Show that $sum^{infty}_{n=1}(-1)^{n+1}frac{1}{n+x^4}$ is uniformly convergent on $Bbb{R}$












1












$begingroup$


Show that the following series is uniformly convergent on $Bbb{R}$



begin{align}sum^{infty}_{n=1}(-1)^{n+1}dfrac{1}{n+x^4}end{align}



MY TRIAL



I tried using the alternating series test before the $beta_n$ approach.



Let $f_n(x)=dfrac{1}{n+x^4},;forall;xinBbb{R},;ninBbb{N},$
then




  1. $f_n(x)=dfrac{1}{n+x^4}geq 0$

  2. $f_{n+1}(x)leq f_{n}(x)$

  3. $f_n(x)=dfrac{1}{n+x^4}to 0$


Then,
begin{align}beta_n &=suplimits_{xinBbb{R}}left|s_n(x)-sum^{infty}_{i=1}f_i(x)right|\&=suplimits_{xinBbb{R}}left|sum^{n}_{i=1}(-1)^{i+1}f_i(x)-sum^{infty}_{i=1}(-1)^{i+1}f_i(x)right|\&=suplimits_{xinBbb{R}}left|sum^{n}_{i=1}(-1)^{i+1}dfrac{1}{i+x^4}-sum^{infty}_{i=1}(-1)^{i+1}dfrac{1}{i+x^4}right|\&=suplimits_{xinBbb{R}}left|(-1)^{n+2}dfrac{1}{(n+1)+x^4}+(-1)^{n+3}dfrac{1}{(n+2)+x^4}+(-1)^{n+4}dfrac{1}{(n+3)+x^4}cdotsright|\&=suplimits_{xinBbb{R}}left|dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}+dfrac{1}{(n+3)+x^4}-dfrac{1}{(n+4)+x^4}cdotsright|\&=suplimits_{xinBbb{R}}left|dfrac{1}{(n+1)+x^4}-left(dfrac{1}{(n+2)+x^4}-dfrac{1}{(n+3)+x^4}right)-left(dfrac{1}{(n+4)+x^4}-dfrac{1}{(n+5)+x^4}right)cdotsright|\&leq suplimits_{xinBbb{R}}left|dfrac{1}{(n+1)+x^4}right|to 0,;;text{as};ntoinftyend{align}
and we are done!



Kindly help me check if I'm correct! Constructive criticisms will be highly welcome! I'll also love to see other approaches to this problem. Thanks!






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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: Abel-Dirichlet tests.
    $endgroup$
    – xbh
    Sep 5 '18 at 10:34










  • $begingroup$
    Maybe you could use the same technique to estimate the $sup$.
    $endgroup$
    – xbh
    Sep 5 '18 at 10:36
















1












$begingroup$


Show that the following series is uniformly convergent on $Bbb{R}$



begin{align}sum^{infty}_{n=1}(-1)^{n+1}dfrac{1}{n+x^4}end{align}



MY TRIAL



I tried using the alternating series test before the $beta_n$ approach.



Let $f_n(x)=dfrac{1}{n+x^4},;forall;xinBbb{R},;ninBbb{N},$
then




  1. $f_n(x)=dfrac{1}{n+x^4}geq 0$

  2. $f_{n+1}(x)leq f_{n}(x)$

  3. $f_n(x)=dfrac{1}{n+x^4}to 0$


Then,
begin{align}beta_n &=suplimits_{xinBbb{R}}left|s_n(x)-sum^{infty}_{i=1}f_i(x)right|\&=suplimits_{xinBbb{R}}left|sum^{n}_{i=1}(-1)^{i+1}f_i(x)-sum^{infty}_{i=1}(-1)^{i+1}f_i(x)right|\&=suplimits_{xinBbb{R}}left|sum^{n}_{i=1}(-1)^{i+1}dfrac{1}{i+x^4}-sum^{infty}_{i=1}(-1)^{i+1}dfrac{1}{i+x^4}right|\&=suplimits_{xinBbb{R}}left|(-1)^{n+2}dfrac{1}{(n+1)+x^4}+(-1)^{n+3}dfrac{1}{(n+2)+x^4}+(-1)^{n+4}dfrac{1}{(n+3)+x^4}cdotsright|\&=suplimits_{xinBbb{R}}left|dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}+dfrac{1}{(n+3)+x^4}-dfrac{1}{(n+4)+x^4}cdotsright|\&=suplimits_{xinBbb{R}}left|dfrac{1}{(n+1)+x^4}-left(dfrac{1}{(n+2)+x^4}-dfrac{1}{(n+3)+x^4}right)-left(dfrac{1}{(n+4)+x^4}-dfrac{1}{(n+5)+x^4}right)cdotsright|\&leq suplimits_{xinBbb{R}}left|dfrac{1}{(n+1)+x^4}right|to 0,;;text{as};ntoinftyend{align}
and we are done!



