How can I study the convergence of the following series $sum_{n = 1}^{infty} frac {n(-1)^n}{3n -1} $?












0












$begingroup$


How can I study the convergence of the following series $sum_{n = 1}^{infty} frac {n(-1)^n}{3n -1} $?



My trial:
I thought about the alternating series as I found that $a_{n}$ is decreasing but lim $a_{n}$ for me was equal to 1/3 not zero, so the test fails .... Do you have any other suggestions?






calculus sequences-and-series







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    0












    $begingroup$


    How can I study the convergence of the following series $sum_{n = 1}^{infty} frac {n(-1)^n}{3n -1} $?



    My trial:
    I thought about the alternating series as I found that $a_{n}$ is decreasing but lim $a_{n}$ for me was equal to 1/3 not zero, so the test fails .... Do you have any other suggestions?






    calculus sequences-and-series







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      How can I study the convergence of the following series $sum_{n = 1}^{infty} frac {n(-1)^n}{3n -1} $?



      My trial:
      I thought about the alternating series as I found that $a_{n}$ is decreasing but lim $a_{n}$ for me was equal to 1/3 not zero, so the test fails .... Do you have any other suggestions?






      calculus sequences-and-series







      share|cite|improve this question









      $endgroup$




      How can I study the convergence of the following series $sum_{n = 1}^{infty} frac {n(-1)^n}{3n -1} $?



      My trial:
      I thought about the alternating series as I found that $a_{n}$ is decreasing but lim $a_{n}$ for me was equal to 1/3 not zero, so the test fails .... Do you have any other suggestions?






      calculus sequences-and-series




      calculus sequences-and-series






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 24 at 10:42









      hopefullyhopefully

      239114




      239114






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          $a_n = dfrac{n(-1)^n}{3n-1};$



          $lim_{n rightarrow }a_n not =0$.



          Hence?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does not any absolutely convergent series is convergent?
            $endgroup$
            – hopefully
            Jan 24 at 10:52










          • $begingroup$
            hopefully. Every abs. convergent. series is convergent. Not every convergent series is abs. convergent: altern. harm. series: (sum (-1)^n/n converges , the harmonic series does not.
            $endgroup$
            – Peter Szilas
            Jan 24 at 10:56










          • $begingroup$
            so what is the value of the limit of $ a_n $ in our case?
            $endgroup$
            – hopefully
            Jan 24 at 11:36












          • $begingroup$
            hopefully.a_n is not convergent. Subsequence a_{2n} goes to +1/3.Subsequence a_{2n+1} goes to -1/3.So what can you say about the sum?
            $endgroup$
            – Peter Szilas
            Jan 24 at 12:06



















          2












          $begingroup$

          It's just that: $a_n$ = $n(-1)^n/(3n-1)$ does not go to zero, so the series could not converge.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does not any absolutely convergent series is convergent?
            $endgroup$
            – hopefully
            Jan 24 at 10:51






          • 1




            $begingroup$
            I don't understand your question. Any series that converges must have it's sequence going to zero, that's the very first fact about infinite series that you should know.
            $endgroup$
            – Jakub Andruszkiewicz
            Jan 24 at 10:55











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          $a_n = dfrac{n(-1)^n}{3n-1};$



          $lim_{n rightarrow }a_n not =0$.



          Hence?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does not any absolutely convergent series is convergent?
            $endgroup$
            – hopefully
            Jan 24 at 10:52










          • $begingroup$
            hopefully. Every abs. convergent. series is convergent. Not every convergent series is abs. convergent: altern. harm. series: (sum (-1)^n/n converges , the harmonic series does not.
            $endgroup$
            – Peter Szilas
            Jan 24 at 10:56










          • $begingroup$
            so what is the value of the limit of $ a_n $ in our case?
            $endgroup$
            – hopefully
            Jan 24 at 11:36












          • $begingroup$
            hopefully.a_n is not convergent. Subsequence a_{2n} goes to +1/3.Subsequence a_{2n+1} goes to -1/3.So what can you say about the sum?
            $endgroup$
            – Peter Szilas
            Jan 24 at 12:06
















          1












          $begingroup$

          $a_n = dfrac{n(-1)^n}{3n-1};$



          $lim_{n rightarrow }a_n not =0$.



