Proving a limit exists under some conditions












1












$begingroup$



Suppose that $f(x) > 0$ is integrable and monotone decreasing on $[0,
infty)$. Let $F_{n} = int_{0}^{n} f(t) mathop{dt}$, $n = 1, 2, 3,
ldots$. Prove that



$$lim_{ntoinfty} F_{n} $$



exists if and only if $sum_{n = 1}^{infty} f(n) < infty$.



Hint: Consider $F(n + 1) - F(n)$




I'm not sure about how to approach this problem. This is a practice problem that I have for a final exam coming soon. I think that the hint helps us show that the sequence is monotone decreasing because if we can show that quantity is less than $0$, it would imply that the terms are getting smaller and smaller.






real-analysis calculus sequences-and-series limits analysis







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    Suppose that $f(x) > 0$ is integrable and monotone decreasing on $[0,
    infty)$    . Let $F_{n} = int_{0}^{n} f(t) mathop{dt}$, $n = 1, 2, 3,
    ldots$    . Prove that



    $$lim_{ntoinfty} F_{n} $$



    exists if and only if $sum_{n = 1}^{infty} f(n) < infty$.



    Hint: Consider $F(n + 1) - F(n)$




    I'm not sure about how to approach this problem. This is a practice problem that I have for a final exam coming soon. I think that the hint helps us show that the sequence is monotone decreasing because if we can show that quantity is less than $0$, it would imply that the terms are getting smaller and smaller.






    real-analysis calculus sequences-and-series limits analysis







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Suppose that $f(x) > 0$ is integrable and monotone decreasing on $[0,
      infty)$      . Let $F_{n} = int_{0}^{n} f(t) mathop{dt}$, $n = 1, 2, 3,
      ldots$      . Prove that



      $$lim_{ntoinfty} F_{n} $$



      exists if and only if $sum_{n = 1}^{infty} f(n) < infty$.



      Hint: Consider $F(n + 1) - F(n)$




      I'm not sure about how to approach this problem. This is a practice problem that I have for a final exam coming soon. I think that the hint helps us show that the sequence is monotone decreasing because if we can show that quantity is less than $0$, it would imply that the terms are getting smaller and smaller.






      real-analysis calculus sequences-and-series limits analysis







      share|cite|improve this question











      $endgroup$





      Suppose that $f(x) > 0$ is integrable and monotone decreasing on $[0,
      infty)$      . Let $F_{n} = int_{0}^{n} f(t) mathop{dt}$, $n = 1, 2, 3,
      ldots$      . Prove that



      $$lim_{ntoinfty} F_{n} $$



      exists if and only if $sum_{n = 1}^{infty} f(n) < infty$.



      Hint: Consider $F(n + 1) - F(n)$




      I'm not sure about how to approach this problem. This is a practice problem that I have for a final exam coming soon. I think that the hint helps us show that the sequence is monotone decreasing because if we can show that quantity is less than $0$, it would imply that the terms are getting smaller and smaller.






      real-analysis calculus sequences-and-series limits analysis




      real-analysis calculus sequences-and-series limits analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 14 '18 at 18:04







      Dillain Smith

















      asked Dec 14 '18 at 17:54









      Dillain SmithDillain Smith

      567




      567






















          3 Answers
          3






          active

          oldest

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          0












          $begingroup$

          Hint



          Note that
          $$
          sum_{j=1}^n f(j)leq int_0^n f(t), dt=sum_{j=1}^nint_{j-1}^{j} f(t) , dtle sum_{j=1}^n f(j-1)
          $$
          where we used linearity of the integral for the middle equality and the fact that $f$ is monotone decreasing for the outer inequalities.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            hint



            For $nge 0,$



            $$F(n+1)-F(n)=int_n^{n+1}f(t)dt$$



            $$(forall tin[n,n+1]);$$
            $$f(n+1)le f(t)le f(n)$$
            and
            $$;f(n+1)leint_n^{n+1}f(t)dtle f(n)$$






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              I am also studying for my real analysis exam, so if this proof is wrong please let me know so I know what to look over :)




              If $sum_{n=1}^infty f(n) < infty$ $implies$ $lim_{nrightarrow infty} F_n < infty$




              Since $f$ is monotonically decreasing on $[0,infty)$, we know that $f(n+1) < f(n)$ $forall n in mathbb{N}$, and we know $f(0)$ exists.