Kindly help me check if I'm correct! Constructive criticisms will be highly welcome! I'll also love to see other approaches to this problem. Thanks!






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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: Abel-Dirichlet tests.
    $endgroup$
    – xbh
    Sep 5 '18 at 10:34










  • $begingroup$
    Maybe you could use the same technique to estimate the $sup$.
    $endgroup$
    – xbh
    Sep 5 '18 at 10:36














1












1








1


2



$begingroup$


Show that the following series is uniformly convergent on $Bbb{R}$



begin{align}sum^{infty}_{n=1}(-1)^{n+1}dfrac{1}{n+x^4}end{align}



MY TRIAL



I tried using the alternating series test before the $beta_n$ approach.



Let $f_n(x)=dfrac{1}{n+x^4},;forall;xinBbb{R},;ninBbb{N},$
then




  1. $f_n(x)=dfrac{1}{n+x^4}geq 0$

  2. $f_{n+1}(x)leq f_{n}(x)$

  3. $f_n(x)=dfrac{1}{n+x^4}to 0$


Then,
begin{align}beta_n &=suplimits_{xinBbb{R}}left|s_n(x)-sum^{infty}_{i=1}f_i(x)right|\&=suplimits_{xinBbb{R}}left|sum^{n}_{i=1}(-1)^{i+1}f_i(x)-sum^{infty}_{i=1}(-1)^{i+1}f_i(x)right|\&=suplimits_{xinBbb{R}}left|sum^{n}_{i=1}(-1)^{i+1}dfrac{1}{i+x^4}-sum^{infty}_{i=1}(-1)^{i+1}dfrac{1}{i+x^4}right|\&=suplimits_{xinBbb{R}}left|(-1)^{n+2}dfrac{1}{(n+1)+x^4}+(-1)^{n+3}dfrac{1}{(n+2)+x^4}+(-1)^{n+4}dfrac{1}{(n+3)+x^4}cdotsright|\&=suplimits_{xinBbb{R}}left|dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}+dfrac{1}{(n+3)+x^4}-dfrac{1}{(n+4)+x^4}cdotsright|\&=suplimits_{xinBbb{R}}left|dfrac{1}{(n+1)+x^4}-left(dfrac{1}{(n+2)+x^4}-dfrac{1}{(n+3)+x^4}right)-left(dfrac{1}{(n+4)+x^4}-dfrac{1}{(n+5)+x^4}right)cdotsright|\&leq suplimits_{xinBbb{R}}left|dfrac{1}{(n+1)+x^4}right|to 0,;;text{as};ntoinftyend{align}
and we are done!



Kindly help me check if I'm correct! Constructive criticisms will be highly welcome! I'll also love to see other approaches to this problem. Thanks!






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







share|cite|improve this question











$endgroup$




Show that the following series is uniformly convergent on $Bbb{R}$



begin{align}sum^{infty}_{n=1}(-1)^{n+1}dfrac{1}{n+x^4}end{align}



MY TRIAL



I tried using the alternating series test before the $beta_n$ approach.



