Uniform convergence of $ U_n(x) = sum_{n=0}^{+ infty} (-1)^n ln ( 1 + frac{x}{1+ nx} ) $.












0












$begingroup$



We consider the series of functions:



$$U_n(x) = sum_{k=0}^{n} (-1)^k ln left( 1 + frac{x}{1+ kx} right) ,~ x geq0.$$




  1. Prove that $U_n$ is convergent.


  2. Study the uniform convergence of $U_n$.


  3. Study the normal convergence.


  4. We consider $U(x) = sum_{n=0}^{+ infty} (-1)^n ln ( 1 + frac{x}{1+ nx} )$. Prove that $U(x)$ is of class $C^1$.


  5. Compute $U'(x)$.






I have problem with question 2 and 4.



For question 2. I do not know how to answer this question since I do not know the value of the sum $U_n (x)$ to compute : $lim sup |U_n(x) - l| $. How to know if the series have uniform convergence?



For question 4. $U(x)$ is of class $C^1$ means it is differentiable and its derivative is continuous. If $ sum U_n$ converges uniformaly, $U$ will have the same properties, but how can I can prove that $U$ is of $C^1$ without computing $U'(x)$ first? I am confused because in the last question I am asked to compute $U'(x)$. Is there a way to deduce that from uniform convergence?






real-analysis sequences-and-series







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$endgroup$

















    0












    $begingroup$



    We consider the series of functions:



    $$U_n(x) = sum_{k=0}^{n} (-1)^k ln left( 1 + frac{x}{1+ kx} right) ,~ x geq0.$$




    1. Prove that $U_n$ is convergent.


    2. Study the uniform convergence of $U_n$.


    3. Study the normal convergence.


    4. We consider $U(x) = sum_{n=0}^{+ infty} (-1)^n ln ( 1 + frac{x}{1+ nx} )$. Prove that $U(x)$ is of class $C^1$.


    5. Compute $U'(x)$.






    I have problem with question 2 and 4.



    For question 2. I do not know how to answer this question since I do not know the value of the sum $U_n (x)$ to compute : $lim sup |U_n(x) - l| $. How to know if the series have uniform convergence?



    For question 4. $U(x)$ is of class $C^1$ means it is differentiable and its derivative is continuous. If $ sum U_n$ converges uniformaly, $U$ will have the same properties, but how can I can prove that $U$ is of $C^1$ without computing $U'(x)$ first? I am confused because in the last question I am asked to compute $U'(x)$. Is there a way to deduce that from uniform convergence?






    real-analysis sequences-and-series







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      We consider the series of functions:



      $$U_n(x) = sum_{k=0}^{n} (-1)^k ln left( 1 + frac{x}{1+ kx} right) ,~ x geq0.$$




      1. Prove that $U_n$ is convergent.


      2. Study the uniform convergence of $U_n$.


      3. Study the normal convergence.


      4. We consider $U(x) = sum_{n=0}^{+ infty} (-1)^n ln ( 1 + frac{x}{1+ nx} )$. Prove that $U(x)$ is of class $C^1$.


      5. Compute $U'(x)$.






      I have problem with question 2 and 4.



      For question 2. I do not know how to answer this question since I do not know the value of the sum $U_n (x)$ to compute : $lim sup |U_n(x) - l| $. How to know if the series have uniform convergence?



      For question 4. $U(x)$ is of class $C^1$ means it is differentiable and its derivative is continuous. If $ sum U_n$ converges uniformaly, $U$ will have the same properties, but how can I can prove that $U$ is of $C^1$ without computing $U'(x)$ first? I am confused because in the last question I am asked to compute $U'(x)$. Is there a way to deduce that from uniform convergence?






      real-analysis sequences-and-series







      share|cite|improve this question











      $endgroup$





      We consider the series of functions:



      $$U_n(x) = sum_{k=0}^{n} (-1)^k ln left( 1 + frac{x}{1+ kx} right) ,~ x geq0.$$




      1. Prove that $U_n$ is convergent.


      2. Study the uniform convergence of $U_n$.


      3. Study the normal convergence.


      4. We consider $U(x) = sum_{n=0}^{+ infty} (-1)^n ln ( 1 + frac{x}{1+ nx} )$. Prove that $U(x)$ is of class $C^1$.


      5. Compute $U'(x)$.






      I have problem with question 2 and 4.



      For question 2. I do not know how to answer this question since I do not know the value of the sum $U_n (x)$ to compute : $lim sup |U_n(x) - l| $. How to know if the series have uniform convergence?



