If $f : [a, infty) → Bbb R$ is monotonically decreasing and the integral $int_0^infty f(x) ,dx$ is convergent, then $lim_x→infty f(x) = 0$.

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If $f : [a, infty) rightarrow Bbb R$ is monotonically decreasing and the integral $int_0^infty f(x) ,dx$ is convergent, then $lim_xrightarrowinfty f(x) = 0$. I don't really understand why this is true.



I know that if $f$ is monotonically decreasing then $f$ is bounded ($exists Min Bbb R s.t. f(x)leq M , forall ngeq k)$.



I also know that $$int_0^infty f(x) ,dx = lim_brightarrowinftyint_0^b f(x) ,dx < infty$$
I guess I can say something about the the monotonicity of limits, but I don't know how to continue from here. I would love for some help (it doesn't need to be formal proof, I just want to understand it).



Thank you.




calculus limits




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    $sum_n=0^infty f(n) geq int_0^infty f(x)dx geq sum_n=1^infty f(n)$
    – Michael
    Jul 26 at 21:32















up vote
1
down vote

favorite












If $f : [a, infty) rightarrow Bbb R$ is monotonically decreasing and the integral $int_0^infty f(x) ,dx$ is convergent, then $lim_xrightarrowinfty f(x) = 0$. I don't really understand why this is true.



I know that if $f$ is monotonically decreasing then $f$ is bounded ($exists Min Bbb R s.t. f(x)leq M , forall ngeq k)$.



I also know that $$int_0^infty f(x) ,dx = lim_brightarrowinftyint_0^b f(x) ,dx < infty$$
I guess I can say something about the the monotonicity of limits, but I don't know how to continue from here. I would love for some help (it doesn't need to be formal proof, I just want to understand it).



Thank you.




calculus limits




share|cite|improve this question

















  • 3




    $sum_n=0^infty f(n) geq int_0^infty f(x)dx geq sum_n=1^infty f(n)$
    – Michael
    Jul 26 at 21:32













up vote
1
down vote

favorite









up vote
1
down vote

favorite











If $f : [a, infty) rightarrow Bbb R$ is monotonically decreasing and the integral $int_0^infty f(x) ,dx$ is convergent, then $lim_xrightarrowinfty f(x) = 0$. I don't really understand why this is true.



I know that if $f$ is monotonically decreasing then $f$ is bounded ($exists Min Bbb R s.t. f(x)leq M , forall ngeq k)$.



I also know that $$int_0^infty f(x) ,dx = lim_brightarrowinftyint_0^b f(x) ,dx < infty$$
I guess I can say something about the the monotonicity of limits, but I don't know how to continue from here. I would love for some help (it doesn't need to be formal proof, I just want to understand it).



Thank you.




calculus limits




share|cite|improve this question













If $f : [a, infty) rightarrow Bbb R$ is monotonically decreasing and the integral $int_0^infty f(x) ,dx$ is convergent, then $lim_xrightarrowinfty f(x) = 0$. I don't really understand why this is true.



I know that if $f$ is monotonically decreasing then $f$ is bounded ($exists Min Bbb R s.t. f(x)leq M , forall ngeq k)$.



I also know that $$int_0^infty f(x) ,dx = lim_brightarrowinftyint_0^b f(x) ,dx < infty$$
I guess I can say something about the the monotonicity of limits, but I don't know how to continue from here. I would love for some help (it doesn't need to be formal proof, I just want to understand it).



Thank you.





calculus limits





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edited Jul 26 at 21:33
























asked Jul 26 at 21:30









rose12

1246




1246







  • 3




    $sum_n=0^infty f(n) geq int_0^infty f(x)dx geq sum_n=1^infty f(n)$
    – Michael
    Jul 26 at 21:32













  • 3




    $sum_n=0^infty f(n) geq int_0^infty f(x)dx geq sum_n=1^infty f(n)$
    – Michael
    Jul 26 at 21:32








3




3




$sum_n=0^infty f(n) geq int_0^infty f(x)dx geq sum_n=1^infty f(n)$
– Michael
Jul 26 at 21:32





$sum_n=0^infty f(n) geq int_0^infty f(x)dx geq sum_n=1^infty f(n)$
– Michael
Jul 26 at 21:32











2 Answers
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accepted










Fix $x_0 > 0$. Since $f$ is decreasing, for any $x in [nx_0, (n+1)x_0]$ we have $f((n+1)x_0) le f(x)$.



Therefore



beginalign
x_0 sum_n=1^infty f(nx_0) &= x_0 sum_n=0^infty f((n+1)x_0)\
&= sum_n=0^infty int_nx_0^(n+1)x_0f((n+1)x_0),dx \
&le sum_n=0^infty int_nx_0^(n+1)x_0f(x),dx\
&= int_0^infty f(x),dx
endalign



so the series $sum_n=1^infty f(nx_0)$ converges and hence $lim_ntoinfty f(nx_0) = 0$.



