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Prove uniform convergence of $sumlimits_{n=1}^infty frac{-1}{n(nx+1)^2}$ on $[a,infty)$
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I need to prove that the limit function of a function series is differentiable on $[a,infty)$, where $a>0$. I wanted to use the theorem that the function series has a derivative if the function is continuously differentiable, the function series is pointwise convergent and and the function series with the derivative of the function is uniformly convergent. The function is
$sumlimits_{n=1}^infty frac{1}{n^2+n^3x}.$
I have proven that it is continuous and uniformly convergent, hence also pointwise convergent.
In order to prove that
$sumlimits_{n=1}^infty frac{-1}{n(nx+1)^2}$ is uniformly convergent on $[a,infty)$, I wanted to prove that $sumlimits_{n=1}^infty frac{-1}{n(na+1)^2}$ is pointwise convergent and then use the Weierstrass M-test to conclude that $sumlimits_{n=1}^infty frac{-1}{n(nx+1)^2}$ is uniformly convergent on $[a,infty)$. However I can prove that it is uniformly convergent for $a>1$ and $a=1$, as $frac{1}{n^2}$ is a convergent majorant, but I got stuck if $0<a<1$. So how could I prove that it is then also convergent? The ratio test fails.
real-analysis sequences-and-series convergence pointwise-convergence
asked Jan 25 at 16:49
C. EliasC. Elias
223
$endgroup$
add a comment |
$begingroup$
I need to prove that the limit function of a function series is differentiable on $[a,infty)$, where $a>0$. I wanted to use the theorem that the function series has a derivative if the function is continuously differentiable, the function series is pointwise convergent and and the function series with the derivative of the function is uniformly convergent. The function is
$sumlimits_{n=1}^infty frac{1}{n^2+n^3x}.$
I have proven that it is continuous and uniformly convergent, hence also pointwise convergent.
In order to prove that
$sumlimits_{n=1}^infty frac{-1}{n(nx+1)^2}$ is uniformly convergent on $[a,infty)$, I wanted to prove that $sumlimits_{n=1}^infty frac{-1}{n(na+1)^2}$ is pointwise convergent and then use the Weierstrass M-test to conclude that $sumlimits_{n=1}^infty frac{-1}{n(nx+1)^2}$ is uniformly convergent on $[a,infty)$. However I can prove that it is uniformly convergent for $a>1$ and $a=1$, as $frac{1}{n^2}$ is a convergent majorant, but I got stuck if $0<a<1$. So how could I prove that it is then also convergent? The ratio test fails.
real-analysis sequences-and-series convergence pointwise-convergence
asked Jan 25 at 16:49
C. EliasC. Elias
223
$endgroup$
$begingroup$
Can you use the fact that $anle an+1$?
$endgroup$
– kimchi lover
Jan 25 at 16:58
$begingroup$
I don't think so, because then you get $n(na)^2=n^3a$, but you can't be sure that this is smaller than $n^2$
$endgroup$
– C. Elias
Jan 25 at 17:40
add a comment |
$begingroup$
I need to prove that the limit function of a function series is differentiable on $[a,infty)$, where $a>0$. I wanted to use the theorem that the function series has a derivative if the function is continuously differentiable, the function series is pointwise convergent and and the function series with the derivative of the function is uniformly convergent. The function is
$sumlimits_{n=1}^infty frac{1}{n^2+n^3x}.$
I have proven that it is continuous and uniformly convergent, hence also pointwise convergent.
