Show that following series converges uniformly on $[0,1]$ if and only if $α > 1$
1
$begingroup$
$$sum_{n=1}^inftyfrac{{x^α}}{1+n^2x^2}$$$α>0$
I tried solving it using Weierstrass M test but i didn't got $M_n$ as function of $n$.
real-analysis power-series uniform-convergence
asked Dec 2 '18 at 11:05
BhowmickBhowmick
1438
$endgroup$
add a comment |
1
$begingroup$
$$sum_{n=1}^inftyfrac{{x^α}}{1+n^2x^2}$$$α>0$
I tried solving it using Weierstrass M test but i didn't got $M_n$ as function of $n$.
real-analysis power-series uniform-convergence
asked Dec 2 '18 at 11:05
BhowmickBhowmick
1438
$endgroup$
$begingroup$
Comparing the series with an integral, one sees that, when $Ntoinfty$, $$sum_{n=N}^inftyfrac{x^a}{1+n^2x^2}sim x^aint_N^inftyfrac{dt}{t^2x^2}=frac1Nx^{a-2}$$ hence the convergence should be uniform on $[0,1]$ for $$ageqslant2$$ Are you sure about the condition $$a>1 ?$$
$endgroup$
– Did
Dec 2 '18 at 11:28
$begingroup$
@Did It appears that's O.K. $$sumlimits_{n=1}^infty frac{x^alpha}{1+n^2x^2}le sumlimits_{n=1}^inftyfrac12 (2-alpha)^{1-alpha/2}alpha^{alpha/2}frac1{n^alpha}$$ for $alphain(1,2]$.
$endgroup$
– user539887
Dec 2 '18 at 11:54
$begingroup$
@user539887 Hmmm... nifty. And even, for every $ain(1,2]$, $$frac{x^a}{1+n^2x^2}leqslantfrac1{n^a}$$
$endgroup$
– Did
Dec 2 '18 at 11:58
add a comment |
1
1
1
2
$begingroup$
$$sum_{n=1}^inftyfrac{{x^α}}{1+n^2x^2}$$$α>0$
I tried solving it using Weierstrass M test but i didn't got $M_n$ as function of $n$.
real-analysis power-series uniform-convergence
asked Dec 2 '18 at 11:05
BhowmickBhowmick
1438
$endgroup$
$$sum_{n=1}^inftyfrac{{x^α}}{1+n^2x^2}$$$α>0$
I tried solving it using Weierstrass M test but i didn't got $M_n$ as function of $n$.
real-analysis power-series uniform-convergence
real-analysis power-series uniform-convergence
asked Dec 2 '18 at 11:05
BhowmickBhowmick
1438
asked Dec 2 '18 at 11:05
BhowmickBhowmick
1438
asked Dec 2 '18 at 11:05
BhowmickBhowmick
1438
asked Dec 2 '18 at 11:05
BhowmickBhowmick
1438
asked Dec 2 '18 at 11:05
BhowmickBhowmick
1438
1438
$begingroup$
Comparing the series with an integral, one sees that, when $Ntoinfty$, $$sum_{n=N}^inftyfrac{x^a}{1+n^2x^2}sim x^aint_N^inftyfrac{dt}{t^2x^2}=frac1Nx^{a-2}$$ hence the convergence should be uniform on $[0,1]$ for $$ageqslant2$$ Are you sure about the condition $$a>1 ?$$
$endgroup$
– Did
Dec 2 '18 at 11:28
$begingroup$
@Did It appears that's O.K. $$sumlimits_{n=1}^infty frac{x^alpha}{1+n^2x^2}le sumlimits_{n=1}^inftyfrac12 (2-alpha)^{1-alpha/2}alpha^{alpha/2}frac1{n^alpha}$$ for $alphain(1,2]$.
