cite|improve this answer









          $endgroup$



          Yes it is true, indeed by $log(1+x)<x$ as $x>0$ we have that



          $$ln{left(frac{n^3+n+1}{n^3-n}right)}=ln{left(frac{n^3-n+2n+1}{n^3-n}right)}=lnleft(1+frac{2n+1}{n^3-n}right)<frac{2n+1}{n^3-n}sim frac{2}{n^2}$$



          then refer to limit comparison test.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 21:49









          gimusigimusi

          92.9k84594




          92.9k84594












          • $begingroup$
            Thank you very much for your answer. Could you also tell me if Dirichlet's test can be used for proving divergence? Or it works only one way? For example if I multiply this logarithm by 5^n, can I use Dirichlet's test to prove the divergence?
            $endgroup$
            – user609637
            Dec 12 '18 at 22:29








          • 1




            $begingroup$
            @iforgotmypass Here we are using that both direct comparison test that is $$lnleft(1+frac{2n+1}{n^3-n}right)<frac{2n+1}{n^3-n} implies sum lnleft(1+frac{2n+1}{n^3-n}right)<sum frac{2n+1}{n^3-n}$$ and limit comparison test that is $$frac{frac{2n+1}{n^3-n}}{frac{2}{n^2}}to 1$$
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:35












          • $begingroup$
            I see, but it won't work when I will multiply it by (-5)^n, am I right? Then we have to prove that the partial sums aren't limited?
            $endgroup$
            – user609637
            Dec 12 '18 at 22:36








          • 1




            $begingroup$
            @iforgotmypass Yes of course we can also use that for a divergent case. What about for example $$sum_{n=2}^{infty}ln{left(frac{n^3+n^2+1}{n^3-n}right)}$$ Here we can use limit comaprison test.
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:37






          • 1




            $begingroup$
            @iforgotmypass For $$sum_{n=2}^{infty}(-1)^nln{left(frac{n^3+n+1}{n^3-n}right)}$$ we can use absolute convergence and for $$sum_{n=2}^{infty}(-5)^nln{left(frac{n^3+n+1}{n^3-n}right)}$$ just check that $$(-5)^nln{left(frac{n^3+n+1}{n^3-n}right)}not to 0$$
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:39


















          • $begingroup$
            Thank you very much for your answer. Could you also tell me if Dirichlet's test can be used for proving divergence? Or it works only one way? For example if I multiply this logarithm by 5^n, can I use Dirichlet's test to prove the divergence?
            $endgroup$
            – user609637
            Dec 12 '18 at 22:29








          • 1




            $begingroup$
            @iforgotmypass Here we are using that both direct comparison test that is $$lnleft(1+frac{2n+1}{n^3-n}right)<frac{2n+1}{n^3-n} implies sum lnleft(1+frac{2n+1}{n^3-n}right)<sum frac{2n+1}{n^3-n}$$ and limit comparison test that is $$frac{frac{2n+1}{n^3-n}}{frac{2}{n^2}}to 1$$
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:35












          • $begingroup$
            I see, but it won't work when I will multiply it by (-5)^n, am I right? Then we have to prove that the partial sums aren't limited?
            $endgroup$
            – user609637
            Dec 12 '18 at 22:36








          • 1




            $begingroup$
            @iforgotmypass Yes of course we can also use that for a divergent case. What about for example $$sum_{n=2}^{infty}ln{left(frac{n^3+n^2+1}{n^3-n}right)}$$ Here we can use limit comaprison test.
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:37






          • 1




            $begingroup$
            @iforgotmypass For $$sum_{n=2}^{infty}(-1)^nln{left(frac{n^3+n+1}{n^3-n}right)}$$ we can use absolute convergence and for $$sum_{n=2}^{infty}(-5)^nln{left(frac{n^3+n+1}{n^3-n}right)}$$ just check that $$(-5)^nln{left(frac{n^3+n+1}{n^3-n}right)}not to 0$$
            $endgroup$
            – gimusi
            Dec 12 '18 at 22:39
