In order to prove that
$sumlimits_{n=1}^infty frac{-1}{n(nx+1)^2}$ is uniformly convergent on $[a,infty)$, I wanted to prove that $sumlimits_{n=1}^infty frac{-1}{n(na+1)^2}$ is pointwise convergent and then use the Weierstrass M-test to conclude that $sumlimits_{n=1}^infty frac{-1}{n(nx+1)^2}$ is uniformly convergent on $[a,infty)$. However I can prove that it is uniformly convergent for $a>1$ and $a=1$, as $frac{1}{n^2}$ is a convergent majorant, but I got stuck if $0<a<1$. So how could I prove that it is then also convergent? The ratio test fails.
real-analysis sequences-and-series convergence pointwise-convergence
asked Jan 25 at 16:49
C. EliasC. Elias
223
$endgroup$
I need to prove that the limit function of a function series is differentiable on $[a,infty)$, where $a>0$. I wanted to use the theorem that the function series has a derivative if the function is continuously differentiable, the function series is pointwise convergent and and the function series with the derivative of the function is uniformly convergent. The function is
$sumlimits_{n=1}^infty frac{1}{n^2+n^3x}.$
I have proven that it is continuous and uniformly convergent, hence also pointwise convergent.
In order to prove that
$sumlimits_{n=1}^infty frac{-1}{n(nx+1)^2}$ is uniformly convergent on $[a,infty)$, I wanted to prove that $sumlimits_{n=1}^infty frac{-1}{n(na+1)^2}$ is pointwise convergent and then use the Weierstrass M-test to conclude that $sumlimits_{n=1}^infty frac{-1}{n(nx+1)^2}$ is uniformly convergent on $[a,infty)$. However I can prove that it is uniformly convergent for $a>1$ and $a=1$, as $frac{1}{n^2}$ is a convergent majorant, but I got stuck if $0<a<1$. So how could I prove that it is then also convergent? The ratio test fails.
real-analysis sequences-and-series convergence pointwise-convergence
real-analysis sequences-and-series convergence pointwise-convergence
asked Jan 25 at 16:49
C. EliasC. Elias
223
asked Jan 25 at 16:49
C. EliasC. Elias
223
edited Jan 25 at 17:39
C. Elias
asked Jan 25 at 16:49
C. EliasC. Elias
223
asked Jan 25 at 16:49
C. EliasC. Elias
223
asked Jan 25 at 16:49
C. EliasC. Elias
223
223
$begingroup$
Can you use the fact that $anle an+1$?
$endgroup$
– kimchi lover
Jan 25 at 16:58
$begingroup$
I don't think so, because then you get $n(na)^2=n^3a$, but you can't be sure that this is smaller than $n^2$
$endgroup$
– C. Elias
Jan 25 at 17:40
add a comment |
$begingroup$
Can you use the fact that $anle an+1$?
$endgroup$
– kimchi lover
Jan 25 at 16:58
$begingroup$
I don't think so, because then you get $n(na)^2=n^3a$, but you can't be sure that this is smaller than $n^2$
$endgroup$
– C. Elias
Jan 25 at 17:40
$begingroup$
Can you use the fact that $anle an+1$?
$endgroup$
– kimchi lover
Jan 25 at 16:58
$begingroup$
Can you use the fact that $anle an+1$?
$endgroup$
– kimchi lover
Jan 25 at 16:58
$begingroup$
I don't think so, because then you get $n(na)^2=n^3a$, but you can't be sure that this is smaller than $n^2$
$endgroup$
– C. Elias
Jan 25 at 17:40
$begingroup$
I don't think so, because then you get $n(na)^2=n^3a$, but you can't be sure that this is smaller than $n^2$
$endgroup$
– C. Elias
Jan 25 at 17:40
add a comment |
1 Answer
1
active
oldest
votes
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Using the notation of the Wikipedia article on the Weierstrass M test: You want to know if $sum_{nge1} frac{-1}{n(nx+1)^2}$ converges uniformly on $A=[a,infty)$,where $a$ is a positive real constant. Note that for $xin A$
you have $|f_n(x)|le frac 1 {n(na+1)^2}$. But $$sum_{nge1} M_n le frac 1 {a^2} sum_{nge1}frac 1 {n^3}<infty,$$ which you should recognize as a convergent series.
answered Jan 25 at 18:10
kimchi loverkimchi lover
11.3k31229
$endgroup$
$begingroup$
Thank you very much!!