Let $f_n(x)=dfrac{1}{n+x^4},;forall;xinBbb{R},;ninBbb{N},$
then




  1. $f_n(x)=dfrac{1}{n+x^4}geq 0$

  2. $f_{n+1}(x)leq f_{n}(x)$

  3. $f_n(x)=dfrac{1}{n+x^4}to 0$


Then,
begin{align}beta_n &=suplimits_{xinBbb{R}}left|s_n(x)-sum^{infty}_{i=1}f_i(x)right|\&=suplimits_{xinBbb{R}}left|sum^{n}_{i=1}(-1)^{i+1}f_i(x)-sum^{infty}_{i=1}(-1)^{i+1}f_i(x)right|\&=suplimits_{xinBbb{R}}left|sum^{n}_{i=1}(-1)^{i+1}dfrac{1}{i+x^4}-sum^{infty}_{i=1}(-1)^{i+1}dfrac{1}{i+x^4}right|\&=suplimits_{xinBbb{R}}left|(-1)^{n+2}dfrac{1}{(n+1)+x^4}+(-1)^{n+3}dfrac{1}{(n+2)+x^4}+(-1)^{n+4}dfrac{1}{(n+3)+x^4}cdotsright|\&=suplimits_{xinBbb{R}}left|dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}+dfrac{1}{(n+3)+x^4}-dfrac{1}{(n+4)+x^4}cdotsright|\&=suplimits_{xinBbb{R}}left|dfrac{1}{(n+1)+x^4}-left(dfrac{1}{(n+2)+x^4}-dfrac{1}{(n+3)+x^4}right)-left(dfrac{1}{(n+4)+x^4}-dfrac{1}{(n+5)+x^4}right)cdotsright|\&leq suplimits_{xinBbb{R}}left|dfrac{1}{(n+1)+x^4}right|to 0,;;text{as};ntoinftyend{align}
and we are done!



Kindly help me check if I'm correct! Constructive criticisms will be highly welcome! I'll also love to see other approaches to this problem. Thanks!






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




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 21:44









Lorenzo B.

1,8602520




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asked Sep 5 '18 at 10:29









MichealMicheal

26511




26511












  • $begingroup$
    Hint: Abel-Dirichlet tests.
    $endgroup$
    – xbh
    Sep 5 '18 at 10:34










  • $begingroup$
    Maybe you could use the same technique to estimate the $sup$.
    $endgroup$
    – xbh
    Sep 5 '18 at 10:36


















  • $begingroup$
    Hint: Abel-Dirichlet tests.
    $endgroup$
    – xbh
    Sep 5 '18 at 10:34










  • $begingroup$
    Maybe you could use the same technique to estimate the $sup$.
    $endgroup$
    – xbh
    Sep 5 '18 at 10:36
















$begingroup$
Hint: Abel-Dirichlet tests.
$endgroup$
– xbh
Sep 5 '18 at 10:34




$begingroup$
Hint: Abel-Dirichlet tests.
$endgroup$
– xbh
Sep 5 '18 at 10:34












$begingroup$
Maybe you could use the same technique to estimate the $sup$.
$endgroup$
– xbh
Sep 5 '18 at 10:36




$begingroup$
Maybe you could use the same technique to estimate the $sup$.
$endgroup$
– xbh
Sep 5 '18 at 10:36










1 Answer
1






active

oldest

votes


















0












$begingroup$

For future readers, I present two proofs:



METHOD I: DIRICHLET'S TEST



The Dirichlet's test for uniform convergence can be found here.



Let $;epsilon>0,; xinBbb{R}$ and $ninBbb{N}$ be fixed but arbitrary. Define



begin{align}f_n(x)=(-1)^{n+1};;text{and };;g_{n}(x)=dfrac{1}{n+x^4}end{align}
Then, uniform boundedness is implied, since
begin{align}
&left|F_n(x)right|=left|sum^{n}_{m}(-1)^{m+1} right|leq 1,end{align}
The sequence ${g_n(x)}_{nin Bbb{N}}$ is decreasing, since
begin{align}
dfrac{n+x^4}{(n+1)+x^4}leq dfrac{(n+1)+x^4}{(n+1)+x^4}=1implies g_{n+1}(x)leq g_{n}(x)end{align}
By Archmidean principle, there exists $N(epsilon)$ such that $N(epsilon)>epsilon$ and
begin{align}
&left|g_n(x)-0right|=left|dfrac{1}{n+x^4}-0 right|=dfrac{1}{n+x^4}leq dfrac{1}{n}leq dfrac{1}{N}<epsilon,;forall;ngeq N(epsilon)end{align}
Hence, ${g_n(x)}_{nin Bbb{N}}$ converges uniformly on $Bbb{R}$ and, by Dirichlet's test, $sum^{infty}_{n=1}f_n(x)g_n(x)$ converges uniformly on $Bbb{R}$.



