Mixing
Uniform convergence of this series?
$begingroup$
Let us consider : $f_n : mathbb{R}ni x mapsto frac{x}{(x^2+n^2)log(n)}inmathbb{R}$ for $n> 1$.
I need to prove that $sum _{nge 0}f_n$ is uniformly convergent.
I've already proved that it is not normally convergent by studying the term $(f_n)'$.
Thanks in advance !
real-analysis sequences-and-series convergence uniform-convergence
edited Jan 15 at 22:47
Song
11.2k628
asked Jan 9 at 1:31
MamanMaman
1,174722
$endgroup$
add a comment |
$begingroup$
Let us consider : $f_n : mathbb{R}ni x mapsto frac{x}{(x^2+n^2)log(n)}inmathbb{R}$ for $n> 1$.
I need to prove that $sum _{nge 0}f_n$ is uniformly convergent.
I've already proved that it is not normally convergent by studying the term $(f_n)'$.
Thanks in advance !
real-analysis sequences-and-series convergence uniform-convergence
edited Jan 15 at 22:47
Song
11.2k628
asked Jan 9 at 1:31
MamanMaman
1,174722
$endgroup$
$begingroup$
@Dzoooks Indeed it's $n>1$, my mistake. Your hint seems to be linked with normal convergence of the series ?
$endgroup$
– Maman
Jan 9 at 1:46
$begingroup$
Indeed, uniform convergence basically means it is independent of $x$.
$endgroup$
– Dzoooks
Jan 9 at 1:50
$begingroup$
@Dzoooks Ok I know what you meant ! It depends on the references, but some people use two notions : "uniform convergence" and "normal convergence" and the second one implies the first. And the goal of the exercise is to prove that there is only uniform convergence with no normal convergence.
$endgroup$
– Maman
Jan 9 at 2:02
2
$begingroup$
@Dzoooks your inequality isn't true. Take $x=n$.
$endgroup$
– mouthetics
Jan 9 at 2:08
add a comment |
$begingroup$
Let us consider : $f_n : mathbb{R}ni x mapsto frac{x}{(x^2+n^2)log(n)}inmathbb{R}$ for $n> 1$.
I need to prove that $sum _{nge 0}f_n$ is uniformly convergent.
I've already proved that it is not normally convergent by studying the term $(f_n)'$.
Thanks in advance !
real-analysis sequences-and-series convergence uniform-convergence
edited Jan 15 at 22:47
Song
11.2k628
asked Jan 9 at 1:31
MamanMaman
1,174722
$endgroup$
Let us consider : $f_n : mathbb{R}ni x mapsto frac{x}{(x^2+n^2)log(n)}inmathbb{R}$ for $n> 1$.
I need to prove that $sum _{nge 0}f_n$ is uniformly convergent.
I've already proved that it is not normally convergent by studying the term $(f_n)'$.
Thanks in advance !
real-analysis sequences-and-series convergence uniform-convergence
real-analysis sequences-and-series convergence uniform-convergence
edited Jan 15 at 22:47
Song
11.2k628
asked Jan 9 at 1:31
MamanMaman
1,174722
edited Jan 15 at 22:47
Song
11.2k628
asked Jan 9 at 1:31
MamanMaman
1,174722
edited Jan 15 at 22:47
Song
11.2k628
edited Jan 15 at 22:47
Song
11.2k628
edited Jan 15 at 22:47
Song
11.2k628
11.2k628
asked Jan 9 at 1:31
MamanMaman
1,174722
asked Jan 9 at 1:31
MamanMaman
1,174722
asked Jan 9 at 1:31
MamanMaman
1,174722
1,174722
$begingroup$
@Dzoooks Indeed it's $n>1$, my mistake. Your hint seems to be linked with normal convergence of the series ?
$endgroup$
– Maman
Jan 9 at 1:46
$begingroup$
Indeed, uniform convergence basically means it is independent of $x$.
$endgroup$
– Dzoooks
Jan 9 at 1:50
$begingroup$
@Dzoooks Ok I know what you meant ! It depends on the references, but some people use two notions : "uniform convergence" and "normal convergence" and the second one implies the first. And the goal of the exercise is to prove that there is only uniform convergence with no normal convergence.
$endgroup$
– Maman
Jan 9 at 2:02
2
$begingroup$
@Dzoooks your inequality isn't true. Take $x=n$.
