Convergence of ${a_n}$ given that $b_n = a_n + 1/a_n$ converges












5














I am given a sequence ${a_n}$ with $a_n > 0$ and $b_n = a_n + 1/a_n$.



I am first asked to assume that $a_n ge 1$ and show that the convergence of ${b_n}$ implies the convergence of ${a_n}$.



My argument is:



Since ${b_n}$ converges, ${b_n}$ is Cauchy, hence for every $epsilon > 0$ some $N$ such that for $n_i,n_j > N$, $|b_{n_i} - b_{n_j}| < epsilon$. Then $|b_{n_i} - b_{n_j}| = |a_{n_i} - a_{n_j} + 1/a_{n_i} - 1/a_{n_j}| < epsilon$. Suppose that $a_{n_i} = a_{n_j}$. Then clearly $|a_{n_i} - a_{n_j}| < epsilon$. Alternatively, suppose without loss of generality that $a_{n_i} > a_{n_j}$. Then $1/a_{n_i} - 1/a_{n_j} < 0$ while $a_{n_i} - a_{n_j} > 0$. It is straightforward to observe that $-(1/a_{n_i} - 1/a_{n_j}) le a_{n_i} - a_{n_j}$ because of the condition $a_n ge 0$, so it follows that $0 < |a_{n_i} - a_{n_j}| < |a_{n_i} - a_{n_j}|$, so then $|a_{n_i} - a_{n_j}| < |a_{n_i} - a_{n_j} + 1/a_{n_i} - 1/a_{n_j}| < epsilon$, hence ${a_n}$ is Cauchy and therefore converges.



However, I am next asked to assume that ${b_n}$ converges but only that $a_n > 0$ and demonstrate, through construction of a counterexample, that it does not necessarily follow that ${a_n}$ converges. I am not entirely sure how to do this.






real-analysis







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    5














    I am given a sequence ${a_n}$ with $a_n > 0$ and $b_n = a_n + 1/a_n$.



    I am first asked to assume that $a_n ge 1$ and show that the convergence of ${b_n}$ implies the convergence of ${a_n}$.



    My argument is:



    Since ${b_n}$ converges, ${b_n}$ is Cauchy, hence for every $epsilon > 0$ some $N$ such that for $n_i,n_j > N$, $|b_{n_i} - b_{n_j}| < epsilon$. Then $|b_{n_i} - b_{n_j}| = |a_{n_i} - a_{n_j} + 1/a_{n_i} - 1/a_{n_j}| < epsilon$. Suppose that $a_{n_i} = a_{n_j}$. Then clearly $|a_{n_i} - a_{n_j}| < epsilon$. Alternatively, suppose without loss of generality that $a_{n_i} > a_{n_j}$. Then $1/a_{n_i} - 1/a_{n_j} < 0$ while $a_{n_i} - a_{n_j} > 0$. It is straightforward to observe that $-(1/a_{n_i} - 1/a_{n_j}) le a_{n_i} - a_{n_j}$ because of the condition $a_n ge 0$, so it follows that $0 < |a_{n_i} - a_{n_j}| < |a_{n_i} - a_{n_j}|$, so then $|a_{n_i} - a_{n_j}| < |a_{n_i} - a_{n_j} + 1/a_{n_i} - 1/a_{n_j}| < epsilon$, hence ${a_n}$ is Cauchy and therefore converges.



    However, I am next asked to assume that ${b_n}$ converges but only that $a_n > 0$ and demonstrate, through construction of a counterexample, that it does not necessarily follow that ${a_n}$ converges. I am not entirely sure how to do this.






    real-analysis







    share|cite|improve this question

























      5












      5








      5







      I am given a sequence ${a_n}$ with $a_n > 0$ and $b_n = a_n + 1/a_n$.



      I am first asked to assume that $a_n ge 1$ and show that the convergence of ${b_n}$ implies the convergence of ${a_n}$.



      My argument is:



      Since ${b_n}$ converges, ${b_n}$ is Cauchy, hence for every $epsilon > 0$ some $N$ such that for $n_i,n_j > N$, $|b_{n_i} - b_{n_j}| < epsilon$. Then $|b_{n_i} - b_{n_j}| = |a_{n_i} - a_{n_j} + 1/a_{n_i} - 1/a_{n_j}| < epsilon$. Suppose that $a_{n_i} = a_{n_j}$. Then clearly $|a_{n_i} - a_{n_j}| < epsilon$. Alternatively, suppose without loss of generality that $a_{n_i} > a_{n_j}$. Then $1/a_{n_i} - 1/a_{n_j} < 0$ while $a_{n_i} - a_{n_j} > 0$. It is straightforward to observe that $-(1/a_{n_i} - 1/a_{n_j}) le a_{n_i} - a_{n_j}$ because of the condition $a_n ge 0$, so it follows that $0 < |a_{n_i} - a_{n_j}| < |a_{n_i} - a_{n_j}|$, so then $|a_{n_i} - a_{n_j}| < |a_{n_i} - a_{n_j} + 1/a_{n_i} - 1/a_{n_j}| < epsilon$, hence ${a_n}$ is Cauchy and therefore converges.



