Show with the direct Comparison Test that $sqrt[n]{|a_n|}leqtheta$ converges absolutely












0












$begingroup$


Let $sum_{n=1}^{n=infty}{a_n}$ be an infinite series of real numbers. There is a $theta$ with $0<theta<1$ and a $n_0 > 0$, so $$sqrt[n]{|a_n|}leqtheta$$ for all $n ge n_0$. Show that the series converges absolutely.



So I need verification here. Quite funny, because this is the first task out of many and I am certain that the other ones are correct, but this one I am not sure of.



My proof feels wrong, it feels like as if it is way too short.



My proof here:



Since $lvert a_{n} rvert leq theta$ and $theta in (0,1)$, the series $sum_{n=1}^{n=infty}{a_n}$ converges. And also since $sqrt[n]{|a_n|}≤a_n$, the series converges absolutely.






real-analysis proof-verification convergence proof-writing absolute-convergence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $|a_n| le theta^n$
    $endgroup$
    – gammatester
    Dec 1 '18 at 21:30






  • 1




    $begingroup$
    I don't see why $0le |a_n| le theta <1$ implies the series converges. Anyway, the hypothesis is $sqrt[n]{a_n}le theta$, not $|a_n|le theta$.
    $endgroup$
    – Bernard
    Dec 1 '18 at 21:32
















0












$begingroup$


Let $sum_{n=1}^{n=infty}{a_n}$ be an infinite series of real numbers. There is a $theta$ with $0<theta<1$ and a $n_0 > 0$, so $$sqrt[n]{|a_n|}leqtheta$$ for all $n ge n_0$. Show that the series converges absolutely.



So I need verification here. Quite funny, because this is the first task out of many and I am certain that the other ones are correct, but this one I am not sure of.



My proof feels wrong, it feels like as if it is way too short.



My proof here:



Since $lvert a_{n} rvert leq theta$ and $theta in (0,1)$, the series $sum_{n=1}^{n=infty}{a_n}$ converges. And also since $sqrt[n]{|a_n|}≤a_n$, the series converges absolutely.






real-analysis proof-verification convergence proof-writing absolute-convergence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $|a_n| le theta^n$
    $endgroup$
    – gammatester
    Dec 1 '18 at 21:30






  • 1




    $begingroup$
    I don't see why $0le |a_n| le theta <1$ implies the series converges. Anyway, the hypothesis is $sqrt[n]{a_n}le theta$, not $|a_n|le theta$.
    $endgroup$
    – Bernard
    Dec 1 '18 at 21:32














0












0








0





$begingroup$


Let $sum_{n=1}^{n=infty}{a_n}$ be an infinite series of real numbers. There is a $theta$ with $0<theta<1$ and a $n_0 > 0$, so $$sqrt[n]{|a_n|}leqtheta$$ for all $n ge n_0$. Show that the series converges absolutely.



So I need verification here. Quite funny, because this is the first task out of many and I am certain that the other ones are correct, but this one I am not sure of.



My proof feels wrong, it feels like as if it is way too short.



My proof here:



Since $lvert a_{n} rvert leq theta$ and $theta in (0,1)$, the series $sum_{n=1}^{n=infty}{a_n}$ converges. And also since $sqrt[n]{|a_n|}≤a_n$, the series converges absolutely.






real-analysis proof-verification convergence proof-writing absolute-convergence







share|cite|improve this question











$endgroup$




Let $sum_{n=1}^{n=infty}{a_n}$ be an infinite series of real numbers. There is a $theta$ with $0<theta<1$ and a $n_0 > 0$, so $$sqrt[n]{|a_n|}leqtheta$$ for all $n ge n_0$. Show that the series converges absolutely.



So I need verification here. Quite funny, because this is the first task out of many and I am certain that the other ones are correct, but this one I am not sure of.



My proof feels wrong, it feels like as if it is way too short.