Now conclude that $lim_xtoinfty f(x) = 0$.






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    It might be best to approach this by contradiction. In particular, if we assume $lim_xtoinftyf(x)neq 0$, then either the limit is nonzero or $f$ decreases without bound. It should be possible to show in each case that the integral does not converge.






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      2 Answers
      2






      active

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      2 Answers
      2






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      active

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      up vote
      2
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      accepted










      Fix $x_0 > 0$. Since $f$ is decreasing, for any $x in [nx_0, (n+1)x_0]$ we have $f((n+1)x_0) le f(x)$.



      Therefore



      beginalign
      x_0 sum_n=1^infty f(nx_0) &= x_0 sum_n=0^infty f((n+1)x_0)\
      &= sum_n=0^infty int_nx_0^(n+1)x_0f((n+1)x_0),dx \
      &le sum_n=0^infty int_nx_0^(n+1)x_0f(x),dx\
      &= int_0^infty f(x),dx
      endalign



      so the series $sum_n=1^infty f(nx_0)$ converges and hence $lim_ntoinfty f(nx_0) = 0$.



      Now conclude that $lim_xtoinfty f(x) = 0$.






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted










        Fix $x_0 > 0$. Since $f$ is decreasing, for any $x in [nx_0, (n+1)x_0]$ we have $f((n+1)x_0) le f(x)$.



        Therefore



        beginalign
        x_0 sum_n=1^infty f(nx_0) &= x_0 sum_n=0^infty f((n+1)x_0)\
        &= sum_n=0^infty int_nx_0^(n+1)x_0f((n+1)x_0),dx \
        &le sum_n=0^infty int_nx_0^(n+1)x_0f(x),dx\
        &= int_0^infty f(x),dx
        endalign



        so the series $sum_n=1^infty f(nx_0)$ converges and hence $lim_ntoinfty f(nx_0) = 0$.



        Now conclude that $lim_xtoinfty f(x) = 0$.






        share|cite|improve this answer























          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          Fix $x_0 > 0$. Since $f$ is decreasing, for any $x in [nx_0, (n+1)x_0]$ we have $f((n+1)x_0) le f(x)$.



          Therefore



          beginalign
          x_0 sum_n=1^infty f(nx_0) &= x_0 sum_n=0^infty f((n+1)x_0)\
          &= sum_n=0^infty int_nx_0^(n+1)x_0f((n+1)x_0),dx \
          &le sum_n=0^infty int_nx_0^(n+1)x_0f(x),dx\
          &= int_0^infty f(x),dx
          endalign



          so the series $sum_n=1^infty f(nx_0)$ converges and hence $lim_ntoinfty f(nx_0) = 0$.



          Now conclude that $lim_xtoinfty f(x) = 0$.






          share|cite|improve this answer













          Fix $x_0 > 0$. Since $f$ is decreasing, for any $x in [nx_0, (n+1)x_0]$ we have $f((n+1)x_0) le f(x)$.



          Therefore



          beginalign
          x_0 sum_n=1^infty f(nx_0) &= x_0 sum_n=0^infty f((n+1)x_0)\
          &= sum_n=0^infty int_nx_0^(n+1)x_0f((n+1)x_0),dx \
          &le sum_n=0^infty int_nx_0^(n+1)x_0f(x),dx\
          &= int_0^infty f(x),dx
          endalign



          so the series $sum_n=1^infty f(nx_0)$ converges and hence $lim_ntoinfty f(nx_0) = 0$.



          Now conclude that $lim_xtoinfty f(x) = 0$.







          share|cite|improve this answer













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          answered Jul 26 at 22:14









          mechanodroid

          22.2k52041




          22.2k52041




















              up vote
              1
              down vote













              It might be best to approach this by contradiction. In particular, if we assume $lim_xtoinftyf(x)neq 0$, then either the limit is nonzero or $f$ decreases without bound. It should be possible to show in each case that the integral does not converge.






              share|cite|improve this answer

























                up vote
                1
                down vote













                It might be best to approach this by contradiction. In particular, if we assume $lim_xtoinftyf(x)neq 0$, then either the limit is nonzero or $f$ decreases without bound. It should be possible to show in each case that the integral does not converge.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  It might be best to approach this by contradiction. In particular, if we assume $lim_xtoinftyf(x)neq 0$, then either the limit is nonzero or $f$ decreases without bound. It should be possible to show in each case that the integral does not converge.






                  share|cite|improve this answer













                  It might be best to approach this by contradiction. In particular, if we assume $lim_xtoinftyf(x)neq 0$, then either the limit is nonzero or $f$ decreases without bound. It should be possible to show in each case that the integral does not converge.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Jul 26 at 21:54









                  Kajelad

                  1,848619




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