Uniform Convergence using Abel's test for a series based on convergence of a series The Next CEO of Stack OverflowShow absolute and uniform convergence of a Fourier seriesTest the uniform convergency of a series of functionTesting a series for uniform convergence using Weierstrass' M testUniform convergence of power series $sum_n=1^infty frac x^n(n+1)(n+2)$Confusion about Uniform Convergence of SeriesUniform Convergence of Power SeriesAre uniform or normal convergence on all $mathbbC$ possible for a power series?Uniform convergence of a series correct?Uniform convergence for series of functionsHow to test the uniform convergence of a function series, without Weierstrass M-test
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Uniform Convergence using Abel's test for a series based on convergence of a series
The Next CEO of Stack OverflowShow absolute and uniform convergence of a Fourier seriesTest the uniform convergency of a series of functionTesting a series for uniform convergence using Weierstrass' M testUniform convergence of power series $sum_n=1^infty frac x^n(n+1)(n+2)$Confusion about Uniform Convergence of SeriesUniform Convergence of Power SeriesAre uniform or normal convergence on all $mathbbC$ possible for a power series?Uniform convergence of a series correct?Uniform convergence for series of functionsHow to test the uniform convergence of a function series, without Weierstrass M-test
$begingroup$
A problem from uniform convergence of series:$$sum_i=1^infty a_n$$ is convergent then show that $$sum_i=1^infty frac nx^n(1-x)1+x^n a_n$$ and $$sum_i=1^infty frac 2nx^n(1-x)1+x^2n a_n$$ are uniformly convergent when $x in [0,1]$.
real-analysis sequences-and-series convergence power-series uniform-convergence
asked Oct 14 '15 at 9:18
LinalgLinalg
13
$endgroup$
add a comment |
$begingroup$
A problem from uniform convergence of series:$$sum_i=1^infty a_n$$ is convergent then show that $$sum_i=1^infty frac nx^n(1-x)1+x^n a_n$$ and $$sum_i=1^infty frac 2nx^n(1-x)1+x^2n a_n$$ are uniformly convergent when $x in [0,1]$.
real-analysis sequences-and-series convergence power-series uniform-convergence
asked Oct 14 '15 at 9:18
LinalgLinalg
13
$endgroup$
$begingroup$
I have tried using Abel's theorem and Dirichlet's theorem. I have not been able to make a headway even after lot of attempts. @Tom-Tom in case you can help, I will be able to save on my study time. Regards
$endgroup$
– Linalg
Oct 14 '15 at 9:31
$begingroup$
What do you mean by uniform convergence of "function $sum a_n$"? Do you mean that $sum a_nx^n$ converges uniformly somewhere and then the other two series also converge uniformly exactly where the first one does?
$endgroup$
– uniquesolution
Oct 14 '15 at 9:41
$begingroup$
@uniquesolution : Edited as per comment. No. $sum a_n$ is given convergent. We have to show the other two series are uniformly convergent. They shall converge to different functions it seems. The question is just to show if the two series are convergent.
$endgroup$
– Linalg
Oct 14 '15 at 9:48
$begingroup$
What do you about the radius of convergence ? Try to see what it means for $sum_na_n=sum_na_n1^n$.
$endgroup$
– Tom-Tom
Oct 14 '15 at 10:53
$begingroup$
@Tom-Tom: I am unable to follow. Can you please elaborate? The link to the problem is here. I am preparing for an exam and am stuck on it since today morning. Requesting your guidance. link
$endgroup$
– Linalg
Oct 14 '15 at 10:56
add a comment |
$begingroup$
A problem from uniform convergence of series:$$sum_i=1^infty a_n$$ is convergent then show that $$sum_i=1^infty frac nx^n(1-x)1+x^n a_n$$ and $$sum_i=1^infty frac 2nx^n(1-x)1+x^2n a_n$$ are uniformly convergent when $x in [0,1]$.
real-analysis sequences-and-series convergence power-series uniform-convergence
asked Oct 14 '15 at 9:18
LinalgLinalg
13
$endgroup$
A problem from uniform convergence of series:$$sum_i=1^infty a_n$$ is convergent then show that $$sum_i=1^infty frac nx^n(1-x)1+x^n a_n$$ and $$sum_i=1^infty frac 2nx^n(1-x)1+x^2n a_n$$ are uniformly convergent when $x in [0,1]$.