          Hence?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does not any absolutely convergent series is convergent?
            $endgroup$
            – hopefully
            Jan 24 at 10:52










          • $begingroup$
            hopefully. Every abs. convergent. series is convergent. Not every convergent series is abs. convergent: altern. harm. series: (sum (-1)^n/n converges , the harmonic series does not.
            $endgroup$
            – Peter Szilas
            Jan 24 at 10:56










          • $begingroup$
            so what is the value of the limit of $ a_n $ in our case?
            $endgroup$
            – hopefully
            Jan 24 at 11:36












          • $begingroup$
            hopefully.a_n is not convergent. Subsequence a_{2n} goes to +1/3.Subsequence a_{2n+1} goes to -1/3.So what can you say about the sum?
            $endgroup$
            – Peter Szilas
            Jan 24 at 12:06














          1












          1








          1





          $begingroup$

          $a_n = dfrac{n(-1)^n}{3n-1};$



          $lim_{n rightarrow }a_n not =0$.



          Hence?






          share|cite|improve this answer









          $endgroup$



          $a_n = dfrac{n(-1)^n}{3n-1};$



          $lim_{n rightarrow }a_n not =0$.



          Hence?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 24 at 10:48









          Peter SzilasPeter Szilas

          11.4k2822




          11.4k2822












          • $begingroup$
            Does not any absolutely convergent series is convergent?
            $endgroup$
            – hopefully
            Jan 24 at 10:52










          • $begingroup$
            hopefully. Every abs. convergent. series is convergent. Not every convergent series is abs. convergent: altern. harm. series: (sum (-1)^n/n converges , the harmonic series does not.
            $endgroup$
            – Peter Szilas
            Jan 24 at 10:56










          • $begingroup$
            so what is the value of the limit of $ a_n $ in our case?
            $endgroup$
            – hopefully
            Jan 24 at 11:36












          • $begingroup$
            hopefully.a_n is not convergent. Subsequence a_{2n} goes to +1/3.Subsequence a_{2n+1} goes to -1/3.So what can you say about the sum?
            $endgroup$
            – Peter Szilas
            Jan 24 at 12:06


















          • $begingroup$
            Does not any absolutely convergent series is convergent?
            $endgroup$
            – hopefully
            Jan 24 at 10:52










          • $begingroup$
            hopefully. Every abs. convergent. series is convergent. Not every convergent series is abs. convergent: altern. harm. series: (sum (-1)^n/n converges , the harmonic series does not.
            $endgroup$
            – Peter Szilas
            Jan 24 at 10:56










          • $begingroup$
            so what is the value of the limit of $ a_n $ in our case?
            $endgroup$
            – hopefully
            Jan 24 at 11:36












          • $begingroup$
            hopefully.a_n is not convergent. Subsequence a_{2n} goes to +1/3.Subsequence a_{2n+1} goes to -1/3.So what can you say about the sum?
            $endgroup$
            – Peter Szilas
            Jan 24 at 12:06
















          $begingroup$
          Does not any absolutely convergent series is convergent?
          $endgroup$
          – hopefully
          Jan 24 at 10:52




          $begingroup$
          Does not any absolutely convergent series is convergent?
          $endgroup$
          – hopefully
          Jan 24 at 10:52












          $begingroup$
          hopefully. Every abs. convergent. series is convergent. Not every convergent series is abs. convergent: altern. harm. series: (sum (-1)^n/n converges , the harmonic series does not.
          $endgroup$
          – Peter Szilas
          Jan 24 at 10:56




