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







share|cite|improve this question









$endgroup$

















    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











      share|cite|improve this question




      share|cite|improve this question










      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





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085723%2fhow-can-i-study-the-convergence-of-the-following-series-sum-n-1-infty%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          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


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085723%2fhow-can-i-study-the-convergence-of-the-following-series-sum-n-1-infty%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Mario Kart Wii

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
          Read more

          The Binding of Isaac: Rebirth/Afterbirth

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
          Read more

          What does “Dominus providebit” mean?

          Image
          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
          Read more