$sum a_n b_n$ when $sum a_n$ convergent and ${b_n}$ nonnegative












1












$begingroup$


Let $sum_{i=0}^infty a_n$ be a conditionally convergent series, and ${b_n}$ be a nonnegative and convergent sequence of real or complex numbers. Does $sum_{i=0}^infty a_n b_n$ converge?



Do we actually need convergence of ${b_n}$ for convergence of $sum_{i=0}^infty a_n b_n$ or is it sufficient that ${b_n}$ is nonnegative and bounded?






real-analysis sequences-and-series convergence







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $sum_{i=0}^infty a_n$ be a conditionally convergent series, and ${b_n}$ be a nonnegative and convergent sequence of real or complex numbers. Does $sum_{i=0}^infty a_n b_n$ converge?



    Do we actually need convergence of ${b_n}$ for convergence of $sum_{i=0}^infty a_n b_n$ or is it sufficient that ${b_n}$ is nonnegative and bounded?






    real-analysis sequences-and-series convergence







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $sum_{i=0}^infty a_n$ be a conditionally convergent series, and ${b_n}$ be a nonnegative and convergent sequence of real or complex numbers. Does $sum_{i=0}^infty a_n b_n$ converge?



      Do we actually need convergence of ${b_n}$ for convergence of $sum_{i=0}^infty a_n b_n$ or is it sufficient that ${b_n}$ is nonnegative and bounded?






      real-analysis sequences-and-series convergence







      share|cite|improve this question









      $endgroup$




      Let $sum_{i=0}^infty a_n$ be a conditionally convergent series, and ${b_n}$ be a nonnegative and convergent sequence of real or complex numbers. Does $sum_{i=0}^infty a_n b_n$ converge?



      Do we actually need convergence of ${b_n}$ for convergence of $sum_{i=0}^infty a_n b_n$ or is it sufficient that ${b_n}$ is nonnegative and bounded?






      real-analysis sequences-and-series convergence




      real-analysis sequences-and-series convergence






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 14 '18 at 16:40









      Solicitous WookieeSolicitous Wookiee

      61




      61






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          Consider $a_n = frac{(-1)^n}{sqrt{n}}$ and $b_n = 2018+frac{(-1)^n}{sqrt{n}}$. Then all the conditions are met, although we have



          $$sum_{n=1}^{infty} a_n b_n = sum_{n=1}^{infty} left( 2018 frac{(-1)^n}{sqrt{n}} + frac{1}{n}right), $$



          which diverges.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Look good, thanks!
            $endgroup$
            – Solicitous Wookiee
            Dec 14 '18 at 16:54



















          1












          $begingroup$

          Bounded and non-negative is not sufficient. Consider $a_n=frac{(-1)^n}n$ and $b_n=1+(-1)^n$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $b_n$ is not convergent
            $endgroup$
            – gimusi
            Dec 14 '18 at 16:52










          • $begingroup$
            That is the point...
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:53










          • $begingroup$
            @SmileyCraft The question assumes that $b_n$ is a convergent sequence.
            $endgroup$
            – BigbearZzz
            Dec 14 '18 at 16:56










          • $begingroup$
            The OP literally asks "is it sufficient that ${b_n}$ is nonnegative and bounded?" and my example answers this question.
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:59





















          0












          $begingroup$

          Assume



          $$a_n = frac{(-1)^n}{sqrt n}$$



          $$b_n =begin{cases}=0quad n,text{odd}\\=frac{1}{sqrt n}quad n,text{even} end{cases}$$



          and therefore



          $$sum_{n=1}^{2N} a_n b_n=sum_{n=1}^{N} frac1{2n} toinfty$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Your sequence $;b_n;$ isn't convergent...
            $endgroup$
            – DonAntonio
            Dec 14 '18 at 16:44










          • $begingroup$
            Opsss...thanks I fix
            $endgroup$
            – gimusi
            Dec 14 '18 at 16:44











          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%2f3039591%2fsum-a-n-b-n-when-sum-a-n-convergent-and-b-n-nonnegative%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Consider $a_n = frac{(-1)^n}{sqrt{n}}$ and $b_n = 2018+frac{(-1)^n}{sqrt{n}}$. Then all the conditions are met, although we have



          $$sum_{n=1}^{infty} a_n b_n = sum_{n=1}^{infty} left( 2018 frac{(-1)^n}{sqrt{n}} + frac{1}{n}right), $$



          which diverges.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Look good, thanks!
            $endgroup$
            – Solicitous Wookiee
            Dec 14 '18 at 16:54
















          4












          $begingroup$

          Consider $a_n = frac{(-1)^n}{sqrt{n}}$ and $b_n = 2018+frac{(-1)^n}{sqrt{n}}$. Then all the conditions are met, although we have



          $$sum_{n=1}^{infty} a_n b_n = sum_{n=1}^{infty} left( 2018 frac{(-1)^n}{sqrt{n}} + frac{1}{n}right), $$



          which diverges.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Look good, thanks!
            $endgroup$
            – Solicitous Wookiee
            Dec 14 '18 at 16:54