real-analysis sequences-and-series convergence power-series uniform-convergence
real-analysis sequences-and-series convergence power-series uniform-convergence
asked Oct 14 '15 at 9:18
LinalgLinalg
13
asked Oct 14 '15 at 9:18
LinalgLinalg
13
edited Oct 14 '15 at 9:51
Linalg
asked Oct 14 '15 at 9:18
LinalgLinalg
13
asked Oct 14 '15 at 9:18
LinalgLinalg
13
asked Oct 14 '15 at 9:18
LinalgLinalg
13
13
$begingroup$
I have tried using Abel's theorem and Dirichlet's theorem. I have not been able to make a headway even after lot of attempts. @Tom-Tom in case you can help, I will be able to save on my study time. Regards
$endgroup$
– Linalg
Oct 14 '15 at 9:31
$begingroup$
What do you mean by uniform convergence of "function $sum a_n$"? Do you mean that $sum a_nx^n$ converges uniformly somewhere and then the other two series also converge uniformly exactly where the first one does?
$endgroup$
– uniquesolution
Oct 14 '15 at 9:41
$begingroup$
@uniquesolution : Edited as per comment. No. $sum a_n$ is given convergent. We have to show the other two series are uniformly convergent. They shall converge to different functions it seems. The question is just to show if the two series are convergent.
$endgroup$
– Linalg
Oct 14 '15 at 9:48
$begingroup$
What do you about the radius of convergence ? Try to see what it means for $sum_na_n=sum_na_n1^n$.
$endgroup$
– Tom-Tom
Oct 14 '15 at 10:53
$begingroup$
@Tom-Tom: I am unable to follow. Can you please elaborate? The link to the problem is here. I am preparing for an exam and am stuck on it since today morning. Requesting your guidance. link
$endgroup$
– Linalg
Oct 14 '15 at 10:56
add a comment |
$begingroup$
I have tried using Abel's theorem and Dirichlet's theorem. I have not been able to make a headway even after lot of attempts. @Tom-Tom in case you can help, I will be able to save on my study time. Regards
$endgroup$
– Linalg
Oct 14 '15 at 9:31
$begingroup$
What do you mean by uniform convergence of "function $sum a_n$"? Do you mean that $sum a_nx^n$ converges uniformly somewhere and then the other two series also converge uniformly exactly where the first one does?
$endgroup$
– uniquesolution
Oct 14 '15 at 9:41
$begingroup$
@uniquesolution : Edited as per comment. No. $sum a_n$ is given convergent. We have to show the other two series are uniformly convergent. They shall converge to different functions it seems. The question is just to show if the two series are convergent.
$endgroup$
– Linalg
Oct 14 '15 at 9:48
$begingroup$
What do you about the radius of convergence ? Try to see what it means for $sum_na_n=sum_na_n1^n$.
$endgroup$
– Tom-Tom
Oct 14 '15 at 10:53
$begingroup$
@Tom-Tom: I am unable to follow. Can you please elaborate? The link to the problem is here. I am preparing for an exam and am stuck on it since today morning. Requesting your guidance. link
$endgroup$
– Linalg
Oct 14 '15 at 10:56
$begingroup$
I have tried using Abel's theorem and Dirichlet's theorem. I have not been able to make a headway even after lot of attempts. @Tom-Tom in case you can help, I will be able to save on my study time. Regards
$endgroup$
– Linalg
Oct 14 '15 at 9:31
$begingroup$
I have tried using Abel's theorem and Dirichlet's theorem. I have not been able to make a headway even after lot of attempts. @Tom-Tom in case you can help, I will be able to save on my study time. Regards
$endgroup$
– Linalg
Oct 14 '15 at 9:31
$begingroup$
What do you mean by uniform convergence of "function $sum a_n$"? Do you mean that $sum a_nx^n$ converges uniformly somewhere and then the other two series also converge uniformly exactly where the first one does?
$endgroup$
– uniquesolution
Oct 14 '15 at 9:41
$begingroup$
What do you mean by uniform convergence of "function $sum a_n$"? Do you mean that $sum a_nx^n$ converges uniformly somewhere and then the other two series also converge uniformly exactly where the first one does?
$endgroup$
– uniquesolution
Oct 14 '15 at 9:41
$begingroup$
@uniquesolution : Edited as per comment. No. $sum a_n$ is given convergent. We have to show the other two series are uniformly convergent. They shall converge to different functions it seems. The question is just to show if the two series are convergent.
$endgroup$
– Linalg
Oct 14 '15 at 9:48
$begingroup$
@uniquesolution : Edited as per comment. No. $sum a_n$ is given convergent. We have to show the other two series are uniformly convergent. They shall converge to different functions it seems. The question is just to show if the two series are convergent.
$endgroup$
– Linalg
Oct 14 '15 at 9:48
$begingroup$
What do you about the radius of convergence ? Try to see what it means for $sum_na_n=sum_na_n1^n$.