      However, I am next asked to assume that ${b_n}$ converges but only that $a_n > 0$ and demonstrate, through construction of a counterexample, that it does not necessarily follow that ${a_n}$ converges. I am not entirely sure how to do this.






      real-analysis







      share|cite|improve this question













      I am given a sequence ${a_n}$ with $a_n > 0$ and $b_n = a_n + 1/a_n$.



      I am first asked to assume that $a_n ge 1$ and show that the convergence of ${b_n}$ implies the convergence of ${a_n}$.



      My argument is:



      Since ${b_n}$ converges, ${b_n}$ is Cauchy, hence for every $epsilon > 0$ some $N$ such that for $n_i,n_j > N$, $|b_{n_i} - b_{n_j}| < epsilon$. Then $|b_{n_i} - b_{n_j}| = |a_{n_i} - a_{n_j} + 1/a_{n_i} - 1/a_{n_j}| < epsilon$. Suppose that $a_{n_i} = a_{n_j}$. Then clearly $|a_{n_i} - a_{n_j}| < epsilon$. Alternatively, suppose without loss of generality that $a_{n_i} > a_{n_j}$. Then $1/a_{n_i} - 1/a_{n_j} < 0$ while $a_{n_i} - a_{n_j} > 0$. It is straightforward to observe that $-(1/a_{n_i} - 1/a_{n_j}) le a_{n_i} - a_{n_j}$ because of the condition $a_n ge 0$, so it follows that $0 < |a_{n_i} - a_{n_j}| < |a_{n_i} - a_{n_j}|$, so then $|a_{n_i} - a_{n_j}| < |a_{n_i} - a_{n_j} + 1/a_{n_i} - 1/a_{n_j}| < epsilon$, hence ${a_n}$ is Cauchy and therefore converges.



      However, I am next asked to assume that ${b_n}$ converges but only that $a_n > 0$ and demonstrate, through construction of a counterexample, that it does not necessarily follow that ${a_n}$ converges. I am not entirely sure how to do this.






      real-analysis




      real-analysis






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      asked Oct 19 '14 at 23:33









      Elizabeth Lin

      815613




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          3 Answers
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          1














          Hint: draw a graph of the function $f(x)= x + frac{1}{x}, x>0$, what are solutions of $f(x) = b$? with $b = dfrac{5}{2}$ for example






          share|cite|improve this answer





















          • Oh, I think I see--is it correct to say that $f(x) = f(1/x)$, so if I take one of the convergent sequences ${a_n}$ where $a_n ge 1$ and I change it so that every other term becomes $1/a_n$, that doesn't change ${b_n}$ but now ${a_n}$ doesn't converge?
            – Elizabeth Lin
            Oct 19 '14 at 23:46








          • 1




            @ElizabethLin Yeah you get it!
            – Petite Etincelle
            Oct 19 '14 at 23:47



















          1














          You are really overthinking the second part.



          Take the sequence $a_n = a, 1/a, a, 1/a, a, 1/a, ... $, then $b_n = a + 1/a$ and the sequence $b_n$ is convergent, but $a_n$ is not convergent unless a = 1.






          share|cite|improve this answer





























            0














            Let $$a_n=2+sqrt 3(-1)^n$$which is not convergent while $b_n=4$ is convergent to $4$.






            share|cite|improve this answer





















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              3 Answers
              3






              active

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              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1














              Hint: draw a graph of the function $f(x)= x + frac{1}{x}, x>0$, what are solutions of $f(x) = b$? with $b = dfrac{5}{2}$ for example






              share|cite|improve this answer





















              • Oh, I think I see--is it correct to say that $f(x) = f(1/x)$, so if I take one of the convergent sequences ${a_n}$ where $a_n ge 1$ and I change it so that every other term becomes $1/a_n$, that doesn't change ${b_n}$ but now ${a_n}$ doesn't converge?
                – Elizabeth Lin
                Oct 19 '14 at 23:46








              • 1




                @ElizabethLin Yeah you get it!
                – Petite Etincelle
                Oct 19 '14 at 23:47
















              1














              Hint: draw a graph of the function $f(x)= x + frac{1}{x}, x>0$, what are solutions of $f(x) = b$? with $b = dfrac{5}{2}$ for example






              share|cite|improve this answer





















              • Oh, I think I see--is it correct to say that $f(x) = f(1/x)$, so if I take one of the convergent sequences ${a_n}$ where $a_n ge 1$ and I change it so that every other term becomes $1/a_n$, that doesn't change ${b_n}$ but now ${a_n}$ doesn't converge?
                – Elizabeth Lin
                Oct 19 '14 at 23:46