          $begingroup$
          Thank you very much for your answer. Could you also tell me if Dirichlet's test can be used for proving divergence? Or it works only one way? For example if I multiply this logarithm by 5^n, can I use Dirichlet's test to prove the divergence?
          $endgroup$
          – user609637
          Dec 12 '18 at 22:29






          $begingroup$
          Thank you very much for your answer. Could you also tell me if Dirichlet's test can be used for proving divergence? Or it works only one way? For example if I multiply this logarithm by 5^n, can I use Dirichlet's test to prove the divergence?
          $endgroup$
          – user609637
          Dec 12 '18 at 22:29






          1




          1




          $begingroup$
          @iforgotmypass Here we are using that both direct comparison test that is $$lnleft(1+frac{2n+1}{n^3-n}right)<frac{2n+1}{n^3-n} implies sum lnleft(1+frac{2n+1}{n^3-n}right)<sum frac{2n+1}{n^3-n}$$ and limit comparison test that is $$frac{frac{2n+1}{n^3-n}}{frac{2}{n^2}}to 1$$
          $endgroup$
          – gimusi
          Dec 12 '18 at 22:35






          $begingroup$
          @iforgotmypass Here we are using that both direct comparison test that is $$lnleft(1+frac{2n+1}{n^3-n}right)<frac{2n+1}{n^3-n} implies sum lnleft(1+frac{2n+1}{n^3-n}right)<sum frac{2n+1}{n^3-n}$$ and limit comparison test that is $$frac{frac{2n+1}{n^3-n}}{frac{2}{n^2}}to 1$$
          $endgroup$
          – gimusi
          Dec 12 '18 at 22:35














          $begingroup$
          I see, but it won't work when I will multiply it by (-5)^n, am I right? Then we have to prove that the partial sums aren't limited?
          $endgroup$
          – user609637
          Dec 12 '18 at 22:36






          $begingroup$
          I see, but it won't work when I will multiply it by (-5)^n, am I right? Then we have to prove that the partial sums aren't limited?
          $endgroup$
          – user609637
          Dec 12 '18 at 22:36






          1




          1




          $begingroup$
          @iforgotmypass Yes of course we can also use that for a divergent case. What about for example $$sum_{n=2}^{infty}ln{left(frac{n^3+n^2+1}{n^3-n}right)}$$ Here we can use limit comaprison test.
          $endgroup$
          – gimusi
          Dec 12 '18 at 22:37




          $begingroup$
          @iforgotmypass Yes of course we can also use that for a divergent case. What about for example $$sum_{n=2}^{infty}ln{left(frac{n^3+n^2+1}{n^3-n}right)}$$ Here we can use limit comaprison test.
          $endgroup$
          – gimusi
          Dec 12 '18 at 22:37




          1




          1




          $begingroup$
          @iforgotmypass For $$sum_{n=2}^{infty}(-1)^nln{left(frac{n^3+n+1}{n^3-n}right)}$$ we can use absolute convergence and for $$sum_{n=2}^{infty}(-5)^nln{left(frac{n^3+n+1}{n^3-n}right)}$$ just check that $$(-5)^nln{left(frac{n^3+n+1}{n^3-n}right)}not to 0$$
          $endgroup$
          – gimusi
          Dec 12 '18 at 22:39




          $begingroup$
          @iforgotmypass For $$sum_{n=2}^{infty}(-1)^nln{left(frac{n^3+n+1}{n^3-n}right)}$$ we can use absolute convergence and for $$sum_{n=2}^{infty}(-5)^nln{left(frac{n^3+n+1}{n^3-n}right)}$$ just check that $$(-5)^nln{left(frac{n^3+n+1}{n^3-n}right)}not to 0$$
          $endgroup$
          – gimusi
          Dec 12 '18 at 22:39


















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