METHOD II: UNIFORM CAUCHY CRITERION



Let $;epsilon>0$, $m,ninBbb{N}$ be given such that $mgeq n$ and $xinBbb{R}$ be arbitrary. Define ${f_n(x)}_{nin Bbb{N}}$ and ${g_n(x)}_{nin Bbb{N}}$ as before. Then,



begin{align}left|sum^{m}_{k=1}f_k(x)g_k(x)-sum^{n}_{k=1}f_k(x)g_k(x) right|&=left|sum^{m}_{k=n+1}f_k(x)g_k(x)right|\&=left|sum^{m}_{k=n+1}(-1)^{k+1}dfrac{1}{k+x^4}right|end{align}
Expanding, we get
begin{align}left|sum^{m}_{k=n+1}(-1)^{k+1}dfrac{1}{k+x^4}right|&=left|(-1)^{n+2}dfrac{1}{(n+1)+x^4}+(-1)^{n+3}dfrac{1}{(n+2)+x^4}+cdots +(-1)^{n+1}dfrac{1}{n+x^4}right|\&=left|dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}+cdots +(-1)^{m-n-2}dfrac{1}{(m-1)+x^4}+(-1)^{m-n-1}dfrac{1}{m+x^4}right|\&leq left|dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}right|+cdots +left|dfrac{1}{(m-1)+x^4}-dfrac{1}{m+x^4}right|\&= dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}+cdots +dfrac{1}{(m-1)+x^4}-dfrac{1}{m+x^4}\&= dfrac{1}{(n+1)+x^4}-left[dfrac{1}{(n+2)+x^4}-dfrac{1}{(n+2)+x^4}right]-cdots \&;-left[dfrac{1}{(m-1)+x^4}-dfrac{1}{(m-1)+x^4}right]-dfrac{1}{m+x^4}\&leq dfrac{1}{(n+1)+x^4}.end{align}
As shown in Method I, there exists $N(epsilon)$ such that $,;forall;ngeq N(epsilon)$,
begin{align}
dfrac{1}{(n+1)+x^4}leq dfrac{1}{n+x^4}leq dfrac{1}{n}leq dfrac{1}{N}<epsilon,end{align}
and we're done!






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    oldest

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    0












    $begingroup$

    For future readers, I present two proofs:



    METHOD I: DIRICHLET'S TEST



    The Dirichlet's test for uniform convergence can be found here.



    Let $;epsilon>0,; xinBbb{R}$ and $ninBbb{N}$ be fixed but arbitrary. Define



    begin{align}f_n(x)=(-1)^{n+1};;text{and };;g_{n}(x)=dfrac{1}{n+x^4}end{align}
    Then, uniform boundedness is implied, since
    begin{align}
    &left|F_n(x)right|=left|sum^{n}_{m}(-1)^{m+1} right|leq 1,end{align}
    The sequence ${g_n(x)}_{nin Bbb{N}}$ is decreasing, since
    begin{align}
    dfrac{n+x^4}{(n+1)+x^4}leq dfrac{(n+1)+x^4}{(n+1)+x^4}=1implies g_{n+1}(x)leq g_{n}(x)end{align}
    By Archmidean principle, there exists $N(epsilon)$ such that $N(epsilon)>epsilon$ and
    begin{align}
    &left|g_n(x)-0right|=left|dfrac{1}{n+x^4}-0 right|=dfrac{1}{n+x^4}leq dfrac{1}{n}leq dfrac{1}{N}<epsilon,;forall;ngeq N(epsilon)end{align}
    Hence, ${g_n(x)}_{nin Bbb{N}}$ converges uniformly on $Bbb{R}$ and, by Dirichlet's test, $sum^{infty}_{n=1}f_n(x)g_n(x)$ converges uniformly on $Bbb{R}$.



    METHOD II: UNIFORM CAUCHY CRITERION



    Let $;epsilon>0$, $m,ninBbb{N}$ be given such that $mgeq n$ and $xinBbb{R}$ be arbitrary. Define ${f_n(x)}_{nin Bbb{N}}$ and ${g_n(x)}_{nin Bbb{N}}$ as before. Then,