$endgroup$
– mouthetics
Jan 9 at 2:08
add a comment |
$begingroup$
@Dzoooks Indeed it's $n>1$, my mistake. Your hint seems to be linked with normal convergence of the series ?
$endgroup$
– Maman
Jan 9 at 1:46
$begingroup$
Indeed, uniform convergence basically means it is independent of $x$.
$endgroup$
– Dzoooks
Jan 9 at 1:50
$begingroup$
@Dzoooks Ok I know what you meant ! It depends on the references, but some people use two notions : "uniform convergence" and "normal convergence" and the second one implies the first. And the goal of the exercise is to prove that there is only uniform convergence with no normal convergence.
$endgroup$
– Maman
Jan 9 at 2:02
2
$begingroup$
@Dzoooks your inequality isn't true. Take $x=n$.
$endgroup$
– mouthetics
Jan 9 at 2:08
$begingroup$
@Dzoooks Indeed it's $n>1$, my mistake. Your hint seems to be linked with normal convergence of the series ?
$endgroup$
– Maman
Jan 9 at 1:46
$begingroup$
@Dzoooks Indeed it's $n>1$, my mistake. Your hint seems to be linked with normal convergence of the series ?
$endgroup$
– Maman
Jan 9 at 1:46
$begingroup$
Indeed, uniform convergence basically means it is independent of $x$.
$endgroup$
– Dzoooks
Jan 9 at 1:50
$begingroup$
Indeed, uniform convergence basically means it is independent of $x$.
$endgroup$
– Dzoooks
Jan 9 at 1:50
$begingroup$
@Dzoooks Ok I know what you meant ! It depends on the references, but some people use two notions : "uniform convergence" and "normal convergence" and the second one implies the first. And the goal of the exercise is to prove that there is only uniform convergence with no normal convergence.
$endgroup$
– Maman
Jan 9 at 2:02
$begingroup$
@Dzoooks Ok I know what you meant ! It depends on the references, but some people use two notions : "uniform convergence" and "normal convergence" and the second one implies the first. And the goal of the exercise is to prove that there is only uniform convergence with no normal convergence.
$endgroup$
– Maman
Jan 9 at 2:02
2
2
$begingroup$
@Dzoooks your inequality isn't true. Take $x=n$.
$endgroup$
– mouthetics
Jan 9 at 2:08
$begingroup$
@Dzoooks your inequality isn't true. Take $x=n$.
$endgroup$
– mouthetics
Jan 9 at 2:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We will use Cauchy's test. Let us consider
$$
Big|sum_{k=n+1}^m f_k(x)Big|=sum_{k=n+1}^m frac{|x|}{(x^2+k^2)log k}
$$ for $m>n$. We can see that
$$begin{eqnarray}
sum_{k=n+1}^m frac{|x|}{(x^2+k^2)log k}&le& frac{1}{log n}sum_{k=n+1}^m frac{|x|}{x^2+k^2}\
&le&frac{1}{log n}int_n^m frac{|x|}{x^2+y^2}dy\
&=&frac{1}{log n}int_{frac{n}{|x|}}^{frac{m}{|x|}} frac{1}{1+z^2}dz\
&le&frac{1}{log n}int_{0}^{infty} frac{1}{1+z^2}dz\
&=&frac{pi}{2log n}
end{eqnarray}$$ for all $xinmathbb{R}$. Thus, we have
$$
lim_{n,mtoinfty}Big|sum_{k=n+1}^m f_kBig|_infty lelim_{ntoinfty}frac{pi}{2log n}=0
$$ and Cauchy's test gives the result.
answered Jan 9 at 6:15
SongSong
11.2k628
$endgroup$
$begingroup$
Thank you but my mistake, it's the "series" not the "sequence" of functions. And with your argument it proves that this series is not normal convergent !
$endgroup$
– Maman
Jan 9 at 14:48
$begingroup$
@Maman I've edited my answer.
$endgroup$
– Song
Jan 9 at 16:36
$begingroup$
Thank you but can you justify the inequality : $sumlimits_{k=n+1}^{m} frac{mid x mid}{x^2 +k^2}le int limits_{n}^{m} frac{mid x mid}{x^2+y^2}mathrm{d}y$ ?