$endgroup$
– C. Elias
Jan 25 at 19:07
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
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active
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$begingroup$
Using the notation of the Wikipedia article on the Weierstrass M test: You want to know if $sum_{nge1} frac{-1}{n(nx+1)^2}$ converges uniformly on $A=[a,infty)$,where $a$ is a positive real constant. Note that for $xin A$
you have $|f_n(x)|le frac 1 {n(na+1)^2}$. But $$sum_{nge1} M_n le frac 1 {a^2} sum_{nge1}frac 1 {n^3}<infty,$$ which you should recognize as a convergent series.
answered Jan 25 at 18:10
kimchi loverkimchi lover
11.3k31229
$endgroup$
$begingroup$
Thank you very much!!
$endgroup$
– C. Elias
Jan 25 at 19:07
add a comment |
$begingroup$
Using the notation of the Wikipedia article on the Weierstrass M test: You want to know if $sum_{nge1} frac{-1}{n(nx+1)^2}$ converges uniformly on $A=[a,infty)$,where $a$ is a positive real constant. Note that for $xin A$
you have $|f_n(x)|le frac 1 {n(na+1)^2}$. But $$sum_{nge1} M_n le frac 1 {a^2} sum_{nge1}frac 1 {n^3}<infty,$$ which you should recognize as a convergent series.
answered Jan 25 at 18:10
kimchi loverkimchi lover
11.3k31229
$endgroup$
$begingroup$
Thank you very much!!
$endgroup$
– C. Elias
Jan 25 at 19:07
add a comment |
$begingroup$
Using the notation of the Wikipedia article on the Weierstrass M test: You want to know if $sum_{nge1} frac{-1}{n(nx+1)^2}$ converges uniformly on $A=[a,infty)$,where $a$ is a positive real constant. Note that for $xin A$
you have $|f_n(x)|le frac 1 {n(na+1)^2}$. But $$sum_{nge1} M_n le frac 1 {a^2} sum_{nge1}frac 1 {n^3}<infty,$$ which you should recognize as a convergent series.
answered Jan 25 at 18:10
kimchi loverkimchi lover
11.3k31229
$endgroup$
Using the notation of the Wikipedia article on the Weierstrass M test: You want to know if $sum_{nge1} frac{-1}{n(nx+1)^2}$ converges uniformly on $A=[a,infty)$,where $a$ is a positive real constant. Note that for $xin A$
you have $|f_n(x)|le frac 1 {n(na+1)^2}$. But $$sum_{nge1} M_n le frac 1 {a^2} sum_{nge1}frac 1 {n^3}<infty,$$ which you should recognize as a convergent series.
answered Jan 25 at 18:10
kimchi loverkimchi lover
11.3k31229
answered Jan 25 at 18:10
kimchi loverkimchi lover
11.3k31229
answered Jan 25 at 18:10
kimchi loverkimchi lover
11.3k31229
answered Jan 25 at 18:10
kimchi loverkimchi lover
11.3k31229
11.3k31229
$begingroup$
Thank you very much!!
$endgroup$
– C. Elias
Jan 25 at 19:07
add a comment |
$begingroup$
Thank you very much!!
$endgroup$
– C. Elias
Jan 25 at 19:07
$begingroup$
Thank you very much!!
$endgroup$
– C. Elias
Jan 25 at 19:07
$begingroup$
Thank you very much!!
$endgroup$
– C. Elias
Jan 25 at 19:07
add a comment |
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$begingroup$
Can you use the fact that $anle an+1$?
$endgroup$
– kimchi lover
Jan 25 at 16:58
$begingroup$
I don't think so, because then you get $n(na)^2=n^3a$, but you can't be sure that this is smaller than $n^2$
$endgroup$
– C. Elias
Jan 25 at 17:40