          4












          4








          4





          $begingroup$

          Consider $a_n = frac{(-1)^n}{sqrt{n}}$ and $b_n = 2018+frac{(-1)^n}{sqrt{n}}$. Then all the conditions are met, although we have



          $$sum_{n=1}^{infty} a_n b_n = sum_{n=1}^{infty} left( 2018 frac{(-1)^n}{sqrt{n}} + frac{1}{n}right), $$



          which diverges.






          share|cite|improve this answer









          $endgroup$



          Consider $a_n = frac{(-1)^n}{sqrt{n}}$ and $b_n = 2018+frac{(-1)^n}{sqrt{n}}$. Then all the conditions are met, although we have



          $$sum_{n=1}^{infty} a_n b_n = sum_{n=1}^{infty} left( 2018 frac{(-1)^n}{sqrt{n}} + frac{1}{n}right), $$



          which diverges.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 16:49









          Sangchul LeeSangchul Lee

          94k12167272




          94k12167272












          • $begingroup$
            Look good, thanks!
            $endgroup$
            – Solicitous Wookiee
            Dec 14 '18 at 16:54


















          • $begingroup$
            Look good, thanks!
            $endgroup$
            – Solicitous Wookiee
            Dec 14 '18 at 16:54
















          $begingroup$
          Look good, thanks!
          $endgroup$
          – Solicitous Wookiee
          Dec 14 '18 at 16:54




          $begingroup$
          Look good, thanks!
          $endgroup$
          – Solicitous Wookiee
          Dec 14 '18 at 16:54











          1












          $begingroup$

          Bounded and non-negative is not sufficient. Consider $a_n=frac{(-1)^n}n$ and $b_n=1+(-1)^n$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $b_n$ is not convergent
            $endgroup$
            – gimusi
            Dec 14 '18 at 16:52










          • $begingroup$
            That is the point...
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:53










          • $begingroup$
            @SmileyCraft The question assumes that $b_n$ is a convergent sequence.
            $endgroup$
            – BigbearZzz
            Dec 14 '18 at 16:56










          • $begingroup$
            The OP literally asks "is it sufficient that ${b_n}$ is nonnegative and bounded?" and my example answers this question.
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:59


















          1












          $begingroup$

          Bounded and non-negative is not sufficient. Consider $a_n=frac{(-1)^n}n$ and $b_n=1+(-1)^n$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $b_n$ is not convergent
            $endgroup$
            – gimusi
            Dec 14 '18 at 16:52










          • $begingroup$
            That is the point...
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:53










          • $begingroup$
            @SmileyCraft The question assumes that $b_n$ is a convergent sequence.
            $endgroup$
            – BigbearZzz
            Dec 14 '18 at 16:56










          • $begingroup$
            The OP literally asks "is it sufficient that ${b_n}$ is nonnegative and bounded?" and my example answers this question.
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:59
















          1












          1








          1





          $begingroup$

          Bounded and non-negative is not sufficient. Consider $a_n=frac{(-1)^n}n$ and $b_n=1+(-1)^n$.






          share|cite|improve this answer









          $endgroup$



          Bounded and non-negative is not sufficient. Consider $a_n=frac{(-1)^n}n$ and $b_n=1+(-1)^n$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 16:41









          SmileyCraftSmileyCraft

          3,591517




          3,591517












          • $begingroup$
            $b_n$ is not convergent
            $endgroup$
            – gimusi
            Dec 14 '18 at 16:52










          • $begingroup$
            That is the point...
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:53










          • $begingroup$
            @SmileyCraft The question assumes that $b_n$ is a convergent sequence.
            $endgroup$
            – BigbearZzz
            Dec 14 '18 at 16:56










          • $begingroup$
            The OP literally asks "is it sufficient that ${b_n}$ is nonnegative and bounded?" and my example answers this question.
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:59




















          • $begingroup$
            $b_n$ is not convergent
            $endgroup$
            – gimusi
            Dec 14 '18 at 16:52










          • $begingroup$
            That is the point...
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:53










          • $begingroup$
            @SmileyCraft The question assumes that $b_n$ is a convergent sequence.
            $endgroup$
            – BigbearZzz
            Dec 14 '18 at 16:56










          • $begingroup$
            The OP literally asks "is it sufficient that ${b_n}$ is nonnegative and bounded?" and my example answers this question.
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:59


















          $begingroup$
          $b_n$ is not convergent
          $endgroup$
          – gimusi
          Dec 14 '18 at 16:52




          $begingroup$
          $b_n$ is not convergent
          $endgroup$
          – gimusi
          Dec 14 '18 at 16:52