              From this we can see that the following comparison is true
              $$
              F_n = int_0^n f(t)dt < sum_{k=0}^n f(k) = f(0) + sum_{k=1}^n f(k)
              $$
              now we can take the limit of both sides and see that
              $$
              lim_{nrightarrow infty} F_n < f(0) + sum_{k=1}^infty f(k)
              $$
              Therofore, the limit must exist because we know that the sum exists, and $f(0)$ exists.



              To prove the other direction the method is the same, except we say that



              $$
              sum_{k=1}^n f(k) < int_0^n f(t)dt
              $$



              since it is not a "right-Riemann Sum"






              share|cite|improve this answer









              $endgroup$













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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                0












                $begingroup$

                Hint



                Note that
                $$
                sum_{j=1}^n f(j)leq int_0^n f(t), dt=sum_{j=1}^nint_{j-1}^{j} f(t) , dtle sum_{j=1}^n f(j-1)
                $$
                where we used linearity of the integral for the middle equality and the fact that $f$ is monotone decreasing for the outer inequalities.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Hint



                  Note that
                  $$
                  sum_{j=1}^n f(j)leq int_0^n f(t), dt=sum_{j=1}^nint_{j-1}^{j} f(t) , dtle sum_{j=1}^n f(j-1)
                  $$
                  where we used linearity of the integral for the middle equality and the fact that $f$ is monotone decreasing for the outer inequalities.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Hint



                    Note that
                    $$
                    sum_{j=1}^n f(j)leq int_0^n f(t), dt=sum_{j=1}^nint_{j-1}^{j} f(t) , dtle sum_{j=1}^n f(j-1)
                    $$
                    where we used linearity of the integral for the middle equality and the fact that $f$ is monotone decreasing for the outer inequalities.






                    share|cite|improve this answer









                    $endgroup$



                    Hint



                    Note that
                    $$
                    sum_{j=1}^n f(j)leq int_0^n f(t), dt=sum_{j=1}^nint_{j-1}^{j} f(t) , dtle sum_{j=1}^n f(j-1)
                    $$
                    where we used linearity of the integral for the middle equality and the fact that $f$ is monotone decreasing for the outer inequalities.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 14 '18 at 18:02









                    Foobaz JohnFoobaz John

                    22.1k41352




                    22.1k41352























                        0












                        $begingroup$

                        hint



                        For $nge 0,$



                        $$F(n+1)-F(n)=int_n^{n+1}f(t)dt$$



                        $$(forall tin[n,n+1]);$$
                        $$f(n+1)le f(t)le f(n)$$
                        and
                        $$;f(n+1)leint_n^{n+1}f(t)dtle f(n)$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          hint



                          For $nge 0,$



                          $$F(n+1)-F(n)=int_n^{n+1}f(t)dt$$



                          $$(forall tin[n,n+1]);$$
                          $$f(n+1)le f(t)le f(n)$$
                          and
                          $$;f(n+1)leint_n^{n+1}f(t)dtle f(n)$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            hint



                            For $nge 0,$



                            $$F(n+1)-F(n)=int_n^{n+1}f(t)dt$$



                            $$(forall tin[n,n+1]);$$
                            $$f(n+1)le f(t)le f(n)$$
                            and
                            $$;f(n+1)leint_n^{n+1}f(t)dtle f(n)$$






                            share|cite|improve this answer









                            $endgroup$



                            hint



                            For $nge 0,$



                            $$F(n+1)-F(n)=int_n^{n+1}f(t)dt$$



                            $$(forall tin[n,n+1]);$$
                            $$f(n+1)le f(t)le f(n)$$
                            and
                            $$;f(n+1)leint_n^{n+1}f(t)dtle f(n)$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 14 '18 at 18:25









                            hamam_Abdallahhamam_Abdallah

                            38.1k21634




                            38.1k21634























                                0












                                $begingroup$

                                I am also studying for my real analysis exam, so if this proof is wrong please let me know so I know what to look over :)




                                If $sum_{n=1}^infty f(n) < infty$ $implies$ $lim_{nrightarrow infty} F_n < infty$




                                Since $f$ is monotonically decreasing on $[0,infty)$, we know that $f(n+1) < f(n)$ $forall n in mathbb{N}$, and we know $f(0)$ exists.