$endgroup$
– Tom-Tom
Oct 14 '15 at 10:53
$begingroup$
What do you about the radius of convergence ? Try to see what it means for $sum_na_n=sum_na_n1^n$.
$endgroup$
– Tom-Tom
Oct 14 '15 at 10:53
$begingroup$
@Tom-Tom: I am unable to follow. Can you please elaborate? The link to the problem is here. I am preparing for an exam and am stuck on it since today morning. Requesting your guidance. link
$endgroup$
– Linalg
Oct 14 '15 at 10:56
$begingroup$
@Tom-Tom: I am unable to follow. Can you please elaborate? The link to the problem is here. I am preparing for an exam and am stuck on it since today morning. Requesting your guidance. link
$endgroup$
– Linalg
Oct 14 '15 at 10:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As per second thoughts @tom-tom, requesting you to look into this
$ 0le x le 1 Rightarrow x^r le 1 quad forall r ge 0 $
$Rightarrow 0le x^n le x^r quad forall rle n & r ge 0$
$therefore sum_r=0^n-1 x^r ge n.x^n ge 0$
$ Rightarrow frac 1-x^n1-x ge nx^n ge 0$
$Rightarrow 0lefrac nx^n(1-x)1-x^n le 1$
$$Rightarrow 0lefrac nx^n(1-x)1+x^n le 1$$
$ <b_n(x)> = frac nx^n(1-x)1+x^n$ is bounded.
$sum a_n$ is given as convergent. As it is free from $x$, $therefore$ it is uniformly convergent.
Now, $ 1. <a_n(x)> $ is uniformly convergent,
$ 2. <b_n(x)> $ is bounded for $ x in [0,1] $, $quad$ 3. $ <b_n(x)> $ shall be piecewise monotonic.
$therefore$ $ quad sum a_n(x) b_n(x)$ shall be uniformly convergent.
$$ Rightarrow sum_i=1^infty frac nx^n(1-x)1+x^n a_n $$ is convergent.
Similarly other one can be proved. Please let me know of mistakes, if any.
answered Oct 14 '15 at 11:38
LinalgLinalg
13
$endgroup$
add a comment |
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1 Answer
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$begingroup$
As per second thoughts @tom-tom, requesting you to look into this
$ 0le x le 1 Rightarrow x^r le 1 quad forall r ge 0 $
$Rightarrow 0le x^n le x^r quad forall rle n & r ge 0$
$therefore sum_r=0^n-1 x^r ge n.x^n ge 0$
$ Rightarrow frac 1-x^n1-x ge nx^n ge 0$
$Rightarrow 0lefrac nx^n(1-x)1-x^n le 1$
$$Rightarrow 0lefrac nx^n(1-x)1+x^n le 1$$
$ <b_n(x)> = frac nx^n(1-x)1+x^n$ is bounded.
$sum a_n$ is given as convergent. As it is free from $x$, $therefore$ it is uniformly convergent.
Now, $ 1. <a_n(x)> $ is uniformly convergent,
$ 2. <b_n(x)> $ is bounded for $ x in [0,1] $, $quad$ 3. $ <b_n(x)> $ shall be piecewise monotonic.
$therefore$ $ quad sum a_n(x) b_n(x)$ shall be uniformly convergent.
$$ Rightarrow sum_i=1^infty frac nx^n(1-x)1+x^n a_n $$ is convergent.
Similarly other one can be proved. Please let me know of mistakes, if any.
answered Oct 14 '15 at 11:38
LinalgLinalg
13
$endgroup$
add a comment |
$begingroup$
As per second thoughts @tom-tom, requesting you to look into this
$ 0le x le 1 Rightarrow x^r le 1 quad forall r ge 0 $
$Rightarrow 0le x^n le x^r quad forall rle n & r ge 0$
$therefore sum_r=0^n-1 x^r ge n.x^n ge 0$
$ Rightarrow frac 1-x^n1-x ge nx^n ge 0$
$Rightarrow 0lefrac nx^n(1-x)1-x^n le 1$
$$Rightarrow 0lefrac nx^n(1-x)1+x^n le 1$$
$ <b_n(x)> = frac nx^n(1-x)1+x^n$ is bounded.
$sum a_n$ is given as convergent. As it is free from $x$, $therefore$ it is uniformly convergent.
Now, $ 1. <a_n(x)> $ is uniformly convergent,
$ 2. <b_n(x)> $ is bounded for $ x in [0,1] $, $quad$ 3. $ <b_n(x)> $ shall be piecewise monotonic.