          $begingroup$
          hopefully. Every abs. convergent. series is convergent. Not every convergent series is abs. convergent: altern. harm. series: (sum (-1)^n/n converges , the harmonic series does not.
          $endgroup$
          – Peter Szilas
          Jan 24 at 10:56












          $begingroup$
          so what is the value of the limit of $ a_n $ in our case?
          $endgroup$
          – hopefully
          Jan 24 at 11:36






          $begingroup$
          so what is the value of the limit of $ a_n $ in our case?
          $endgroup$
          – hopefully
          Jan 24 at 11:36














          $begingroup$
          hopefully.a_n is not convergent. Subsequence a_{2n} goes to +1/3.Subsequence a_{2n+1} goes to -1/3.So what can you say about the sum?
          $endgroup$
          – Peter Szilas
          Jan 24 at 12:06




          $begingroup$
          hopefully.a_n is not convergent. Subsequence a_{2n} goes to +1/3.Subsequence a_{2n+1} goes to -1/3.So what can you say about the sum?
          $endgroup$
          – Peter Szilas
          Jan 24 at 12:06











          2












          $begingroup$

          It's just that: $a_n$ = $n(-1)^n/(3n-1)$ does not go to zero, so the series could not converge.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does not any absolutely convergent series is convergent?
            $endgroup$
            – hopefully
            Jan 24 at 10:51






          • 1




            $begingroup$
            I don't understand your question. Any series that converges must have it's sequence going to zero, that's the very first fact about infinite series that you should know.
            $endgroup$
            – Jakub Andruszkiewicz
            Jan 24 at 10:55
















          2












          $begingroup$

          It's just that: $a_n$ = $n(-1)^n/(3n-1)$ does not go to zero, so the series could not converge.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does not any absolutely convergent series is convergent?
            $endgroup$
            – hopefully
            Jan 24 at 10:51






          • 1




            $begingroup$
            I don't understand your question. Any series that converges must have it's sequence going to zero, that's the very first fact about infinite series that you should know.
            $endgroup$
            – Jakub Andruszkiewicz
            Jan 24 at 10:55














          2












          2








          2





          $begingroup$

          It's just that: $a_n$ = $n(-1)^n/(3n-1)$ does not go to zero, so the series could not converge.






          share|cite|improve this answer









          $endgroup$



          It's just that: $a_n$ = $n(-1)^n/(3n-1)$ does not go to zero, so the series could not converge.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 24 at 10:49









          Jakub AndruszkiewiczJakub Andruszkiewicz

          2116




          2116












          • $begingroup$
            Does not any absolutely convergent series is convergent?
            $endgroup$
            – hopefully
            Jan 24 at 10:51






          • 1




            $begingroup$
            I don't understand your question. Any series that converges must have it's sequence going to zero, that's the very first fact about infinite series that you should know.
            $endgroup$
            – Jakub Andruszkiewicz
            Jan 24 at 10:55


















          • $begingroup$
            Does not any absolutely convergent series is convergent?
            $endgroup$
            – hopefully
            Jan 24 at 10:51






          • 1




            $begingroup$
            I don't understand your question. Any series that converges must have it's sequence going to zero, that's the very first fact about infinite series that you should know.
            $endgroup$
            – Jakub Andruszkiewicz
            Jan 24 at 10:55
















          $begingroup$
          Does not any absolutely convergent series is convergent?
          $endgroup$
          – hopefully
          Jan 24 at 10:51




          $begingroup$
          Does not any absolutely convergent series is convergent?
          $endgroup$
          – hopefully
          Jan 24 at 10:51




          1




          1




          $begingroup$
          I don't understand your question. Any series that converges must have it's sequence going to zero, that's the very first fact about infinite series that you should know.
          $endgroup$
          – Jakub Andruszkiewicz
          Jan 24 at 10:55




          $begingroup$
          I don't understand your question. Any series that converges must have it's sequence going to zero, that's the very first fact about infinite series that you should know.
          $endgroup$
          – Jakub Andruszkiewicz
          Jan 24 at 10:55


















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