                                From this we can see that the following comparison is true
                                $$
                                F_n = int_0^n f(t)dt < sum_{k=0}^n f(k) = f(0) + sum_{k=1}^n f(k)
                                $$
                                now we can take the limit of both sides and see that
                                $$
                                lim_{nrightarrow infty} F_n < f(0) + sum_{k=1}^infty f(k)
                                $$
                                Therofore, the limit must exist because we know that the sum exists, and $f(0)$ exists.



                                To prove the other direction the method is the same, except we say that



                                $$
                                sum_{k=1}^n f(k) < int_0^n f(t)dt
                                $$



                                since it is not a "right-Riemann Sum"






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  I am also studying for my real analysis exam, so if this proof is wrong please let me know so I know what to look over :)




                                  If $sum_{n=1}^infty f(n) < infty$ $implies$ $lim_{nrightarrow infty} F_n < infty$




                                  Since $f$ is monotonically decreasing on $[0,infty)$, we know that $f(n+1) < f(n)$ $forall n in mathbb{N}$, and we know $f(0)$ exists.



                                  From this we can see that the following comparison is true
                                  $$
                                  F_n = int_0^n f(t)dt < sum_{k=0}^n f(k) = f(0) + sum_{k=1}^n f(k)
                                  $$
                                  now we can take the limit of both sides and see that
                                  $$
                                  lim_{nrightarrow infty} F_n < f(0) + sum_{k=1}^infty f(k)
                                  $$
                                  Therofore, the limit must exist because we know that the sum exists, and $f(0)$ exists.



                                  To prove the other direction the method is the same, except we say that



                                  $$
                                  sum_{k=1}^n f(k) < int_0^n f(t)dt
                                  $$



                                  since it is not a "right-Riemann Sum"






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    I am also studying for my real analysis exam, so if this proof is wrong please let me know so I know what to look over :)




                                    If $sum_{n=1}^infty f(n) < infty$ $implies$ $lim_{nrightarrow infty} F_n < infty$




                                    Since $f$ is monotonically decreasing on $[0,infty)$, we know that $f(n+1) < f(n)$ $forall n in mathbb{N}$, and we know $f(0)$ exists.



                                    From this we can see that the following comparison is true
                                    $$
                                    F_n = int_0^n f(t)dt < sum_{k=0}^n f(k) = f(0) + sum_{k=1}^n f(k)
                                    $$
                                    now we can take the limit of both sides and see that
                                    $$
                                    lim_{nrightarrow infty} F_n < f(0) + sum_{k=1}^infty f(k)
                                    $$
                                    Therofore, the limit must exist because we know that the sum exists, and $f(0)$ exists.



                                    To prove the other direction the method is the same, except we say that



                                    $$
                                    sum_{k=1}^n f(k) < int_0^n f(t)dt
                                    $$



                                    since it is not a "right-Riemann Sum"






                                    share|cite|improve this answer









                                    $endgroup$



                                    I am also studying for my real analysis exam, so if this proof is wrong please let me know so I know what to look over :)




                                    If $sum_{n=1}^infty f(n) < infty$ $implies$ $lim_{nrightarrow infty} F_n < infty$




                                    Since $f$ is monotonically decreasing on $[0,infty)$, we know that $f(n+1) < f(n)$ $forall n in mathbb{N}$, and we know $f(0)$ exists.



                                    From this we can see that the following comparison is true
                                    $$
                                    F_n = int_0^n f(t)dt < sum_{k=0}^n f(k) = f(0) + sum_{k=1}^n f(k)
                                    $$
                                    now we can take the limit of both sides and see that
                                    $$
                                    lim_{nrightarrow infty} F_n < f(0) + sum_{k=1}^infty f(k)
                                    $$
                                    Therofore, the limit must exist because we know that the sum exists, and $f(0)$ exists.



                                    To prove the other direction the method is the same, except we say that



                                    $$
                                    sum_{k=1}^n f(k) < int_0^n f(t)dt
                                    $$



                                    since it is not a "right-Riemann Sum"







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 15 '18 at 0:33









                                    wjmccannwjmccann

                                    654118




                                    654118






























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