$endgroup$
– Maman
Jan 9 at 18:32
1
$begingroup$
@Maman It is derived from$$ frac{1}{x^2+k^2}= int_{k-1}^k frac{1}{x^2+k^2}dyle int_{k-1}^k frac{1}{x^2+y^2}dy.$$ Compare the area below the graph of $ymapsto frac{1}{x^2+y^2}$.
$endgroup$
– Song
Jan 9 at 18:44
$begingroup$
Indeed very nice !
$endgroup$
– Maman
Jan 9 at 18:59
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We will use Cauchy's test. Let us consider
$$
Big|sum_{k=n+1}^m f_k(x)Big|=sum_{k=n+1}^m frac{|x|}{(x^2+k^2)log k}
$$ for $m>n$. We can see that
$$begin{eqnarray}
sum_{k=n+1}^m frac{|x|}{(x^2+k^2)log k}&le& frac{1}{log n}sum_{k=n+1}^m frac{|x|}{x^2+k^2}\
&le&frac{1}{log n}int_n^m frac{|x|}{x^2+y^2}dy\
&=&frac{1}{log n}int_{frac{n}{|x|}}^{frac{m}{|x|}} frac{1}{1+z^2}dz\
&le&frac{1}{log n}int_{0}^{infty} frac{1}{1+z^2}dz\
&=&frac{pi}{2log n}
end{eqnarray}$$ for all $xinmathbb{R}$. Thus, we have
$$
lim_{n,mtoinfty}Big|sum_{k=n+1}^m f_kBig|_infty lelim_{ntoinfty}frac{pi}{2log n}=0
$$ and Cauchy's test gives the result.
answered Jan 9 at 6:15
SongSong
11.2k628
$endgroup$
$begingroup$
Thank you but my mistake, it's the "series" not the "sequence" of functions. And with your argument it proves that this series is not normal convergent !
$endgroup$
– Maman
Jan 9 at 14:48
$begingroup$
@Maman I've edited my answer.
$endgroup$
– Song
Jan 9 at 16:36
$begingroup$
Thank you but can you justify the inequality : $sumlimits_{k=n+1}^{m} frac{mid x mid}{x^2 +k^2}le int limits_{n}^{m} frac{mid x mid}{x^2+y^2}mathrm{d}y$ ?
$endgroup$
– Maman
Jan 9 at 18:32
1
$begingroup$
@Maman It is derived from$$ frac{1}{x^2+k^2}= int_{k-1}^k frac{1}{x^2+k^2}dyle int_{k-1}^k frac{1}{x^2+y^2}dy.$$ Compare the area below the graph of $ymapsto frac{1}{x^2+y^2}$.
$endgroup$
– Song
Jan 9 at 18:44
$begingroup$
Indeed very nice !
$endgroup$
– Maman
Jan 9 at 18:59
add a comment |
$begingroup$
We will use Cauchy's test. Let us consider
$$
Big|sum_{k=n+1}^m f_k(x)Big|=sum_{k=n+1}^m frac{|x|}{(x^2+k^2)log k}
$$ for $m>n$. We can see that
$$begin{eqnarray}
sum_{k=n+1}^m frac{|x|}{(x^2+k^2)log k}&le& frac{1}{log n}sum_{k=n+1}^m frac{|x|}{x^2+k^2}\
&le&frac{1}{log n}int_n^m frac{|x|}{x^2+y^2}dy\
&=&frac{1}{log n}int_{frac{n}{|x|}}^{frac{m}{|x|}} frac{1}{1+z^2}dz\
&le&frac{1}{log n}int_{0}^{infty} frac{1}{1+z^2}dz\
&=&frac{pi}{2log n}
end{eqnarray}$$ for all $xinmathbb{R}$. Thus, we have
$$
lim_{n,mtoinfty}Big|sum_{k=n+1}^m f_kBig|_infty lelim_{ntoinfty}frac{pi}{2log n}=0
$$ and Cauchy's test gives the result.
answered Jan 9 at 6:15
SongSong
11.2k628
$endgroup$
$begingroup$
Thank you but my mistake, it's the "series" not the "sequence" of functions. And with your argument it proves that this series is not normal convergent !
$endgroup$
– Maman
Jan 9 at 14:48
$begingroup$
@Maman I've edited my answer.
