On the modes of convergence of the series $sumlimits_{k=1}^{infty} frac{1}{k} sin left(frac{x}{k+1} right)$












2












$begingroup$


Show that




$$f(x) = sum_{k=1}^{infty} dfrac{1}{k} sin left(dfrac{x}{k+1}right).$$




converges pointwise on $mathbb{R}$ and uniformly on each bounded interval in $mathbb{R}$, to a differentiable function $f$ which satisifies $vert f(x) vert leq vert x vert$ and $vert f'(x) vert leq 1$.



This problem comes from a section that covers the Weierstrass M-Test and Dirichlet's Test. I am not sure where to begin with this. Analysis really isn't my thing at this point.






real-analysis sequences-and-series analysis uniform-convergence pointwise-convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you trying to prove those statements about $f$, or something else?
    $endgroup$
    – zibadawa timmy
    Nov 4 '13 at 18:45










  • $begingroup$
    Yes, sorry I left that off
    $endgroup$
    – MFD55012
    Nov 4 '13 at 18:50










  • $begingroup$
    Have you tried using those tests in any fashion?
    $endgroup$
    – zibadawa timmy
    Nov 4 '13 at 18:54










  • $begingroup$
    Note that $|sin t|leq |t|$ for every real number $t$.
    $endgroup$
    – Julien
    Nov 4 '13 at 18:57










  • $begingroup$
    I am trying to use Dirichlet's now.
    $endgroup$
    – MFD55012
    Nov 4 '13 at 19:05
















2












$begingroup$


Show that




$$f(x) = sum_{k=1}^{infty} dfrac{1}{k} sin left(dfrac{x}{k+1}right).$$




converges pointwise on $mathbb{R}$ and uniformly on each bounded interval in $mathbb{R}$, to a differentiable function $f$ which satisifies $vert f(x) vert leq vert x vert$ and $vert f'(x) vert leq 1$.



This problem comes from a section that covers the Weierstrass M-Test and Dirichlet's Test. I am not sure where to begin with this. Analysis really isn't my thing at this point.






real-analysis sequences-and-series analysis uniform-convergence pointwise-convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you trying to prove those statements about $f$, or something else?
    $endgroup$
    – zibadawa timmy
    Nov 4 '13 at 18:45










  • $begingroup$
    Yes, sorry I left that off
    $endgroup$
    – MFD55012
    Nov 4 '13 at 18:50










  • $begingroup$
    Have you tried using those tests in any fashion?
    $endgroup$
    – zibadawa timmy
    Nov 4 '13 at 18:54










  • $begingroup$
    Note that $|sin t|leq |t|$ for every real number $t$.
    $endgroup$
    – Julien
    Nov 4 '13 at 18:57










  • $begingroup$
    I am trying to use Dirichlet's now.
    $endgroup$
    – MFD55012
    Nov 4 '13 at 19:05














2












2








2





$begingroup$


Show that




$$f(x) = sum_{k=1}^{infty} dfrac{1}{k} sin left(dfrac{x}{k+1}right).$$




converges pointwise on $mathbb{R}$ and uniformly on each bounded interval in $mathbb{R}$, to a differentiable function $f$ which satisifies $vert f(x) vert leq vert x vert$ and $vert f'(x) vert leq 1$.



This problem comes from a section that covers the Weierstrass M-Test and Dirichlet's Test. I am not sure where to begin with this. Analysis really isn't my thing at this point.






real-analysis sequences-and-series analysis uniform-convergence pointwise-convergence







share|cite|improve this question











$endgroup$




Show that




$$f(x) = sum_{k=1}^{infty} dfrac{1}{k} sin left(dfrac{x}{k+1}right).$$




converges pointwise on $mathbb{R}$ and uniformly on each bounded interval in $mathbb{R}$, to a differentiable function $f$ which satisifies $vert f(x) vert leq vert x vert$ and $vert f'(x) vert leq 1$.