      For question 4. $U(x)$ is of class $C^1$ means it is differentiable and its derivative is continuous. If $ sum U_n$ converges uniformaly, $U$ will have the same properties, but how can I can prove that $U$ is of $C^1$ without computing $U'(x)$ first? I am confused because in the last question I am asked to compute $U'(x)$. Is there a way to deduce that from uniform convergence?






      real-analysis sequences-and-series




      real-analysis sequences-and-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 28 '18 at 23:56









      Larry

      2,41131129




      2,41131129










      asked Dec 28 '18 at 14:08









      Zouhair El YaagoubiZouhair El Yaagoubi

      538411




      538411






















          1 Answer
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          -1












          $begingroup$

          This is a series with alternating sign terms, so a good idea would be to split it into even and odd terms. For even $n$, say $n=2m$, we have
          $$U_{2m-1}(x) = sum_{k=0}^m ln left( 1+ frac{x}{1+2kx} right) - ln left( 1+ frac{x}{1+(2k+1)x} right)$$ which (after some calculations) turns out to be
          $$sum_{k=0}^m ln left( 1+ frac{x^2}{(1+2kx)(1+2(k+1)x)} right)$$



          As $m to + infty$ this series is normally convergent, thus $U_{2m-1}(x)$ is uniformly convergent on $Bbb R$. Now you should study the series $U_{2m}$ and conclude similarly.



          Thus $U_n$ is uniformly convergent, although it is not normally convergent (since it is not absolutely convergent).



          As for differentiability of $U(x)$, you have to apply this limit .






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            In the order of the problem, I am asked about the uniform convergence first. You have started with the normal convergence. Although, I found that: $sum sup |U_n(x)| = sum ln (1 + frac{1}{n}) = + infty $, which is not normally convergent. For your last sentence, I want to know about the continuity of $U'(x)$ not the differentiability. I need to prove that $U(x) in C^1$
            $endgroup$
            – Zouhair El Yaagoubi
            Dec 28 '18 at 16:45












          • $begingroup$
            Indeed what I proved is that the series converges uniformly. After that, you have to study the uniform convergence of $U'_n(x)$ in a similar way: since those are continuous, they will converge to a continuous $U'(x)$.
            $endgroup$
            – Crostul
            Dec 28 '18 at 18:04











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          1 Answer
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          active

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          -1












          $begingroup$

          This is a series with alternating sign terms, so a good idea would be to split it into even and odd terms. For even $n$, say $n=2m$, we have
          $$U_{2m-1}(x) = sum_{k=0}^m ln left( 1+ frac{x}{1+2kx} right) - ln left( 1+ frac{x}{1+(2k+1)x} right)$$ which (after some calculations) turns out to be
          $$sum_{k=0}^m ln left( 1+ frac{x^2}{(1+2kx)(1+2(k+1)x)} right)$$



          As $m to + infty$ this series is normally convergent, thus $U_{2m-1}(x)$ is uniformly convergent on $Bbb R$. Now you should study the series $U_{2m}$ and conclude similarly.



          Thus $U_n$ is uniformly convergent, although it is not normally convergent (since it is not absolutely convergent).



          As for differentiability of $U(x)$, you have to apply this limit .






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            In the order of the problem, I am asked about the uniform convergence first. You have started with the normal convergence. Although, I found that: $sum sup |U_n(x)| = sum ln (1 + frac{1}{n}) = + infty $, which is not normally convergent. For your last sentence, I want to know about the continuity of $U'(x)$ not the differentiability. I need to prove that $U(x) in C^1$
            $endgroup$
            – Zouhair El Yaagoubi
            Dec 28 '18 at 16:45












          • $begingroup$
            Indeed what I proved is that the series converges uniformly. After that, you have to study the uniform convergence of $U'_n(x)$ in a similar way: since those are continuous, they will converge to a continuous $U'(x)$.
            $endgroup$
            – Crostul
            Dec 28 '18 at 18:04
















          -1












          $begingroup$

          This is a series with alternating sign terms, so a good idea would be to split it into even and odd terms. For even $n$, say $n=2m$, we have
          $$U_{2m-1}(x) = sum_{k=0}^m ln left( 1+ frac{x}{1+2kx} right) - ln left( 1+ frac{x}{1+(2k+1)x} right)$$ which (after some calculations) turns out to be
          $$sum_{k=0}^m ln left( 1+ frac{x^2}{(1+2kx)(1+2(k+1)x)} right)$$



          As $m to + infty$ this series is normally convergent, thus $U_{2m-1}(x)$ is uniformly convergent on $Bbb R$. Now you should study the series $U_{2m}$ and conclude similarly.



          Thus $U_n$ is uniformly convergent, although it is not normally convergent (since it is not absolutely convergent).



          As for differentiability of $U(x)$, you have to apply this limit .