$endgroup$
– user539887
Dec 2 '18 at 11:54
$begingroup$
@user539887 Hmmm... nifty. And even, for every $ain(1,2]$, $$frac{x^a}{1+n^2x^2}leqslantfrac1{n^a}$$
$endgroup$
– Did
Dec 2 '18 at 11:58
add a comment |
$begingroup$
Comparing the series with an integral, one sees that, when $Ntoinfty$, $$sum_{n=N}^inftyfrac{x^a}{1+n^2x^2}sim x^aint_N^inftyfrac{dt}{t^2x^2}=frac1Nx^{a-2}$$ hence the convergence should be uniform on $[0,1]$ for $$ageqslant2$$ Are you sure about the condition $$a>1 ?$$
$endgroup$
– Did
Dec 2 '18 at 11:28
$begingroup$
@Did It appears that's O.K. $$sumlimits_{n=1}^infty frac{x^alpha}{1+n^2x^2}le sumlimits_{n=1}^inftyfrac12 (2-alpha)^{1-alpha/2}alpha^{alpha/2}frac1{n^alpha}$$ for $alphain(1,2]$.
$endgroup$
– user539887
Dec 2 '18 at 11:54
$begingroup$
@user539887 Hmmm... nifty. And even, for every $ain(1,2]$, $$frac{x^a}{1+n^2x^2}leqslantfrac1{n^a}$$
$endgroup$
– Did
Dec 2 '18 at 11:58
$begingroup$
Comparing the series with an integral, one sees that, when $Ntoinfty$, $$sum_{n=N}^inftyfrac{x^a}{1+n^2x^2}sim x^aint_N^inftyfrac{dt}{t^2x^2}=frac1Nx^{a-2}$$ hence the convergence should be uniform on $[0,1]$ for $$ageqslant2$$ Are you sure about the condition $$a>1 ?$$
$endgroup$
– Did
Dec 2 '18 at 11:28
$begingroup$
Comparing the series with an integral, one sees that, when $Ntoinfty$, $$sum_{n=N}^inftyfrac{x^a}{1+n^2x^2}sim x^aint_N^inftyfrac{dt}{t^2x^2}=frac1Nx^{a-2}$$ hence the convergence should be uniform on $[0,1]$ for $$ageqslant2$$ Are you sure about the condition $$a>1 ?$$
$endgroup$
– Did
Dec 2 '18 at 11:28
$begingroup$
@Did It appears that's O.K. $$sumlimits_{n=1}^infty frac{x^alpha}{1+n^2x^2}le sumlimits_{n=1}^inftyfrac12 (2-alpha)^{1-alpha/2}alpha^{alpha/2}frac1{n^alpha}$$ for $alphain(1,2]$.
$endgroup$
– user539887
Dec 2 '18 at 11:54
$begingroup$
@Did It appears that's O.K. $$sumlimits_{n=1}^infty frac{x^alpha}{1+n^2x^2}le sumlimits_{n=1}^inftyfrac12 (2-alpha)^{1-alpha/2}alpha^{alpha/2}frac1{n^alpha}$$ for $alphain(1,2]$.
$endgroup$
– user539887
Dec 2 '18 at 11:54
$begingroup$
@user539887 Hmmm... nifty. And even, for every $ain(1,2]$, $$frac{x^a}{1+n^2x^2}leqslantfrac1{n^a}$$
$endgroup$
– Did
Dec 2 '18 at 11:58
$begingroup$
@user539887 Hmmm... nifty. And even, for every $ain(1,2]$, $$frac{x^a}{1+n^2x^2}leqslantfrac1{n^a}$$
$endgroup$
– Did
Dec 2 '18 at 11:58
add a comment |
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$begingroup$
Comparing the series with an integral, one sees that, when $Ntoinfty$, $$sum_{n=N}^inftyfrac{x^a}{1+n^2x^2}sim x^aint_N^inftyfrac{dt}{t^2x^2}=frac1Nx^{a-2}$$ hence the convergence should be uniform on $[0,1]$ for $$ageqslant2$$ Are you sure about the condition $$a>1 ?$$
$endgroup$
– Did
Dec 2 '18 at 11:28
$begingroup$
@Did It appears that's O.K. $$sumlimits_{n=1}^infty frac{x^alpha}{1+n^2x^2}le sumlimits_{n=1}^inftyfrac12 (2-alpha)^{1-alpha/2}alpha^{alpha/2}frac1{n^alpha}$$ for $alphain(1,2]$.
$endgroup$
– user539887
Dec 2 '18 at 11:54
$begingroup$
@user539887 Hmmm... nifty. And even, for every $ain(1,2]$, $$frac{x^a}{1+n^2x^2}leqslantfrac1{n^a}$$
$endgroup$
– Did
Dec 2 '18 at 11:58