          $begingroup$
          That is the point...
          $endgroup$
          – SmileyCraft
          Dec 14 '18 at 16:53




          $begingroup$
          That is the point...
          $endgroup$
          – SmileyCraft
          Dec 14 '18 at 16:53












          $begingroup$
          @SmileyCraft The question assumes that $b_n$ is a convergent sequence.
          $endgroup$
          – BigbearZzz
          Dec 14 '18 at 16:56




          $begingroup$
          @SmileyCraft The question assumes that $b_n$ is a convergent sequence.
          $endgroup$
          – BigbearZzz
          Dec 14 '18 at 16:56












          $begingroup$
          The OP literally asks "is it sufficient that ${b_n}$ is nonnegative and bounded?" and my example answers this question.
          $endgroup$
          – SmileyCraft
          Dec 14 '18 at 16:59






          $begingroup$
          The OP literally asks "is it sufficient that ${b_n}$ is nonnegative and bounded?" and my example answers this question.
          $endgroup$
          – SmileyCraft
          Dec 14 '18 at 16:59













          0












          $begingroup$

          Assume



          $$a_n = frac{(-1)^n}{sqrt n}$$



          $$b_n =begin{cases}=0quad n,text{odd}\\=frac{1}{sqrt n}quad n,text{even} end{cases}$$



          and therefore



          $$sum_{n=1}^{2N} a_n b_n=sum_{n=1}^{N} frac1{2n} toinfty$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Your sequence $;b_n;$ isn't convergent...
            $endgroup$
            – DonAntonio
            Dec 14 '18 at 16:44










          • $begingroup$
            Opsss...thanks I fix
            $endgroup$
            – gimusi
            Dec 14 '18 at 16:44
















          0












          $begingroup$

          Assume



          $$a_n = frac{(-1)^n}{sqrt n}$$



          $$b_n =begin{cases}=0quad n,text{odd}\\=frac{1}{sqrt n}quad n,text{even} end{cases}$$



          and therefore



          $$sum_{n=1}^{2N} a_n b_n=sum_{n=1}^{N} frac1{2n} toinfty$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Your sequence $;b_n;$ isn't convergent...
            $endgroup$
            – DonAntonio
            Dec 14 '18 at 16:44










          • $begingroup$
            Opsss...thanks I fix
            $endgroup$
            – gimusi
            Dec 14 '18 at 16:44














          0












          0








          0





          $begingroup$

          Assume



          $$a_n = frac{(-1)^n}{sqrt n}$$



          $$b_n =begin{cases}=0quad n,text{odd}\\=frac{1}{sqrt n}quad n,text{even} end{cases}$$



          and therefore



          $$sum_{n=1}^{2N} a_n b_n=sum_{n=1}^{N} frac1{2n} toinfty$$






          share|cite|improve this answer











          $endgroup$



          Assume



          $$a_n = frac{(-1)^n}{sqrt n}$$



          $$b_n =begin{cases}=0quad n,text{odd}\\=frac{1}{sqrt n}quad n,text{even} end{cases}$$



          and therefore



          $$sum_{n=1}^{2N} a_n b_n=sum_{n=1}^{N} frac1{2n} toinfty$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 16:58

























          answered Dec 14 '18 at 16:43









          gimusigimusi

          92.8k84494




          92.8k84494












          • $begingroup$
            Your sequence $;b_n;$ isn't convergent...
            $endgroup$
            – DonAntonio
            Dec 14 '18 at 16:44










          • $begingroup$
            Opsss...thanks I fix
            $endgroup$
            – gimusi
            Dec 14 '18 at 16:44


















          • $begingroup$
            Your sequence $;b_n;$ isn't convergent...
            $endgroup$
            – DonAntonio
            Dec 14 '18 at 16:44










          • $begingroup$
            Opsss...thanks I fix
            $endgroup$
            – gimusi
            Dec 14 '18 at 16:44
















          $begingroup$
          Your sequence $;b_n;$ isn't convergent...
          $endgroup$
          – DonAntonio
          Dec 14 '18 at 16:44




          $begingroup$
          Your sequence $;b_n;$ isn't convergent...
          $endgroup$
          – DonAntonio
          Dec 14 '18 at 16:44












          $begingroup$
          Opsss...thanks I fix
          $endgroup$
          – gimusi
          Dec 14 '18 at 16:44




          $begingroup$
          Opsss...thanks I fix
          $endgroup$
          – gimusi
          Dec 14 '18 at 16:44


















          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%2f3039591%2fsum-a-n-b-n-when-sum-a-n-convergent-and-b-n-nonnegative%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

          Probability when a professor distributes a quiz and homework assignment to a class of n students.

          Image
          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
          Read more

          Aardman Animations

          Image
          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
          Read more

          Are they similar matrix

          Image
          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
          Read more