    begin{align}left|sum^{m}_{k=1}f_k(x)g_k(x)-sum^{n}_{k=1}f_k(x)g_k(x) right|&=left|sum^{m}_{k=n+1}f_k(x)g_k(x)right|\&=left|sum^{m}_{k=n+1}(-1)^{k+1}dfrac{1}{k+x^4}right|end{align}
    Expanding, we get
    begin{align}left|sum^{m}_{k=n+1}(-1)^{k+1}dfrac{1}{k+x^4}right|&=left|(-1)^{n+2}dfrac{1}{(n+1)+x^4}+(-1)^{n+3}dfrac{1}{(n+2)+x^4}+cdots +(-1)^{n+1}dfrac{1}{n+x^4}right|\&=left|dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}+cdots +(-1)^{m-n-2}dfrac{1}{(m-1)+x^4}+(-1)^{m-n-1}dfrac{1}{m+x^4}right|\&leq left|dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}right|+cdots +left|dfrac{1}{(m-1)+x^4}-dfrac{1}{m+x^4}right|\&= dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}+cdots +dfrac{1}{(m-1)+x^4}-dfrac{1}{m+x^4}\&= dfrac{1}{(n+1)+x^4}-left[dfrac{1}{(n+2)+x^4}-dfrac{1}{(n+2)+x^4}right]-cdots \&;-left[dfrac{1}{(m-1)+x^4}-dfrac{1}{(m-1)+x^4}right]-dfrac{1}{m+x^4}\&leq dfrac{1}{(n+1)+x^4}.end{align}
    As shown in Method I, there exists $N(epsilon)$ such that $,;forall;ngeq N(epsilon)$,
    begin{align}
    dfrac{1}{(n+1)+x^4}leq dfrac{1}{n+x^4}leq dfrac{1}{n}leq dfrac{1}{N}<epsilon,end{align}
    and we're done!






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      For future readers, I present two proofs:



      METHOD I: DIRICHLET'S TEST



      The Dirichlet's test for uniform convergence can be found here.



      Let $;epsilon>0,; xinBbb{R}$ and $ninBbb{N}$ be fixed but arbitrary. Define



      begin{align}f_n(x)=(-1)^{n+1};;text{and };;g_{n}(x)=dfrac{1}{n+x^4}end{align}
      Then, uniform boundedness is implied, since
      begin{align}
      &left|F_n(x)right|=left|sum^{n}_{m}(-1)^{m+1} right|leq 1,end{align}
      The sequence ${g_n(x)}_{nin Bbb{N}}$ is decreasing, since
      begin{align}
      dfrac{n+x^4}{(n+1)+x^4}leq dfrac{(n+1)+x^4}{(n+1)+x^4}=1implies g_{n+1}(x)leq g_{n}(x)end{align}
      By Archmidean principle, there exists $N(epsilon)$ such that $N(epsilon)>epsilon$ and
      begin{align}
      &left|g_n(x)-0right|=left|dfrac{1}{n+x^4}-0 right|=dfrac{1}{n+x^4}leq dfrac{1}{n}leq dfrac{1}{N}<epsilon,;forall;ngeq N(epsilon)end{align}
      Hence, ${g_n(x)}_{nin Bbb{N}}$ converges uniformly on $Bbb{R}$ and, by Dirichlet's test, $sum^{infty}_{n=1}f_n(x)g_n(x)$ converges uniformly on $Bbb{R}$.



      METHOD II: UNIFORM CAUCHY CRITERION



      Let $;epsilon>0$, $m,ninBbb{N}$ be given such that $mgeq n$ and $xinBbb{R}$ be arbitrary. Define ${f_n(x)}_{nin Bbb{N}}$ and ${g_n(x)}_{nin Bbb{N}}$ as before. Then,



      begin{align}left|sum^{m}_{k=1}f_k(x)g_k(x)-sum^{n}_{k=1}f_k(x)g_k(x) right|&=left|sum^{m}_{k=n+1}f_k(x)g_k(x)right|\&=left|sum^{m}_{k=n+1}(-1)^{k+1}dfrac{1}{k+x^4}right|end{align}
      Expanding, we get
      begin{align}left|sum^{m}_{k=n+1}(-1)^{k+1}dfrac{1}{k+x^4}right|&=left|(-1)^{n+2}dfrac{1}{(n+1)+x^4}+(-1)^{n+3}dfrac{1}{(n+2)+x^4}+cdots +(-1)^{n+1}dfrac{1}{n+x^4}right|\&=left|dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}+cdots +(-1)^{m-n-2}dfrac{1}{(m-1)+x^4}+(-1)^{m-n-1}dfrac{1}{m+x^4}right|\&leq left|dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}right|+cdots +left|dfrac{1}{(m-1)+x^4}-dfrac{1}{m+x^4}right|\&= dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}+cdots +dfrac{1}{(m-1)+x^4}-dfrac{1}{m+x^4}\&= dfrac{1}{(n+1)+x^4}-left[dfrac{1}{(n+2)+x^4}-dfrac{1}{(n+2)+x^4}right]-cdots \&;-left[dfrac{1}{(m-1)+x^4}-dfrac{1}{(m-1)+x^4}right]-dfrac{1}{m+x^4}\&leq dfrac{1}{(n+1)+x^4}.end{align}
      As shown in Method I, there exists $N(epsilon)$ such that $,;forall;ngeq N(epsilon)$,
      begin{align}
      dfrac{1}{(n+1)+x^4}leq dfrac{1}{n+x^4}leq dfrac{1}{n}leq dfrac{1}{N}<epsilon,end{align}
      and we're done!