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            In the order of the problem, I am asked about the uniform convergence first. You have started with the normal convergence. Although, I found that: $sum sup |U_n(x)| = sum ln (1 + frac{1}{n}) = + infty $, which is not normally convergent. For your last sentence, I want to know about the continuity of $U'(x)$ not the differentiability. I need to prove that $U(x) in C^1$
            $endgroup$
            – Zouhair El Yaagoubi
            Dec 28 '18 at 16:45












          • $begingroup$
            Indeed what I proved is that the series converges uniformly. After that, you have to study the uniform convergence of $U'_n(x)$ in a similar way: since those are continuous, they will converge to a continuous $U'(x)$.
            $endgroup$
            – Crostul
            Dec 28 '18 at 18:04














          -1












          -1








          -1





          $begingroup$

          This is a series with alternating sign terms, so a good idea would be to split it into even and odd terms. For even $n$, say $n=2m$, we have
          $$U_{2m-1}(x) = sum_{k=0}^m ln left( 1+ frac{x}{1+2kx} right) - ln left( 1+ frac{x}{1+(2k+1)x} right)$$ which (after some calculations) turns out to be
          $$sum_{k=0}^m ln left( 1+ frac{x^2}{(1+2kx)(1+2(k+1)x)} right)$$



          As $m to + infty$ this series is normally convergent, thus $U_{2m-1}(x)$ is uniformly convergent on $Bbb R$. Now you should study the series $U_{2m}$ and conclude similarly.



          Thus $U_n$ is uniformly convergent, although it is not normally convergent (since it is not absolutely convergent).



          As for differentiability of $U(x)$, you have to apply this limit .






          share|cite|improve this answer









          $endgroup$



          This is a series with alternating sign terms, so a good idea would be to split it into even and odd terms. For even $n$, say $n=2m$, we have
          $$U_{2m-1}(x) = sum_{k=0}^m ln left( 1+ frac{x}{1+2kx} right) - ln left( 1+ frac{x}{1+(2k+1)x} right)$$ which (after some calculations) turns out to be
          $$sum_{k=0}^m ln left( 1+ frac{x^2}{(1+2kx)(1+2(k+1)x)} right)$$



          As $m to + infty$ this series is normally convergent, thus $U_{2m-1}(x)$ is uniformly convergent on $Bbb R$. Now you should study the series $U_{2m}$ and conclude similarly.



          Thus $U_n$ is uniformly convergent, although it is not normally convergent (since it is not absolutely convergent).



          As for differentiability of $U(x)$, you have to apply this limit .







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 28 '18 at 15:17









          CrostulCrostul

          27.9k22352




          27.9k22352












          • $begingroup$
            In the order of the problem, I am asked about the uniform convergence first. You have started with the normal convergence. Although, I found that: $sum sup |U_n(x)| = sum ln (1 + frac{1}{n}) = + infty $, which is not normally convergent. For your last sentence, I want to know about the continuity of $U'(x)$ not the differentiability. I need to prove that $U(x) in C^1$
            $endgroup$
            – Zouhair El Yaagoubi
            Dec 28 '18 at 16:45












          • $begingroup$
            Indeed what I proved is that the series converges uniformly. After that, you have to study the uniform convergence of $U'_n(x)$ in a similar way: since those are continuous, they will converge to a continuous $U'(x)$.
            $endgroup$
            – Crostul
            Dec 28 '18 at 18:04


















          • $begingroup$
            In the order of the problem, I am asked about the uniform convergence first. You have started with the normal convergence. Although, I found that: $sum sup |U_n(x)| = sum ln (1 + frac{1}{n}) = + infty $, which is not normally convergent. For your last sentence, I want to know about the continuity of $U'(x)$ not the differentiability. I need to prove that $U(x) in C^1$
            $endgroup$
            – Zouhair El Yaagoubi
            Dec 28 '18 at 16:45












          • $begingroup$
            Indeed what I proved is that the series converges uniformly. After that, you have to study the uniform convergence of $U'_n(x)$ in a similar way: since those are continuous, they will converge to a continuous $U'(x)$.
            $endgroup$
            – Crostul
            Dec 28 '18 at 18:04
















          $begingroup$
          In the order of the problem, I am asked about the uniform convergence first. You have started with the normal convergence. Although, I found that: $sum sup |U_n(x)| = sum ln (1 + frac{1}{n}) = + infty $, which is not normally convergent. For your last sentence, I want to know about the continuity of $U'(x)$ not the differentiability. I need to prove that $U(x) in C^1$
          $endgroup$
          – Zouhair El Yaagoubi
          Dec 28 '18 at 16:45






          $begingroup$
          In the order of the problem, I am asked about the uniform convergence first. You have started with the normal convergence. Although, I found that: $sum sup |U_n(x)| = sum ln (1 + frac{1}{n}) = + infty $, which is not normally convergent. For your last sentence, I want to know about the continuity of $U'(x)$ not the differentiability. I need to prove that $U(x) in C^1$
          $endgroup$
          – Zouhair El Yaagoubi
          Dec 28 '18 at 16:45














          $begingroup$
          Indeed what I proved is that the series converges uniformly. After that, you have to study the uniform convergence of $U'_n(x)$ in a similar way: since those are continuous, they will converge to a continuous $U'(x)$.
          $endgroup$
          – Crostul
          Dec 28 '18 at 18:04




          $begingroup$
          Indeed what I proved is that the series converges uniformly. After that, you have to study the uniform convergence of $U'_n(x)$ in a similar way: since those are continuous, they will converge to a continuous $U'(x)$.
          $endgroup$
          – Crostul
          Dec 28 '18 at 18:04


















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