This problem comes from a section that covers the Weierstrass M-Test and Dirichlet's Test. I am not sure where to begin with this. Analysis really isn't my thing at this point.






real-analysis sequences-and-series analysis uniform-convergence pointwise-convergence




real-analysis sequences-and-series analysis 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 Dec 30 '18 at 21:11









Did

247k23222457




247k23222457










asked Nov 4 '13 at 18:37









MFD55012MFD55012

604




604












  • $begingroup$
    Are you trying to prove those statements about $f$, or something else?
    $endgroup$
    – zibadawa timmy
    Nov 4 '13 at 18:45










  • $begingroup$
    Yes, sorry I left that off
    $endgroup$
    – MFD55012
    Nov 4 '13 at 18:50










  • $begingroup$
    Have you tried using those tests in any fashion?
    $endgroup$
    – zibadawa timmy
    Nov 4 '13 at 18:54










  • $begingroup$
    Note that $|sin t|leq |t|$ for every real number $t$.
    $endgroup$
    – Julien
    Nov 4 '13 at 18:57










  • $begingroup$
    I am trying to use Dirichlet's now.
    $endgroup$
    – MFD55012
    Nov 4 '13 at 19:05


















  • $begingroup$
    Are you trying to prove those statements about $f$, or something else?
    $endgroup$
    – zibadawa timmy
    Nov 4 '13 at 18:45










  • $begingroup$
    Yes, sorry I left that off
    $endgroup$
    – MFD55012
    Nov 4 '13 at 18:50










  • $begingroup$
    Have you tried using those tests in any fashion?
    $endgroup$
    – zibadawa timmy
    Nov 4 '13 at 18:54










  • $begingroup$
    Note that $|sin t|leq |t|$ for every real number $t$.
    $endgroup$
    – Julien
    Nov 4 '13 at 18:57










  • $begingroup$
    I am trying to use Dirichlet's now.
    $endgroup$
    – MFD55012
    Nov 4 '13 at 19:05
















$begingroup$
Are you trying to prove those statements about $f$, or something else?
$endgroup$
– zibadawa timmy
Nov 4 '13 at 18:45




$begingroup$
Are you trying to prove those statements about $f$, or something else?
$endgroup$
– zibadawa timmy
Nov 4 '13 at 18:45












$begingroup$
Yes, sorry I left that off
$endgroup$
– MFD55012
Nov 4 '13 at 18:50




$begingroup$
Yes, sorry I left that off
$endgroup$
– MFD55012
Nov 4 '13 at 18:50












$begingroup$
Have you tried using those tests in any fashion?
$endgroup$
– zibadawa timmy
Nov 4 '13 at 18:54




$begingroup$
Have you tried using those tests in any fashion?
$endgroup$
– zibadawa timmy
Nov 4 '13 at 18:54












$begingroup$
Note that $|sin t|leq |t|$ for every real number $t$.
$endgroup$
– Julien
Nov 4 '13 at 18:57




$begingroup$
Note that $|sin t|leq |t|$ for every real number $t$.
$endgroup$
– Julien
Nov 4 '13 at 18:57












$begingroup$
I am trying to use Dirichlet's now.
$endgroup$
– MFD55012
Nov 4 '13 at 19:05




$begingroup$
I am trying to use Dirichlet's now.
$endgroup$
– MFD55012
Nov 4 '13 at 19:05










1 Answer
1






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oldest

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2












$begingroup$

I'm sorry you're just getting your answer in five years. Lets have a look at this;



Let $[a,b]subseteq Bbb{R}$ be any bounded interval, $xin [a,b]$ and $kin Bbb{N},$ then



begin{align} left|dfrac{1}{k}sinleft(dfrac{x}{k+1}right)right|leq dfrac{1}{k}left|dfrac{x}{k+1}right|=dfrac{left|xright|}{k(k+1)} leq dfrac{left|xright|}{k(k+1)} leq dfrac{c}{k(k+1)} , end{align}
where $c:=max{|a|,|b|}quad$ [see If $xin[a,b],$ then $|x|leq max{|a|,|b|}$. Since
begin{align} sum^{infty}_{k=1}dfrac{c}{k(k+1)} , end{align}
converges, then,
begin{align} sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right) end{align}
converges uniformly on $[a,b],$ by Weierstrass-M test. Since $[a,b]subseteq Bbb{R}$ was arbitrary, then uniform convergence is true on each bounded interval in $Bbb{R}$ which also implies pointwise convergence on $Bbb{R}$. Now, for arbitrary $xinBbb{R},$