$therefore$ $ quad sum a_n(x) b_n(x)$ shall be uniformly convergent.
$$ Rightarrow sum_i=1^infty frac nx^n(1-x)1+x^n a_n $$ is convergent.
Similarly other one can be proved. Please let me know of mistakes, if any.
answered Oct 14 '15 at 11:38
LinalgLinalg
13
$endgroup$
add a comment |
$begingroup$
As per second thoughts @tom-tom, requesting you to look into this
$ 0le x le 1 Rightarrow x^r le 1 quad forall r ge 0 $
$Rightarrow 0le x^n le x^r quad forall rle n & r ge 0$
$therefore sum_r=0^n-1 x^r ge n.x^n ge 0$
$ Rightarrow frac 1-x^n1-x ge nx^n ge 0$
$Rightarrow 0lefrac nx^n(1-x)1-x^n le 1$
$$Rightarrow 0lefrac nx^n(1-x)1+x^n le 1$$
$ <b_n(x)> = frac nx^n(1-x)1+x^n$ is bounded.
$sum a_n$ is given as convergent. As it is free from $x$, $therefore$ it is uniformly convergent.
Now, $ 1. <a_n(x)> $ is uniformly convergent,
$ 2. <b_n(x)> $ is bounded for $ x in [0,1] $, $quad$ 3. $ <b_n(x)> $ shall be piecewise monotonic.
$therefore$ $ quad sum a_n(x) b_n(x)$ shall be uniformly convergent.
$$ Rightarrow sum_i=1^infty frac nx^n(1-x)1+x^n a_n $$ is convergent.
Similarly other one can be proved. Please let me know of mistakes, if any.
answered Oct 14 '15 at 11:38
LinalgLinalg
13
$endgroup$
As per second thoughts @tom-tom, requesting you to look into this
$ 0le x le 1 Rightarrow x^r le 1 quad forall r ge 0 $
$Rightarrow 0le x^n le x^r quad forall rle n & r ge 0$
$therefore sum_r=0^n-1 x^r ge n.x^n ge 0$
$ Rightarrow frac 1-x^n1-x ge nx^n ge 0$
$Rightarrow 0lefrac nx^n(1-x)1-x^n le 1$
$$Rightarrow 0lefrac nx^n(1-x)1+x^n le 1$$
$ <b_n(x)> = frac nx^n(1-x)1+x^n$ is bounded.
$sum a_n$ is given as convergent. As it is free from $x$, $therefore$ it is uniformly convergent.
Now, $ 1. <a_n(x)> $ is uniformly convergent,
$ 2. <b_n(x)> $ is bounded for $ x in [0,1] $, $quad$ 3. $ <b_n(x)> $ shall be piecewise monotonic.
$therefore$ $ quad sum a_n(x) b_n(x)$ shall be uniformly convergent.
$$ Rightarrow sum_i=1^infty frac nx^n(1-x)1+x^n a_n $$ is convergent.
Similarly other one can be proved. Please let me know of mistakes, if any.
answered Oct 14 '15 at 11:38
LinalgLinalg
13
answered Oct 14 '15 at 11:38
LinalgLinalg
13
answered Oct 14 '15 at 11:38
LinalgLinalg
13
answered Oct 14 '15 at 11:38
LinalgLinalg
13
13
add a comment |
add a comment |
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$begingroup$
I have tried using Abel's theorem and Dirichlet's theorem. I have not been able to make a headway even after lot of attempts. @Tom-Tom in case you can help, I will be able to save on my study time. Regards
$endgroup$
– Linalg
Oct 14 '15 at 9:31
$begingroup$
What do you mean by uniform convergence of "function $sum a_n$"? Do you mean that $sum a_nx^n$ converges uniformly somewhere and then the other two series also converge uniformly exactly where the first one does?
$endgroup$
– uniquesolution
Oct 14 '15 at 9:41
$begingroup$
@uniquesolution : Edited as per comment. No. $sum a_n$ is given convergent. We have to show the other two series are uniformly convergent. They shall converge to different functions it seems. The question is just to show if the two series are convergent.
$endgroup$
– Linalg
Oct 14 '15 at 9:48
$begingroup$
What do you about the radius of convergence ? Try to see what it means for $sum_na_n=sum_na_n1^n$.
$endgroup$
– Tom-Tom
Oct 14 '15 at 10:53
$begingroup$
@Tom-Tom: I am unable to follow. Can you please elaborate? The link to the problem is here. I am preparing for an exam and am stuck on it since today morning. Requesting your guidance. link
$endgroup$
– Linalg
Oct 14 '15 at 10:56