My proof here:



Since $lvert a_{n} rvert leq theta$ and $theta in (0,1)$, the series $sum_{n=1}^{n=infty}{a_n}$ converges. And also since $sqrt[n]{|a_n|}≤a_n$, the series converges absolutely.






real-analysis proof-verification convergence proof-writing absolute-convergence




real-analysis proof-verification convergence proof-writing absolute-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 21:28









Bernard

119k639112




119k639112










asked Dec 1 '18 at 21:24









SacredScoutSacredScout

477




477








  • 1




    $begingroup$
    $|a_n| le theta^n$
    $endgroup$
    – gammatester
    Dec 1 '18 at 21:30






  • 1




    $begingroup$
    I don't see why $0le |a_n| le theta <1$ implies the series converges. Anyway, the hypothesis is $sqrt[n]{a_n}le theta$, not $|a_n|le theta$.
    $endgroup$
    – Bernard
    Dec 1 '18 at 21:32














  • 1




    $begingroup$
    $|a_n| le theta^n$
    $endgroup$
    – gammatester
    Dec 1 '18 at 21:30






  • 1




    $begingroup$
    I don't see why $0le |a_n| le theta <1$ implies the series converges. Anyway, the hypothesis is $sqrt[n]{a_n}le theta$, not $|a_n|le theta$.
    $endgroup$
    – Bernard
    Dec 1 '18 at 21:32








1




1




$begingroup$
$|a_n| le theta^n$
$endgroup$
– gammatester
Dec 1 '18 at 21:30




$begingroup$
$|a_n| le theta^n$
$endgroup$
– gammatester
Dec 1 '18 at 21:30




1




1




$begingroup$
I don't see why $0le |a_n| le theta <1$ implies the series converges. Anyway, the hypothesis is $sqrt[n]{a_n}le theta$, not $|a_n|le theta$.
$endgroup$
– Bernard
Dec 1 '18 at 21:32




$begingroup$
I don't see why $0le |a_n| le theta <1$ implies the series converges. Anyway, the hypothesis is $sqrt[n]{a_n}le theta$, not $|a_n|le theta$.
$endgroup$
– Bernard
Dec 1 '18 at 21:32










3 Answers
3






active

oldest

votes


















2












$begingroup$

We have $$sqrt[n]{|a_n|}le thetato 0le |a_n|le theta^n$$therefore $$0le sum_{n=1}^{infty}|a_n|le sum_{n=1}^{infty}theta ^n={thetaover 1-theta}$$which completes the proof.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do you conclude that it equals $1/1-theta$ at the end?
    $endgroup$
    – SacredScout
    Dec 1 '18 at 23:04






  • 1




    $begingroup$
    I simply used Geometric Sum (en.wikipedia.org/wiki/Geometric_series)
    $endgroup$
    – Mostafa Ayaz
    Dec 2 '18 at 13:20



















1












$begingroup$

We are told that $|a_n|^{frac1n}leq theta$ whenever $n geq n_0$. So if $n geq n_0$, it then follows that we have $0leq |a_n|leqtheta^n$. Since $theta in (0,1)$, it also follows that $sum_{n=1}^infty theta^n$ is a convergent geometric series. Now the comparison test seems relevant...



Always check to make sure you are using the hypotheses of the problem, otherwise things tend to go agley.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $sqrt[n]{mid a_nmid}ltthetaimplies mid a_nmidletheta ^nimplies sum_{n=0}^infty mid a_nmidlesum_{n=0}^inftytheta^n=frac 1{1-theta}ltinfty $, since $theta lt1$.



    Thus the series converges absolutely by comparison with a geometric series.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How do you conclude that it equals $1/1-theta$ at the end?
      $endgroup$
      – SacredScout
      Dec 1 '18 at 23:03








    • 2




      $begingroup$
      $(1-theta)(1+theta+theta^2+dots+theta ^n)=1-theta ^{n+1}$. Divide both sides by $(1-theta) $ and let $ntoinfty$. This is known as "telescoping", because of all the cancellation.
      $endgroup$
      – Chris Custer
      Dec 1 '18 at 23:25











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    We have $$sqrt[n]{|a_n|}le thetato 0le |a_n|le theta^n$$therefore $$0le sum_{n=1}^{infty}|a_n|le sum_{n=1}^{infty}theta ^n={thetaover 1-theta}$$which completes the proof.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How do you conclude that it equals $1/1-theta$ at the end?
      $endgroup$
      – SacredScout
      Dec 1 '18 at 23:04






    • 1




      $begingroup$
      I simply used Geometric Sum (en.wikipedia.org/wiki/Geometric_series)
      $endgroup$
      – Mostafa Ayaz
      Dec 2 '18 at 13:20
















    2












    $begingroup$

    We have $$sqrt[n]{|a_n|}le thetato 0le |a_n|le theta^n$$therefore $$0le sum_{n=1}^{infty}|a_n|le sum_{n=1}^{infty}theta ^n={thetaover 1-theta}$$which completes the proof.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How do you conclude that it equals $1/1-theta$ at the end?
      $endgroup$
      – SacredScout
      Dec 1 '18 at 23:04






