converge pointwise but not uniformly











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How can I prove that
$$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}}$$
converges pointwise on $[-pi, pi]$ but not uniformly?



For the pointwise part, I tried to prove it by comparison, using
$$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}} leq sum_{n=1}^infty frac{nx}{sqrt{n}} = sum_{n=1}^infty xsqrt{n},$$
which does not converge.



I also tried
$$ sum_{n=1}^infty left| frac{sin(nx)}{sqrt{n}}right| leq sum_{n=1}^infty left| frac{1}{sqrt{n}}right|,$$
which does not converge either.






sequences-and-series convergence uniform-convergence pointwise-convergence







share|cite|improve this question




























    up vote
    1
    down vote

    favorite












    How can I prove that
    $$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}}$$
    converges pointwise on $[-pi, pi]$ but not uniformly?



    For the pointwise part, I tried to prove it by comparison, using
    $$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}} leq sum_{n=1}^infty frac{nx}{sqrt{n}} = sum_{n=1}^infty xsqrt{n},$$
    which does not converge.



    I also tried
    $$ sum_{n=1}^infty left| frac{sin(nx)}{sqrt{n}}right| leq sum_{n=1}^infty left| frac{1}{sqrt{n}}right|,$$
    which does not converge either.






    sequences-and-series convergence uniform-convergence pointwise-convergence







    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      How can I prove that
      $$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}}$$
      converges pointwise on $[-pi, pi]$ but not uniformly?



      For the pointwise part, I tried to prove it by comparison, using
      $$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}} leq sum_{n=1}^infty frac{nx}{sqrt{n}} = sum_{n=1}^infty xsqrt{n},$$
      which does not converge.



      I also tried
      $$ sum_{n=1}^infty left| frac{sin(nx)}{sqrt{n}}right| leq sum_{n=1}^infty left| frac{1}{sqrt{n}}right|,$$
      which does not converge either.






      sequences-and-series convergence uniform-convergence pointwise-convergence







      share|cite|improve this question















      How can I prove that
      $$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}}$$
      converges pointwise on $[-pi, pi]$ but not uniformly?



      For the pointwise part, I tried to prove it by comparison, using
      $$sum_{n=1}^infty frac{sin(nx)}{sqrt{n}} leq sum_{n=1}^infty frac{nx}{sqrt{n}} = sum_{n=1}^infty xsqrt{n},$$
      which does not converge.



      I also tried
      $$ sum_{n=1}^infty left| frac{sin(nx)}{sqrt{n}}right| leq sum_{n=1}^infty left| frac{1}{sqrt{n}}right|,$$
      which does not converge either.






      sequences-and-series convergence uniform-convergence pointwise-convergence




      sequences-and-series convergence uniform-convergence pointwise-convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 21 at 21:40









      rafa11111

      1,040417




      1,040417










      asked Nov 21 at 21:04









      kit kat

      91




      91






















          2 Answers
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          down vote













          For the pointwise convergence, use Dirichlet's test.



          To see that it's not uniform, note that $sin(nx) > 2 n x/pi$ if $0 < n x < pi/2$. Take $x = 1/N$ and consider the $N$'th partial sum



          $$ eqalign{sum_{n=1}^N frac{sin(n/N)}{
          sqrt{n}} &ge frac{2}{pi N} sum_{n=1}^N sqrt{n} cr&ge frac{2}{pi N} int_0^N
          sqrt{t}; dt = frac{4sqrt{N}}{3pi}}$$






          share|cite|improve this answer




























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            Pointwise convergence is granted by Dirichlet's test, but a series of continuous functions cannot converge uniformly to an unbounded function. We may compute $lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}$ through a convolution with an approximate identity:



            $$begin{eqnarray*}lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}int_{0}^{+infty}msin(nx) e^{-mx},dx\&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}cdotfrac{mn}{m^2+n^2}.end{eqnarray*}$$
            The RHS is $+infty$ since by Riemann sums
            $$ lim_{mto +infty}sum_{ngeq 1}frac{sqrt{mn}}{m^2+n^2}=int_{0}^{+infty}frac{sqrt{x}}{1+x^2},dx = frac{pi}{sqrt{2}}.$$
            Much simpler, once $f(x)$ is defined as $sum_{ngeq 1}frac{sin(nx)}{sqrt{n}}$ we have
            $$ int_{-pi}^{pi}f(x)^2,dx = pisum_{ngeq 1}frac{1}{n} = +infty $$
            by Parseval's theorem, hence $f(x)$ cannot be bounded on $[-pi,pi]$.






            share|cite|improve this answer























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              2 Answers
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              2 Answers
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              up vote
              2
              down vote













              For the pointwise convergence, use Dirichlet's test.



              To see that it's not uniform, note that $sin(nx) > 2 n x/pi$ if $0 < n x < pi/2$. Take $x = 1/N$ and consider the $N$'th partial sum



              $$ eqalign{sum_{n=1}^N frac{sin(n/N)}{
              sqrt{n}} &ge frac{2}{pi N} sum_{n=1}^N sqrt{n} cr&ge frac{2}{pi N} int_0^N
              sqrt{t}; dt = frac{4sqrt{N}}{3pi}}$$






              share|cite|improve this answer

























                up vote
                2
                down vote













                For the pointwise convergence, use Dirichlet's test.



                To see that it's not uniform, note that $sin(nx) > 2 n x/pi$ if $0 < n x < pi/2$. Take $x = 1/N$ and consider the $N$'th partial sum



                $$ eqalign{sum_{n=1}^N frac{sin(n/N)}{
                sqrt{n}} &ge frac{2}{pi N} sum_{n=1}^N sqrt{n} cr&ge frac{2}{pi N} int_0^N
                sqrt{t}; dt = frac{4sqrt{N}}{3pi}}$$






                share|cite|improve this answer























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  For the pointwise convergence, use Dirichlet's test.



                  To see that it's not uniform, note that $sin(nx) > 2 n x/pi$ if $0 < n x < pi/2$. Take $x = 1/N$ and consider the $N$'th partial sum



                  $$ eqalign{sum_{n=1}^N frac{sin(n/N)}{
                  sqrt{n}} &ge frac{2}{pi N} sum_{n=1}^N sqrt{n} cr&ge frac{2}{pi N} int_0^N
                  sqrt{t}; dt = frac{4sqrt{N}}{3pi}}$$






                  share|cite|improve this answer












                  For the pointwise convergence, use Dirichlet's test.



                  To see that it's not uniform, note that $sin(nx) > 2 n x/pi$ if $0 < n x < pi/2$. Take $x = 1/N$ and consider the $N$'th partial sum



                  $$ eqalign{sum_{n=1}^N frac{sin(n/N)}{
                  sqrt{n}} &ge frac{2}{pi N} sum_{n=1}^N sqrt{n} cr&ge frac{2}{pi N} int_0^N
                  sqrt{t}; dt = frac{4sqrt{N}}{3pi}}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 21 at 21:20









                  Robert Israel

                  316k23206457




                  316k23206457






















                      up vote
                      1
                      down vote













                      Pointwise convergence is granted by Dirichlet's test, but a series of continuous functions cannot converge uniformly to an unbounded function. We may compute $lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}$ through a convolution with an approximate identity:



                      $$begin{eqnarray*}lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}int_{0}^{+infty}msin(nx) e^{-mx},dx\&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}cdotfrac{mn}{m^2+n^2}.end{eqnarray*}$$
                      The RHS is $+infty$ since by Riemann sums
                      $$ lim_{mto +infty}sum_{ngeq 1}frac{sqrt{mn}}{m^2+n^2}=int_{0}^{+infty}frac{sqrt{x}}{1+x^2},dx = frac{pi}{sqrt{2}}.$$
                      Much simpler, once $f(x)$ is defined as $sum_{ngeq 1}frac{sin(nx)}{sqrt{n}}$ we have
                      $$ int_{-pi}^{pi}f(x)^2,dx = pisum_{ngeq 1}frac{1}{n} = +infty $$
                      by Parseval's theorem, hence $f(x)$ cannot be bounded on $[-pi,pi]$.