              • 1




                @ElizabethLin Yeah you get it!
                – Petite Etincelle
                Oct 19 '14 at 23:47














              1












              1








              1






              Hint: draw a graph of the function $f(x)= x + frac{1}{x}, x>0$, what are solutions of $f(x) = b$? with $b = dfrac{5}{2}$ for example






              share|cite|improve this answer












              Hint: draw a graph of the function $f(x)= x + frac{1}{x}, x>0$, what are solutions of $f(x) = b$? with $b = dfrac{5}{2}$ for example







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Oct 19 '14 at 23:37









              Petite Etincelle

              12.3k12147




              12.3k12147












              • Oh, I think I see--is it correct to say that $f(x) = f(1/x)$, so if I take one of the convergent sequences ${a_n}$ where $a_n ge 1$ and I change it so that every other term becomes $1/a_n$, that doesn't change ${b_n}$ but now ${a_n}$ doesn't converge?
                – Elizabeth Lin
                Oct 19 '14 at 23:46








              • 1




                @ElizabethLin Yeah you get it!
                – Petite Etincelle
                Oct 19 '14 at 23:47


















              • Oh, I think I see--is it correct to say that $f(x) = f(1/x)$, so if I take one of the convergent sequences ${a_n}$ where $a_n ge 1$ and I change it so that every other term becomes $1/a_n$, that doesn't change ${b_n}$ but now ${a_n}$ doesn't converge?
                – Elizabeth Lin
                Oct 19 '14 at 23:46








              • 1




                @ElizabethLin Yeah you get it!
                – Petite Etincelle
                Oct 19 '14 at 23:47
















              Oh, I think I see--is it correct to say that $f(x) = f(1/x)$, so if I take one of the convergent sequences ${a_n}$ where $a_n ge 1$ and I change it so that every other term becomes $1/a_n$, that doesn't change ${b_n}$ but now ${a_n}$ doesn't converge?
              – Elizabeth Lin
              Oct 19 '14 at 23:46






              Oh, I think I see--is it correct to say that $f(x) = f(1/x)$, so if I take one of the convergent sequences ${a_n}$ where $a_n ge 1$ and I change it so that every other term becomes $1/a_n$, that doesn't change ${b_n}$ but now ${a_n}$ doesn't converge?
              – Elizabeth Lin
              Oct 19 '14 at 23:46






              1




              1




              @ElizabethLin Yeah you get it!
              – Petite Etincelle
              Oct 19 '14 at 23:47




              @ElizabethLin Yeah you get it!
              – Petite Etincelle
              Oct 19 '14 at 23:47











              1














              You are really overthinking the second part.



              Take the sequence $a_n = a, 1/a, a, 1/a, a, 1/a, ... $, then $b_n = a + 1/a$ and the sequence $b_n$ is convergent, but $a_n$ is not convergent unless a = 1.






              share|cite|improve this answer


























                1














                You are really overthinking the second part.



                Take the sequence $a_n = a, 1/a, a, 1/a, a, 1/a, ... $, then $b_n = a + 1/a$ and the sequence $b_n$ is convergent, but $a_n$ is not convergent unless a = 1.






                share|cite|improve this answer
























                  1












                  1








                  1






                  You are really overthinking the second part.



                  Take the sequence $a_n = a, 1/a, a, 1/a, a, 1/a, ... $, then $b_n = a + 1/a$ and the sequence $b_n$ is convergent, but $a_n$ is not convergent unless a = 1.






                  share|cite|improve this answer












                  You are really overthinking the second part.



                  Take the sequence $a_n = a, 1/a, a, 1/a, a, 1/a, ... $, then $b_n = a + 1/a$ and the sequence $b_n$ is convergent, but $a_n$ is not convergent unless a = 1.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 19 '14 at 23:58









                  gnasher729

                  5,9671028




                  5,9671028























                      0














                      Let $$a_n=2+sqrt 3(-1)^n$$which is not convergent while $b_n=4$ is convergent to $4$.






                      share|cite|improve this answer


























                        0














                        Let $$a_n=2+sqrt 3(-1)^n$$which is not convergent while $b_n=4$ is convergent to $4$.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          Let $$a_n=2+sqrt 3(-1)^n$$which is not convergent while $b_n=4$ is convergent to $4$.






                          share|cite|improve this answer












                          Let $$a_n=2+sqrt 3(-1)^n$$which is not convergent while $b_n=4$ is convergent to $4$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 27 '18 at 9:19









                          Mostafa Ayaz

                          13.7k3936




                          13.7k3936






























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