begin{align} left|f(x)right|&=left|sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right)right|\&leq sum^{infty}_{k=1} dfrac{1}{k}left|dfrac{x}{k+1}right|\&= sum^{infty}_{k=1}dfrac{left|xright|}{k(k+1)}\&= left|xright|sum^{infty}_{k=1}dfrac{1}{k(k+1)}\&= left|xright|limlimits_{ntoinfty}sum^{n}_{k=1}dfrac{1}{k(k+1)} \&= left|xright|limlimits_{ntoinfty}sum^{n}_{k=1}left(dfrac{1}{k}-dfrac{1}{k+1}right)\&= left|xright|limlimits_{ntoinfty}left(1-dfrac{1}{n+1}right) \&= left|xright|. end{align}
Since the series converges uniformly on $[a,b],$ then we can differentiate term-term on $[a,b].$ Thus, for $xin [a,b]$
begin{align} f'(x)&=dfrac{d}{dx}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right)\&=sum^{infty}_{k=1}dfrac{1}{k}dfrac{d}{dx}left[sinleft(dfrac{x}{k+1}right)right]\&=sum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)cosleft(dfrac{x}{k+1}right) , end{align}
which implies that
begin{align} left|f'(x)right|&=left|sum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)cosleft(dfrac{x}{k+1}right)right|\&leqsum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)left|cosleft(dfrac{x}{k+1}right) right|\&leqsum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)\&= limlimits_{ntoinfty}sum^{n}_{k=1}dfrac{1}{k(k+1)} \&= limlimits_{ntoinfty}sum^{n}_{k=1}left(dfrac{1}{k}-dfrac{1}{k+1}right)\&= limlimits_{ntoinfty}left(1-dfrac{1}{n+1}right) \&= 1 end{align}
as required.






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    2












    $begingroup$

    I'm sorry you're just getting your answer in five years. Lets have a look at this;



    Let $[a,b]subseteq Bbb{R}$ be any bounded interval, $xin [a,b]$ and $kin Bbb{N},$ then



    begin{align} left|dfrac{1}{k}sinleft(dfrac{x}{k+1}right)right|leq dfrac{1}{k}left|dfrac{x}{k+1}right|=dfrac{left|xright|}{k(k+1)} leq dfrac{left|xright|}{k(k+1)} leq dfrac{c}{k(k+1)} , end{align}
    where $c:=max{|a|,|b|}quad$ [see If $xin[a,b],$ then $|x|leq max{|a|,|b|}$. Since
    begin{align} sum^{infty}_{k=1}dfrac{c}{k(k+1)} , end{align}
    converges, then,
    begin{align} sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right) end{align}
    converges uniformly on $[a,b],$ by Weierstrass-M test. Since $[a,b]subseteq Bbb{R}$ was arbitrary, then uniform convergence is true on each bounded interval in $Bbb{R}$ which also implies pointwise convergence on $Bbb{R}$. Now, for arbitrary $xinBbb{R},$



    begin{align} left|f(x)right|&=left|sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right)right|\&leq sum^{infty}_{k=1} dfrac{1}{k}left|dfrac{x}{k+1}right|\&= sum^{infty}_{k=1}dfrac{left|xright|}{k(k+1)}\&= left|xright|sum^{infty}_{k=1}dfrac{1}{k(k+1)}\&= left|xright|limlimits_{ntoinfty}sum^{n}_{k=1}dfrac{1}{k(k+1)} \&= left|xright|limlimits_{ntoinfty}sum^{n}_{k=1}left(dfrac{1}{k}-dfrac{1}{k+1}right)\&= left|xright|limlimits_{ntoinfty}left(1-dfrac{1}{n+1}right) \&= left|xright|. end{align}
    Since the series converges uniformly on $[a,b],$ then we can differentiate term-term on $[a,b].$ Thus, for $xin [a,b]$
    begin{align} f'(x)&=dfrac{d}{dx}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right)\&=sum^{infty}_{k=1}dfrac{1}{k}dfrac{d}{dx}left[sinleft(dfrac{x}{k+1}right)right]\&=sum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)cosleft(dfrac{x}{k+1}right) , end{align}
    which implies that
    begin{align} left|f'(x)right|&=left|sum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)cosleft(dfrac{x}{k+1}right)right|\&leqsum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)left|cosleft(dfrac{x}{k+1}right) right|\&leqsum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)\&= limlimits_{ntoinfty}sum^{n}_{k=1}dfrac{1}{k(k+1)} \&= limlimits_{ntoinfty}sum^{n}_{k=1}left(dfrac{1}{k}-dfrac{1}{k+1}right)\&= limlimits_{ntoinfty}left(1-dfrac{1}{n+1}right) \&= 1 end{align}
    as required.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      I'm sorry you're just getting your answer in five years. Lets have a look at this;