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        For future readers, I present two proofs:



        METHOD I: DIRICHLET'S TEST



        The Dirichlet's test for uniform convergence can be found here.



        Let $;epsilon>0,; xinBbb{R}$ and $ninBbb{N}$ be fixed but arbitrary. Define



        begin{align}f_n(x)=(-1)^{n+1};;text{and };;g_{n}(x)=dfrac{1}{n+x^4}end{align}
        Then, uniform boundedness is implied, since
        begin{align}
        &left|F_n(x)right|=left|sum^{n}_{m}(-1)^{m+1} right|leq 1,end{align}
        The sequence ${g_n(x)}_{nin Bbb{N}}$ is decreasing, since
        begin{align}
        dfrac{n+x^4}{(n+1)+x^4}leq dfrac{(n+1)+x^4}{(n+1)+x^4}=1implies g_{n+1}(x)leq g_{n}(x)end{align}
        By Archmidean principle, there exists $N(epsilon)$ such that $N(epsilon)>epsilon$ and
        begin{align}
        &left|g_n(x)-0right|=left|dfrac{1}{n+x^4}-0 right|=dfrac{1}{n+x^4}leq dfrac{1}{n}leq dfrac{1}{N}<epsilon,;forall;ngeq N(epsilon)end{align}
        Hence, ${g_n(x)}_{nin Bbb{N}}$ converges uniformly on $Bbb{R}$ and, by Dirichlet's test, $sum^{infty}_{n=1}f_n(x)g_n(x)$ converges uniformly on $Bbb{R}$.



        METHOD II: UNIFORM CAUCHY CRITERION



        Let $;epsilon>0$, $m,ninBbb{N}$ be given such that $mgeq n$ and $xinBbb{R}$ be arbitrary. Define ${f_n(x)}_{nin Bbb{N}}$ and ${g_n(x)}_{nin Bbb{N}}$ as before. Then,



        begin{align}left|sum^{m}_{k=1}f_k(x)g_k(x)-sum^{n}_{k=1}f_k(x)g_k(x) right|&=left|sum^{m}_{k=n+1}f_k(x)g_k(x)right|\&=left|sum^{m}_{k=n+1}(-1)^{k+1}dfrac{1}{k+x^4}right|end{align}
        Expanding, we get
        begin{align}left|sum^{m}_{k=n+1}(-1)^{k+1}dfrac{1}{k+x^4}right|&=left|(-1)^{n+2}dfrac{1}{(n+1)+x^4}+(-1)^{n+3}dfrac{1}{(n+2)+x^4}+cdots +(-1)^{n+1}dfrac{1}{n+x^4}right|\&=left|dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}+cdots +(-1)^{m-n-2}dfrac{1}{(m-1)+x^4}+(-1)^{m-n-1}dfrac{1}{m+x^4}right|\&leq left|dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}right|+cdots +left|dfrac{1}{(m-1)+x^4}-dfrac{1}{m+x^4}right|\&= dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}+cdots +dfrac{1}{(m-1)+x^4}-dfrac{1}{m+x^4}\&= dfrac{1}{(n+1)+x^4}-left[dfrac{1}{(n+2)+x^4}-dfrac{1}{(n+2)+x^4}right]-cdots \&;-left[dfrac{1}{(m-1)+x^4}-dfrac{1}{(m-1)+x^4}right]-dfrac{1}{m+x^4}\&leq dfrac{1}{(n+1)+x^4}.end{align}
        As shown in Method I, there exists $N(epsilon)$ such that $,;forall;ngeq N(epsilon)$,
        begin{align}
        dfrac{1}{(n+1)+x^4}leq dfrac{1}{n+x^4}leq dfrac{1}{n}leq dfrac{1}{N}<epsilon,end{align}
        and we're done!






        share|cite|improve this answer











        $endgroup$



        For future readers, I present two proofs:



        METHOD I: DIRICHLET'S TEST



        The Dirichlet's test for uniform convergence can be found here.



