Uniform Convergence of $sum_{n=1}^{infty}frac{x}{[(n-1)x+1][nx+1]}$












-1












$begingroup$


What can be said about the uniform Convergence of $sum_{n=1}^{infty}frac{x}{[(n-1)x+1][nx+1]}$ in the interval $[0,1]$?



The sequence inside the summation bracket doesn't seem to yield to root or ratio tests. The pointwise convergence itself seems doubtful. Should we use Cauchy-Criterion directly here? Maybe some comparison would be useful in this case? And what about uniform convergence? Any hints? Thanks beforehand.






real-analysis sequences-and-series uniform-convergence







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  • $begingroup$
    Uniform convergence where?
    $endgroup$
    – José Carlos Santos
    Jan 24 at 12:53










  • $begingroup$
    @JoséCarlosSantos thanks! edited the question
    $endgroup$
    – vidyarthi
    Jan 24 at 12:54










  • $begingroup$
    What is allowed? Why don't you just compute partial sums directly?
    $endgroup$
    – metamorphy
    Jan 25 at 7:38
















-1












$begingroup$


What can be said about the uniform Convergence of $sum_{n=1}^{infty}frac{x}{[(n-1)x+1][nx+1]}$ in the interval $[0,1]$?



The sequence inside the summation bracket doesn't seem to yield to root or ratio tests. The pointwise convergence itself seems doubtful. Should we use Cauchy-Criterion directly here? Maybe some comparison would be useful in this case? And what about uniform convergence? Any hints? Thanks beforehand.






real-analysis sequences-and-series uniform-convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Uniform convergence where?
    $endgroup$
    – José Carlos Santos
    Jan 24 at 12:53










  • $begingroup$
    @JoséCarlosSantos thanks! edited the question
    $endgroup$
    – vidyarthi
    Jan 24 at 12:54










  • $begingroup$
    What is allowed? Why don't you just compute partial sums directly?
    $endgroup$
    – metamorphy
    Jan 25 at 7:38














-1












-1








-1





$begingroup$


What can be said about the uniform Convergence of $sum_{n=1}^{infty}frac{x}{[(n-1)x+1][nx+1]}$ in the interval $[0,1]$?



The sequence inside the summation bracket doesn't seem to yield to root or ratio tests. The pointwise convergence itself seems doubtful. Should we use Cauchy-Criterion directly here? Maybe some comparison would be useful in this case? And what about uniform convergence? Any hints? Thanks beforehand.






real-analysis sequences-and-series uniform-convergence







share|cite|improve this question











$endgroup$




What can be said about the uniform Convergence of $sum_{n=1}^{infty}frac{x}{[(n-1)x+1][nx+1]}$ in the interval $[0,1]$?



The sequence inside the summation bracket doesn't seem to yield to root or ratio tests. The pointwise convergence itself seems doubtful. Should we use Cauchy-Criterion directly here? Maybe some comparison would be useful in this case? And what about uniform convergence? Any hints? Thanks beforehand.






real-analysis sequences-and-series uniform-convergence




real-analysis sequences-and-series uniform-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 12:57









José Carlos Santos

168k22132236




168k22132236










asked Jan 24 at 12:49









vidyarthividyarthi

3,0341833




3,0341833












  • $begingroup$
    Uniform convergence where?
    $endgroup$
    – José Carlos Santos
    Jan 24 at 12:53










  • $begingroup$
    @JoséCarlosSantos thanks! edited the question
    $endgroup$
    – vidyarthi
    Jan 24 at 12:54










  • $begingroup$
    What is allowed? Why don't you just compute partial sums directly?
    $endgroup$
    – metamorphy
    Jan 25 at 7:38


















  • $begingroup$
    Uniform convergence where?
    $endgroup$
    – José Carlos Santos
    Jan 24 at 12:53










  • $begingroup$
    @JoséCarlosSantos thanks! edited the question
    $endgroup$
    – vidyarthi
    Jan 24 at 12:54










  • $begingroup$
    What is allowed? Why don't you just compute partial sums directly?
    $endgroup$
    – metamorphy
    Jan 25 at 7:38
















$begingroup$
Uniform convergence where?
$endgroup$
– José Carlos Santos
Jan 24 at 12:53




$begingroup$
Uniform convergence where?
$endgroup$
– José Carlos Santos
Jan 24 at 12:53












$begingroup$
@JoséCarlosSantos thanks! edited the question
$endgroup$
– vidyarthi
Jan 24 at 12:54




$begingroup$
@JoséCarlosSantos thanks! edited the question
$endgroup$
– vidyarthi
Jan 24 at 12:54












$begingroup$
What is allowed? Why don't you just compute partial sums directly?
$endgroup$
– metamorphy
Jan 25 at 7:38




$begingroup$
What is allowed? Why don't you just compute partial sums directly?
$endgroup$
– metamorphy
Jan 25 at 7:38










2 Answers
2






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0












$begingroup$

Hint: $displaystylefrac1{bigl((n-1)x+1bigr)(nx+1)}=frac1{(n-1)x+1}-frac1{nx+1}$.






share|cite|improve this answer









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  • $begingroup$
    I just missed it though!
    $endgroup$
    – vidyarthi
    Jan 24 at 13:00










  • $begingroup$
    though this would be biased, but I think my own answer above is more detailed, so you miss $25$ rep!
    $endgroup$
    – vidyarthi
    Jan 25 at 9:58












  • $begingroup$
    yes, done it and you got it
    $endgroup$
    – vidyarthi
    Jan 25 at 10:01



















0












$begingroup$

The series can be computed by using telescoping. We have $frac{x}{[(n-1)x+1][nx+1]}=frac1{(n-1)x+1}-frac1{nx+1}$. Thus, the sequence of partial sums would be $s_n=frac{nx}{nx+1}$. Hence, sum is equal to $$begin{cases}0 text{when} x=0\1 text{when} xneq0end{cases}$$. This easily shows non-uniform convergence, as the limit is discontinuous at $0$.