                      share|cite|improve this answer



























                        up vote
                        1
                        down vote













                        Pointwise convergence is granted by Dirichlet's test, but a series of continuous functions cannot converge uniformly to an unbounded function. We may compute $lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}$ through a convolution with an approximate identity:



                        $$begin{eqnarray*}lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}int_{0}^{+infty}msin(nx) e^{-mx},dx\&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}cdotfrac{mn}{m^2+n^2}.end{eqnarray*}$$
                        The RHS is $+infty$ since by Riemann sums
                        $$ lim_{mto +infty}sum_{ngeq 1}frac{sqrt{mn}}{m^2+n^2}=int_{0}^{+infty}frac{sqrt{x}}{1+x^2},dx = frac{pi}{sqrt{2}}.$$
                        Much simpler, once $f(x)$ is defined as $sum_{ngeq 1}frac{sin(nx)}{sqrt{n}}$ we have
                        $$ int_{-pi}^{pi}f(x)^2,dx = pisum_{ngeq 1}frac{1}{n} = +infty $$
                        by Parseval's theorem, hence $f(x)$ cannot be bounded on $[-pi,pi]$.






                        share|cite|improve this answer

























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          Pointwise convergence is granted by Dirichlet's test, but a series of continuous functions cannot converge uniformly to an unbounded function. We may compute $lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}$ through a convolution with an approximate identity:



                          $$begin{eqnarray*}lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}int_{0}^{+infty}msin(nx) e^{-mx},dx\&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}cdotfrac{mn}{m^2+n^2}.end{eqnarray*}$$
                          The RHS is $+infty$ since by Riemann sums
                          $$ lim_{mto +infty}sum_{ngeq 1}frac{sqrt{mn}}{m^2+n^2}=int_{0}^{+infty}frac{sqrt{x}}{1+x^2},dx = frac{pi}{sqrt{2}}.$$
                          Much simpler, once $f(x)$ is defined as $sum_{ngeq 1}frac{sin(nx)}{sqrt{n}}$ we have
                          $$ int_{-pi}^{pi}f(x)^2,dx = pisum_{ngeq 1}frac{1}{n} = +infty $$
                          by Parseval's theorem, hence $f(x)$ cannot be bounded on $[-pi,pi]$.






                          share|cite|improve this answer














                          Pointwise convergence is granted by Dirichlet's test, but a series of continuous functions cannot converge uniformly to an unbounded function. We may compute $lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}$ through a convolution with an approximate identity:



                          $$begin{eqnarray*}lim_{xto 0^+}sum_{ngeq 1}frac{sin(n x)}{sqrt{n}}&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}int_{0}^{+infty}msin(nx) e^{-mx},dx\&=&lim_{mto +infty}sum_{ngeq 1}frac{1}{sqrt{n}}cdotfrac{mn}{m^2+n^2}.end{eqnarray*}$$
                          The RHS is $+infty$ since by Riemann sums
                          $$ lim_{mto +infty}sum_{ngeq 1}frac{sqrt{mn}}{m^2+n^2}=int_{0}^{+infty}frac{sqrt{x}}{1+x^2},dx = frac{pi}{sqrt{2}}.$$
                          Much simpler, once $f(x)$ is defined as $sum_{ngeq 1}frac{sin(nx)}{sqrt{n}}$ we have
                          $$ int_{-pi}^{pi}f(x)^2,dx = pisum_{ngeq 1}frac{1}{n} = +infty $$
                          by Parseval's theorem, hence $f(x)$ cannot be bounded on $[-pi,pi]$.







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                          edited Nov 21 at 21:49

























                          answered Nov 21 at 21:34









                          Jack D'Aurizio

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