    • 1




      $begingroup$
      I simply used Geometric Sum (en.wikipedia.org/wiki/Geometric_series)
      $endgroup$
      – Mostafa Ayaz
      Dec 2 '18 at 13:20














    2












    2








    2





    $begingroup$

    We have $$sqrt[n]{|a_n|}le thetato 0le |a_n|le theta^n$$therefore $$0le sum_{n=1}^{infty}|a_n|le sum_{n=1}^{infty}theta ^n={thetaover 1-theta}$$which completes the proof.






    share|cite|improve this answer











    $endgroup$



    We have $$sqrt[n]{|a_n|}le thetato 0le |a_n|le theta^n$$therefore $$0le sum_{n=1}^{infty}|a_n|le sum_{n=1}^{infty}theta ^n={thetaover 1-theta}$$which completes the proof.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 1 '18 at 22:16









    Chris Custer

    11.2k3824




    11.2k3824










    answered Dec 1 '18 at 21:34









    Mostafa AyazMostafa Ayaz

    15.2k3939




    15.2k3939












    • $begingroup$
      How do you conclude that it equals $1/1-theta$ at the end?
      $endgroup$
      – SacredScout
      Dec 1 '18 at 23:04






    • 1




      $begingroup$
      I simply used Geometric Sum (en.wikipedia.org/wiki/Geometric_series)
      $endgroup$
      – Mostafa Ayaz
      Dec 2 '18 at 13:20


















    • $begingroup$
      How do you conclude that it equals $1/1-theta$ at the end?
      $endgroup$
      – SacredScout
      Dec 1 '18 at 23:04






    • 1




      $begingroup$
      I simply used Geometric Sum (en.wikipedia.org/wiki/Geometric_series)
      $endgroup$
      – Mostafa Ayaz
      Dec 2 '18 at 13:20
















    $begingroup$
    How do you conclude that it equals $1/1-theta$ at the end?
    $endgroup$
    – SacredScout
    Dec 1 '18 at 23:04




    $begingroup$
    How do you conclude that it equals $1/1-theta$ at the end?
    $endgroup$
    – SacredScout
    Dec 1 '18 at 23:04




    1




    1




    $begingroup$
    I simply used Geometric Sum (en.wikipedia.org/wiki/Geometric_series)
    $endgroup$
    – Mostafa Ayaz
    Dec 2 '18 at 13:20




    $begingroup$
    I simply used Geometric Sum (en.wikipedia.org/wiki/Geometric_series)
    $endgroup$
    – Mostafa Ayaz
    Dec 2 '18 at 13:20











    1












    $begingroup$

    We are told that $|a_n|^{frac1n}leq theta$ whenever $n geq n_0$. So if $n geq n_0$, it then follows that we have $0leq |a_n|leqtheta^n$. Since $theta in (0,1)$, it also follows that $sum_{n=1}^infty theta^n$ is a convergent geometric series. Now the comparison test seems relevant...



    Always check to make sure you are using the hypotheses of the problem, otherwise things tend to go agley.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      We are told that $|a_n|^{frac1n}leq theta$ whenever $n geq n_0$. So if $n geq n_0$, it then follows that we have $0leq |a_n|leqtheta^n$. Since $theta in (0,1)$, it also follows that $sum_{n=1}^infty theta^n$ is a convergent geometric series. Now the comparison test seems relevant...



      Always check to make sure you are using the hypotheses of the problem, otherwise things tend to go agley.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        We are told that $|a_n|^{frac1n}leq theta$ whenever $n geq n_0$. So if $n geq n_0$, it then follows that we have $0leq |a_n|leqtheta^n$. Since $theta in (0,1)$, it also follows that $sum_{n=1}^infty theta^n$ is a convergent geometric series. Now the comparison test seems relevant...



        Always check to make sure you are using the hypotheses of the problem, otherwise things tend to go agley.






        share|cite|improve this answer









        $endgroup$



        We are told that $|a_n|^{frac1n}leq theta$ whenever $n geq n_0$. So if $n geq n_0$, it then follows that we have $0leq |a_n|leqtheta^n$. Since $theta in (0,1)$, it also follows that $sum_{n=1}^infty theta^n$ is a convergent geometric series. Now the comparison test seems relevant...



        Always check to make sure you are using the hypotheses of the problem, otherwise things tend to go agley.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 '18 at 21:36









        Matt A PeltoMatt A Pelto

        2,477620




        2,477620























            1












            $begingroup$

            $sqrt[n]{mid a_nmid}ltthetaimplies mid a_nmidletheta ^nimplies sum_{n=0}^infty mid a_nmidlesum_{n=0}^inftytheta^n=frac 1{1-theta}ltinfty $, since $theta lt1$.