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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    0












    $begingroup$

    Hint: $displaystylefrac1{bigl((n-1)x+1bigr)(nx+1)}=frac1{(n-1)x+1}-frac1{nx+1}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I just missed it though!
      $endgroup$
      – vidyarthi
      Jan 24 at 13:00










    • $begingroup$
      though this would be biased, but I think my own answer above is more detailed, so you miss $25$ rep!
      $endgroup$
      – vidyarthi
      Jan 25 at 9:58












    • $begingroup$
      yes, done it and you got it
      $endgroup$
      – vidyarthi
      Jan 25 at 10:01
















    0












    $begingroup$

    Hint: $displaystylefrac1{bigl((n-1)x+1bigr)(nx+1)}=frac1{(n-1)x+1}-frac1{nx+1}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I just missed it though!
      $endgroup$
      – vidyarthi
      Jan 24 at 13:00










    • $begingroup$
      though this would be biased, but I think my own answer above is more detailed, so you miss $25$ rep!
      $endgroup$
      – vidyarthi
      Jan 25 at 9:58












    • $begingroup$
      yes, done it and you got it
      $endgroup$
      – vidyarthi
      Jan 25 at 10:01














    0












    0








    0





    $begingroup$

    Hint: $displaystylefrac1{bigl((n-1)x+1bigr)(nx+1)}=frac1{(n-1)x+1}-frac1{nx+1}$.






    share|cite|improve this answer









    $endgroup$



    Hint: $displaystylefrac1{bigl((n-1)x+1bigr)(nx+1)}=frac1{(n-1)x+1}-frac1{nx+1}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 24 at 12:54









    José Carlos SantosJosé Carlos Santos

    168k22132236




    168k22132236












    • $begingroup$
      I just missed it though!
      $endgroup$
      – vidyarthi
      Jan 24 at 13:00










    • $begingroup$
      though this would be biased, but I think my own answer above is more detailed, so you miss $25$ rep!
      $endgroup$
      – vidyarthi
      Jan 25 at 9:58












    • $begingroup$
      yes, done it and you got it
      $endgroup$
      – vidyarthi
      Jan 25 at 10:01


















    • $begingroup$
      I just missed it though!
      $endgroup$
      – vidyarthi
      Jan 24 at 13:00










    • $begingroup$
      though this would be biased, but I think my own answer above is more detailed, so you miss $25$ rep!
      $endgroup$
      – vidyarthi
      Jan 25 at 9:58












    • $begingroup$
      yes, done it and you got it
      $endgroup$
      – vidyarthi
      Jan 25 at 10:01
















    $begingroup$
    I just missed it though!
    $endgroup$
    – vidyarthi
    Jan 24 at 13:00




    $begingroup$
    I just missed it though!
    $endgroup$
    – vidyarthi
    Jan 24 at 13:00












    $begingroup$
    though this would be biased, but I think my own answer above is more detailed, so you miss $25$ rep!
    $endgroup$
    – vidyarthi
    Jan 25 at 9:58






    $begingroup$
    though this would be biased, but I think my own answer above is more detailed, so you miss $25$ rep!
    $endgroup$
    – vidyarthi
    Jan 25 at 9:58














    $begingroup$
    yes, done it and you got it
    $endgroup$
    – vidyarthi
    Jan 25 at 10:01




    $begingroup$
    yes, done it and you got it
    $endgroup$
    – vidyarthi
    Jan 25 at 10:01











    0












    $begingroup$

    The series can be computed by using telescoping. We have $frac{x}{[(n-1)x+1][nx+1]}=frac1{(n-1)x+1}-frac1{nx+1}$. Thus, the sequence of partial sums would be $s_n=frac{nx}{nx+1}$. Hence, sum is equal to $$begin{cases}0 text{when} x=0\1 text{when} xneq0end{cases}$$. This easily shows non-uniform convergence, as the limit is discontinuous at $0$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The series can be computed by using telescoping. We have $frac{x}{[(n-1)x+1][nx+1]}=frac1{(n-1)x+1}-frac1{nx+1}$. Thus, the sequence of partial sums would be $s_n=frac{nx}{nx+1}$. Hence, sum is equal to $$begin{cases}0 text{when} x=0\1 text{when} xneq0end{cases}$$. This easily shows non-uniform convergence, as the limit is discontinuous at $0$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The series can be computed by using telescoping. We have $frac{x}{[(n-1)x+1][nx+1]}=frac1{(n-1)x+1}-frac1{nx+1}$. Thus, the sequence of partial sums would be $s_n=frac{nx}{nx+1}$. Hence, sum is equal to $$begin{cases}0 text{when} x=0\1 text{when} xneq0end{cases}$$. This easily shows non-uniform convergence, as the limit is discontinuous at $0$.






        share|cite|improve this answer









        $endgroup$



        The series can be computed by using telescoping. We have $frac{x}{[(n-1)x+1][nx+1]}=frac1{(n-1)x+1}-frac1{nx+1}$. Thus, the sequence of partial sums would be $s_n=frac{nx}{nx+1}$. Hence, sum is equal to $$begin{cases}0 text{when} x=0\1 text{when} xneq0end{cases}$$. This easily shows non-uniform convergence, as the limit is discontinuous at $0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 24 at 12:59









        vidyarthividyarthi

        3,0341833




        3,0341833






























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