        Let $;epsilon>0,; xinBbb{R}$ and $ninBbb{N}$ be fixed but arbitrary. Define



        begin{align}f_n(x)=(-1)^{n+1};;text{and };;g_{n}(x)=dfrac{1}{n+x^4}end{align}
        Then, uniform boundedness is implied, since
        begin{align}
        &left|F_n(x)right|=left|sum^{n}_{m}(-1)^{m+1} right|leq 1,end{align}
        The sequence ${g_n(x)}_{nin Bbb{N}}$ is decreasing, since
        begin{align}
        dfrac{n+x^4}{(n+1)+x^4}leq dfrac{(n+1)+x^4}{(n+1)+x^4}=1implies g_{n+1}(x)leq g_{n}(x)end{align}
        By Archmidean principle, there exists $N(epsilon)$ such that $N(epsilon)>epsilon$ and
        begin{align}
        &left|g_n(x)-0right|=left|dfrac{1}{n+x^4}-0 right|=dfrac{1}{n+x^4}leq dfrac{1}{n}leq dfrac{1}{N}<epsilon,;forall;ngeq N(epsilon)end{align}
        Hence, ${g_n(x)}_{nin Bbb{N}}$ converges uniformly on $Bbb{R}$ and, by Dirichlet's test, $sum^{infty}_{n=1}f_n(x)g_n(x)$ converges uniformly on $Bbb{R}$.



        METHOD II: UNIFORM CAUCHY CRITERION



        Let $;epsilon>0$, $m,ninBbb{N}$ be given such that $mgeq n$ and $xinBbb{R}$ be arbitrary. Define ${f_n(x)}_{nin Bbb{N}}$ and ${g_n(x)}_{nin Bbb{N}}$ as before. Then,



        begin{align}left|sum^{m}_{k=1}f_k(x)g_k(x)-sum^{n}_{k=1}f_k(x)g_k(x) right|&=left|sum^{m}_{k=n+1}f_k(x)g_k(x)right|\&=left|sum^{m}_{k=n+1}(-1)^{k+1}dfrac{1}{k+x^4}right|end{align}
        Expanding, we get
        begin{align}left|sum^{m}_{k=n+1}(-1)^{k+1}dfrac{1}{k+x^4}right|&=left|(-1)^{n+2}dfrac{1}{(n+1)+x^4}+(-1)^{n+3}dfrac{1}{(n+2)+x^4}+cdots +(-1)^{n+1}dfrac{1}{n+x^4}right|\&=left|dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}+cdots +(-1)^{m-n-2}dfrac{1}{(m-1)+x^4}+(-1)^{m-n-1}dfrac{1}{m+x^4}right|\&leq left|dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}right|+cdots +left|dfrac{1}{(m-1)+x^4}-dfrac{1}{m+x^4}right|\&= dfrac{1}{(n+1)+x^4}-dfrac{1}{(n+2)+x^4}+cdots +dfrac{1}{(m-1)+x^4}-dfrac{1}{m+x^4}\&= dfrac{1}{(n+1)+x^4}-left[dfrac{1}{(n+2)+x^4}-dfrac{1}{(n+2)+x^4}right]-cdots \&;-left[dfrac{1}{(m-1)+x^4}-dfrac{1}{(m-1)+x^4}right]-dfrac{1}{m+x^4}\&leq dfrac{1}{(n+1)+x^4}.end{align}
        As shown in Method I, there exists $N(epsilon)$ such that $,;forall;ngeq N(epsilon)$,
        begin{align}
        dfrac{1}{(n+1)+x^4}leq dfrac{1}{n+x^4}leq dfrac{1}{n}leq dfrac{1}{N}<epsilon,end{align}
        and we're done!







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 2 at 20:54

























        answered Jan 2 at 20:49









        Omojola MichealOmojola Micheal

        2,004424




        2,004424






























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