            Thus the series converges absolutely by comparison with a geometric series.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              How do you conclude that it equals $1/1-theta$ at the end?
              $endgroup$
              – SacredScout
              Dec 1 '18 at 23:03








            • 2




              $begingroup$
              $(1-theta)(1+theta+theta^2+dots+theta ^n)=1-theta ^{n+1}$. Divide both sides by $(1-theta) $ and let $ntoinfty$. This is known as "telescoping", because of all the cancellation.
              $endgroup$
              – Chris Custer
              Dec 1 '18 at 23:25
















            1












            $begingroup$

            $sqrt[n]{mid a_nmid}ltthetaimplies mid a_nmidletheta ^nimplies sum_{n=0}^infty mid a_nmidlesum_{n=0}^inftytheta^n=frac 1{1-theta}ltinfty $, since $theta lt1$.



            Thus the series converges absolutely by comparison with a geometric series.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              How do you conclude that it equals $1/1-theta$ at the end?
              $endgroup$
              – SacredScout
              Dec 1 '18 at 23:03








            • 2




              $begingroup$
              $(1-theta)(1+theta+theta^2+dots+theta ^n)=1-theta ^{n+1}$. Divide both sides by $(1-theta) $ and let $ntoinfty$. This is known as "telescoping", because of all the cancellation.
              $endgroup$
              – Chris Custer
              Dec 1 '18 at 23:25














            1












            1








            1





            $begingroup$

            $sqrt[n]{mid a_nmid}ltthetaimplies mid a_nmidletheta ^nimplies sum_{n=0}^infty mid a_nmidlesum_{n=0}^inftytheta^n=frac 1{1-theta}ltinfty $, since $theta lt1$.



            Thus the series converges absolutely by comparison with a geometric series.






            share|cite|improve this answer











            $endgroup$



            $sqrt[n]{mid a_nmid}ltthetaimplies mid a_nmidletheta ^nimplies sum_{n=0}^infty mid a_nmidlesum_{n=0}^inftytheta^n=frac 1{1-theta}ltinfty $, since $theta lt1$.



            Thus the series converges absolutely by comparison with a geometric series.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 1 '18 at 22:14

























            answered Dec 1 '18 at 21:48









            Chris CusterChris Custer

            11.2k3824




            11.2k3824












            • $begingroup$
              How do you conclude that it equals $1/1-theta$ at the end?
              $endgroup$
              – SacredScout
              Dec 1 '18 at 23:03








            • 2




              $begingroup$
              $(1-theta)(1+theta+theta^2+dots+theta ^n)=1-theta ^{n+1}$. Divide both sides by $(1-theta) $ and let $ntoinfty$. This is known as "telescoping", because of all the cancellation.
              $endgroup$
              – Chris Custer
              Dec 1 '18 at 23:25


















            • $begingroup$
              How do you conclude that it equals $1/1-theta$ at the end?
              $endgroup$
              – SacredScout
              Dec 1 '18 at 23:03








            • 2




              $begingroup$
              $(1-theta)(1+theta+theta^2+dots+theta ^n)=1-theta ^{n+1}$. Divide both sides by $(1-theta) $ and let $ntoinfty$. This is known as "telescoping", because of all the cancellation.
              $endgroup$
              – Chris Custer
              Dec 1 '18 at 23:25
















            $begingroup$
            How do you conclude that it equals $1/1-theta$ at the end?
            $endgroup$
            – SacredScout
            Dec 1 '18 at 23:03






            $begingroup$
            How do you conclude that it equals $1/1-theta$ at the end?
            $endgroup$
            – SacredScout
            Dec 1 '18 at 23:03






            2




            2




            $begingroup$
            $(1-theta)(1+theta+theta^2+dots+theta ^n)=1-theta ^{n+1}$. Divide both sides by $(1-theta) $ and let $ntoinfty$. This is known as "telescoping", because of all the cancellation.
            $endgroup$
            – Chris Custer
            Dec 1 '18 at 23:25




            $begingroup$
            $(1-theta)(1+theta+theta^2+dots+theta ^n)=1-theta ^{n+1}$. Divide both sides by $(1-theta) $ and let $ntoinfty$. This is known as "telescoping", because of all the cancellation.
            $endgroup$
            – Chris Custer
            Dec 1 '18 at 23:25


















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