$sum a_n b_n$ 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

          93.4k12167270




          93.4k12167270












          • $begingroup$
            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

          Quarter-circle Tiles

          Image
          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
          Read more

          build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

          Image
          0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
          Read more

          Mont Emei

          Image
          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
          Read more