$endgroup$
– Song
Jan 9 at 16:36
$begingroup$
Thank you but can you justify the inequality : $sumlimits_{k=n+1}^{m} frac{mid x mid}{x^2 +k^2}le int limits_{n}^{m} frac{mid x mid}{x^2+y^2}mathrm{d}y$ ?
$endgroup$
– Maman
Jan 9 at 18:32
1
$begingroup$
@Maman It is derived from$$ frac{1}{x^2+k^2}= int_{k-1}^k frac{1}{x^2+k^2}dyle int_{k-1}^k frac{1}{x^2+y^2}dy.$$ Compare the area below the graph of $ymapsto frac{1}{x^2+y^2}$.
$endgroup$
– Song
Jan 9 at 18:44
$begingroup$
Indeed very nice !
$endgroup$
– Maman
Jan 9 at 18:59
add a comment |
$begingroup$
We will use Cauchy's test. Let us consider
$$
Big|sum_{k=n+1}^m f_k(x)Big|=sum_{k=n+1}^m frac{|x|}{(x^2+k^2)log k}
$$ for $m>n$. We can see that
$$begin{eqnarray}
sum_{k=n+1}^m frac{|x|}{(x^2+k^2)log k}&le& frac{1}{log n}sum_{k=n+1}^m frac{|x|}{x^2+k^2}\
&le&frac{1}{log n}int_n^m frac{|x|}{x^2+y^2}dy\
&=&frac{1}{log n}int_{frac{n}{|x|}}^{frac{m}{|x|}} frac{1}{1+z^2}dz\
&le&frac{1}{log n}int_{0}^{infty} frac{1}{1+z^2}dz\
&=&frac{pi}{2log n}
end{eqnarray}$$ for all $xinmathbb{R}$. Thus, we have
$$
lim_{n,mtoinfty}Big|sum_{k=n+1}^m f_kBig|_infty lelim_{ntoinfty}frac{pi}{2log n}=0
$$ and Cauchy's test gives the result.
answered Jan 9 at 6:15
SongSong
11.2k628
$endgroup$
We will use Cauchy's test. Let us consider
$$
Big|sum_{k=n+1}^m f_k(x)Big|=sum_{k=n+1}^m frac{|x|}{(x^2+k^2)log k}
$$ for $m>n$. We can see that
$$begin{eqnarray}
sum_{k=n+1}^m frac{|x|}{(x^2+k^2)log k}&le& frac{1}{log n}sum_{k=n+1}^m frac{|x|}{x^2+k^2}\
&le&frac{1}{log n}int_n^m frac{|x|}{x^2+y^2}dy\
&=&frac{1}{log n}int_{frac{n}{|x|}}^{frac{m}{|x|}} frac{1}{1+z^2}dz\
&le&frac{1}{log n}int_{0}^{infty} frac{1}{1+z^2}dz\
&=&frac{pi}{2log n}
end{eqnarray}$$ for all $xinmathbb{R}$. Thus, we have
$$
lim_{n,mtoinfty}Big|sum_{k=n+1}^m f_kBig|_infty lelim_{ntoinfty}frac{pi}{2log n}=0
$$ and Cauchy's test gives the result.
answered Jan 9 at 6:15
SongSong
11.2k628
edited Jan 9 at 16:42
answered Jan 9 at 6:15
SongSong
11.2k628
answered Jan 9 at 6:15
SongSong
11.2k628
answered Jan 9 at 6:15
SongSong
11.2k628
11.2k628
$begingroup$
Thank you but my mistake, it's the "series" not the "sequence" of functions. And with your argument it proves that this series is not normal convergent !
$endgroup$
– Maman
Jan 9 at 14:48
$begingroup$
@Maman I've edited my answer.
$endgroup$
– Song
Jan 9 at 16:36
$begingroup$
Thank you but can you justify the inequality : $sumlimits_{k=n+1}^{m} frac{mid x mid}{x^2 +k^2}le int limits_{n}^{m} frac{mid x mid}{x^2+y^2}mathrm{d}y$ ?
$endgroup$
– Maman
Jan 9 at 18:32
1
$begingroup$
@Maman It is derived from$$ frac{1}{x^2+k^2}= int_{k-1}^k frac{1}{x^2+k^2}dyle int_{k-1}^k frac{1}{x^2+y^2}dy.$$ Compare the area below the graph of $ymapsto frac{1}{x^2+y^2}$.