      Let $[a,b]subseteq Bbb{R}$ be any bounded interval, $xin [a,b]$ and $kin Bbb{N},$ then



      begin{align} left|dfrac{1}{k}sinleft(dfrac{x}{k+1}right)right|leq dfrac{1}{k}left|dfrac{x}{k+1}right|=dfrac{left|xright|}{k(k+1)} leq dfrac{left|xright|}{k(k+1)} leq dfrac{c}{k(k+1)} , end{align}
      where $c:=max{|a|,|b|}quad$ [see If $xin[a,b],$ then $|x|leq max{|a|,|b|}$. Since
      begin{align} sum^{infty}_{k=1}dfrac{c}{k(k+1)} , end{align}
      converges, then,
      begin{align} sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right) end{align}
      converges uniformly on $[a,b],$ by Weierstrass-M test. Since $[a,b]subseteq Bbb{R}$ was arbitrary, then uniform convergence is true on each bounded interval in $Bbb{R}$ which also implies pointwise convergence on $Bbb{R}$. Now, for arbitrary $xinBbb{R},$



      begin{align} left|f(x)right|&=left|sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right)right|\&leq sum^{infty}_{k=1} dfrac{1}{k}left|dfrac{x}{k+1}right|\&= sum^{infty}_{k=1}dfrac{left|xright|}{k(k+1)}\&= left|xright|sum^{infty}_{k=1}dfrac{1}{k(k+1)}\&= left|xright|limlimits_{ntoinfty}sum^{n}_{k=1}dfrac{1}{k(k+1)} \&= left|xright|limlimits_{ntoinfty}sum^{n}_{k=1}left(dfrac{1}{k}-dfrac{1}{k+1}right)\&= left|xright|limlimits_{ntoinfty}left(1-dfrac{1}{n+1}right) \&= left|xright|. end{align}
      Since the series converges uniformly on $[a,b],$ then we can differentiate term-term on $[a,b].$ Thus, for $xin [a,b]$
      begin{align} f'(x)&=dfrac{d}{dx}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right)\&=sum^{infty}_{k=1}dfrac{1}{k}dfrac{d}{dx}left[sinleft(dfrac{x}{k+1}right)right]\&=sum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)cosleft(dfrac{x}{k+1}right) , end{align}
      which implies that
      begin{align} left|f'(x)right|&=left|sum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)cosleft(dfrac{x}{k+1}right)right|\&leqsum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)left|cosleft(dfrac{x}{k+1}right) right|\&leqsum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)\&= limlimits_{ntoinfty}sum^{n}_{k=1}dfrac{1}{k(k+1)} \&= limlimits_{ntoinfty}sum^{n}_{k=1}left(dfrac{1}{k}-dfrac{1}{k+1}right)\&= limlimits_{ntoinfty}left(1-dfrac{1}{n+1}right) \&= 1 end{align}
      as required.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        I'm sorry you're just getting your answer in five years. Lets have a look at this;



        Let $[a,b]subseteq Bbb{R}$ be any bounded interval, $xin [a,b]$ and $kin Bbb{N},$ then



        begin{align} left|dfrac{1}{k}sinleft(dfrac{x}{k+1}right)right|leq dfrac{1}{k}left|dfrac{x}{k+1}right|=dfrac{left|xright|}{k(k+1)} leq dfrac{left|xright|}{k(k+1)} leq dfrac{c}{k(k+1)} , end{align}
        where $c:=max{|a|,|b|}quad$ [see If $xin[a,b],$ then $|x|leq max{|a|,|b|}$. Since
        begin{align} sum^{infty}_{k=1}dfrac{c}{k(k+1)} , end{align}
        converges, then,
        begin{align} sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right) end{align}
        converges uniformly on $[a,b],$ by Weierstrass-M test. Since $[a,b]subseteq Bbb{R}$ was arbitrary, then uniform convergence is true on each bounded interval in $Bbb{R}$ which also implies pointwise convergence on $Bbb{R}$. Now, for arbitrary $xinBbb{R},$



