$sum a_n b_n$ 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

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          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





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          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


















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