$endgroup$
– Song
Jan 9 at 18:44
$begingroup$
Indeed very nice !
$endgroup$
– Maman
Jan 9 at 18:59
add a comment |
$begingroup$
Thank you but my mistake, it's the "series" not the "sequence" of functions. And with your argument it proves that this series is not normal convergent !
$endgroup$
– Maman
Jan 9 at 14:48
$begingroup$
@Maman I've edited my answer.
$endgroup$
– Song
Jan 9 at 16:36
$begingroup$
Thank you but can you justify the inequality : $sumlimits_{k=n+1}^{m} frac{mid x mid}{x^2 +k^2}le int limits_{n}^{m} frac{mid x mid}{x^2+y^2}mathrm{d}y$ ?
$endgroup$
– Maman
Jan 9 at 18:32
1
$begingroup$
@Maman It is derived from$$ frac{1}{x^2+k^2}= int_{k-1}^k frac{1}{x^2+k^2}dyle int_{k-1}^k frac{1}{x^2+y^2}dy.$$ Compare the area below the graph of $ymapsto frac{1}{x^2+y^2}$.
$endgroup$
– Song
Jan 9 at 18:44
$begingroup$
Indeed very nice !
$endgroup$
– Maman
Jan 9 at 18:59
$begingroup$
Thank you but my mistake, it's the "series" not the "sequence" of functions. And with your argument it proves that this series is not normal convergent !
$endgroup$
– Maman
Jan 9 at 14:48
$begingroup$
Thank you but my mistake, it's the "series" not the "sequence" of functions. And with your argument it proves that this series is not normal convergent !
$endgroup$
– Maman
Jan 9 at 14:48
$begingroup$
@Maman I've edited my answer.
$endgroup$
– Song
Jan 9 at 16:36
$begingroup$
@Maman I've edited my answer.
$endgroup$
– Song
Jan 9 at 16:36
$begingroup$
Thank you but can you justify the inequality : $sumlimits_{k=n+1}^{m} frac{mid x mid}{x^2 +k^2}le int limits_{n}^{m} frac{mid x mid}{x^2+y^2}mathrm{d}y$ ?
$endgroup$
– Maman
Jan 9 at 18:32
$begingroup$
Thank you but can you justify the inequality : $sumlimits_{k=n+1}^{m} frac{mid x mid}{x^2 +k^2}le int limits_{n}^{m} frac{mid x mid}{x^2+y^2}mathrm{d}y$ ?
$endgroup$
– Maman
Jan 9 at 18:32
1
1
$begingroup$
@Maman It is derived from$$ frac{1}{x^2+k^2}= int_{k-1}^k frac{1}{x^2+k^2}dyle int_{k-1}^k frac{1}{x^2+y^2}dy.$$ Compare the area below the graph of $ymapsto frac{1}{x^2+y^2}$.
$endgroup$
– Song
Jan 9 at 18:44
$begingroup$
@Maman It is derived from$$ frac{1}{x^2+k^2}= int_{k-1}^k frac{1}{x^2+k^2}dyle int_{k-1}^k frac{1}{x^2+y^2}dy.$$ Compare the area below the graph of $ymapsto frac{1}{x^2+y^2}$.
$endgroup$
– Song
Jan 9 at 18:44
$begingroup$
Indeed very nice !
$endgroup$
– Maman
Jan 9 at 18:59
$begingroup$
Indeed very nice !
$endgroup$
– Maman
Jan 9 at 18:59
add a comment |
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$begingroup$
@Dzoooks Indeed it's $n>1$, my mistake. Your hint seems to be linked with normal convergence of the series ?
$endgroup$
– Maman
Jan 9 at 1:46
$begingroup$
Indeed, uniform convergence basically means it is independent of $x$.
$endgroup$
– Dzoooks
Jan 9 at 1:50
$begingroup$
@Dzoooks Ok I know what you meant ! It depends on the references, but some people use two notions : "uniform convergence" and "normal convergence" and the second one implies the first. And the goal of the exercise is to prove that there is only uniform convergence with no normal convergence.
$endgroup$
– Maman
Jan 9 at 2:02
2
$begingroup$
@Dzoooks your inequality isn't true. Take $x=n$.
$endgroup$
– mouthetics
Jan 9 at 2:08