        begin{align} left|f(x)right|&=left|sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right)right|\&leq sum^{infty}_{k=1} dfrac{1}{k}left|dfrac{x}{k+1}right|\&= sum^{infty}_{k=1}dfrac{left|xright|}{k(k+1)}\&= left|xright|sum^{infty}_{k=1}dfrac{1}{k(k+1)}\&= left|xright|limlimits_{ntoinfty}sum^{n}_{k=1}dfrac{1}{k(k+1)} \&= left|xright|limlimits_{ntoinfty}sum^{n}_{k=1}left(dfrac{1}{k}-dfrac{1}{k+1}right)\&= left|xright|limlimits_{ntoinfty}left(1-dfrac{1}{n+1}right) \&= left|xright|. end{align}
        Since the series converges uniformly on $[a,b],$ then we can differentiate term-term on $[a,b].$ Thus, for $xin [a,b]$
        begin{align} f'(x)&=dfrac{d}{dx}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right)\&=sum^{infty}_{k=1}dfrac{1}{k}dfrac{d}{dx}left[sinleft(dfrac{x}{k+1}right)right]\&=sum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)cosleft(dfrac{x}{k+1}right) , end{align}
        which implies that
        begin{align} left|f'(x)right|&=left|sum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)cosleft(dfrac{x}{k+1}right)right|\&leqsum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)left|cosleft(dfrac{x}{k+1}right) right|\&leqsum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)\&= limlimits_{ntoinfty}sum^{n}_{k=1}dfrac{1}{k(k+1)} \&= limlimits_{ntoinfty}sum^{n}_{k=1}left(dfrac{1}{k}-dfrac{1}{k+1}right)\&= limlimits_{ntoinfty}left(1-dfrac{1}{n+1}right) \&= 1 end{align}
        as required.






        share|cite|improve this answer











        $endgroup$



        I'm sorry you're just getting your answer in five years. Lets have a look at this;



        Let $[a,b]subseteq Bbb{R}$ be any bounded interval, $xin [a,b]$ and $kin Bbb{N},$ then



        begin{align} left|dfrac{1}{k}sinleft(dfrac{x}{k+1}right)right|leq dfrac{1}{k}left|dfrac{x}{k+1}right|=dfrac{left|xright|}{k(k+1)} leq dfrac{left|xright|}{k(k+1)} leq dfrac{c}{k(k+1)} , end{align}
        where $c:=max{|a|,|b|}quad$ [see If $xin[a,b],$ then $|x|leq max{|a|,|b|}$. Since
        begin{align} sum^{infty}_{k=1}dfrac{c}{k(k+1)} , end{align}
        converges, then,
        begin{align} sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right) end{align}
        converges uniformly on $[a,b],$ by Weierstrass-M test. Since $[a,b]subseteq Bbb{R}$ was arbitrary, then uniform convergence is true on each bounded interval in $Bbb{R}$ which also implies pointwise convergence on $Bbb{R}$. Now, for arbitrary $xinBbb{R},$



        begin{align} left|f(x)right|&=left|sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right)right|\&leq sum^{infty}_{k=1} dfrac{1}{k}left|dfrac{x}{k+1}right|\&= sum^{infty}_{k=1}dfrac{left|xright|}{k(k+1)}\&= left|xright|sum^{infty}_{k=1}dfrac{1}{k(k+1)}\&= left|xright|limlimits_{ntoinfty}sum^{n}_{k=1}dfrac{1}{k(k+1)} \&= left|xright|limlimits_{ntoinfty}sum^{n}_{k=1}left(dfrac{1}{k}-dfrac{1}{k+1}right)\&= left|xright|limlimits_{ntoinfty}left(1-dfrac{1}{n+1}right) \&= left|xright|. end{align}
        Since the series converges uniformly on $[a,b],$ then we can differentiate term-term on $[a,b].$ Thus, for $xin [a,b]$
        begin{align} f'(x)&=dfrac{d}{dx}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k+1}right)\&=sum^{infty}_{k=1}dfrac{1}{k}dfrac{d}{dx}left[sinleft(dfrac{x}{k+1}right)right]\&=sum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)cosleft(dfrac{x}{k+1}right) , end{align}
        which implies that
        begin{align} left|f'(x)right|&=left|sum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)cosleft(dfrac{x}{k+1}right)right|\&leqsum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)left|cosleft(dfrac{x}{k+1}right) right|\&leqsum^{infty}_{k=1}dfrac{1}{k}left(dfrac{1}{k+1}right)\&= limlimits_{ntoinfty}sum^{n}_{k=1}dfrac{1}{k(k+1)} \&= limlimits_{ntoinfty}sum^{n}_{k=1}left(dfrac{1}{k}-dfrac{1}{k+1}right)\&= limlimits_{ntoinfty}left(1-dfrac{1}{n+1}right) \&= 1 end{align}
        as required.







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        edited Dec 30 '18 at 21:50

























        answered Dec 30 '18 at 20:54









        Omojola MichealOmojola Micheal

        1,802324




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