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end3r7
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I'm posting this again because the other was plagued with errors in the first post. My fault and I apologize. I'll do a better job this time, I hope.
Let f(n,x) = <br /> \sum\limits_{n = 1}^{\inf } {( - 1)^n (1-x^{2})x^{n}} <br />
a) Test for absolutely convergence on [0,1]
b) Test for uniformly convergence on [0,1]
c) Is <br /> \sum\limits_{n = 1}^{\inf } {|f(n,x)|} <br /> uniformly convergent on [0,1]?
N/A
Alright, a) is simple actually, a simple application of the root/ratio test will do, but here comes the catch.
I know that b) is true, since a power series is uniformly convergent inside its radius of convergence and the endpoint in this case is trivial. It just seems, however, that with a comparison with a geometric series, I can use the Weierstrass M-Test to prove all 3 at once.
For any 'x' in [0,1), I can choose an a such that 0 <= x <= a <1 by the density of R. Therefore
<br /> |\sum\limits_{n = 1}^{\inf } {f(n,x)} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}<br />
and for x = 1 and any a > 0
<br /> |\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} = 0 < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}
I have just chosen the function to be uniformally cauchy on [0,1) U {1} (have I?) by a basic comparison test with a series which is uniformly cauchy on (0,1) and by considering the end point separately.
But it also seems that the Weierstrass M-Test gives me "Absolutely Uniformly Convergence" (that is, the absolute value of the function is also uniformly convergent) for free. Am I right? Is that always going to hold true, or is this a property of power series?
Homework Statement
Let f(n,x) = <br /> \sum\limits_{n = 1}^{\inf } {( - 1)^n (1-x^{2})x^{n}} <br />
a) Test for absolutely convergence on [0,1]
b) Test for uniformly convergence on [0,1]
c) Is <br /> \sum\limits_{n = 1}^{\inf } {|f(n,x)|} <br /> uniformly convergent on [0,1]?
Homework Equations
N/A
The Attempt at a Solution
Alright, a) is simple actually, a simple application of the root/ratio test will do, but here comes the catch.
I know that b) is true, since a power series is uniformly convergent inside its radius of convergence and the endpoint in this case is trivial. It just seems, however, that with a comparison with a geometric series, I can use the Weierstrass M-Test to prove all 3 at once.
For any 'x' in [0,1), I can choose an a such that 0 <= x <= a <1 by the density of R. Therefore
<br /> |\sum\limits_{n = 1}^{\inf } {f(n,x)} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}<br />
and for x = 1 and any a > 0
<br /> |\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} = 0 < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}
I have just chosen the function to be uniformally cauchy on [0,1) U {1} (have I?) by a basic comparison test with a series which is uniformly cauchy on (0,1) and by considering the end point separately.
But it also seems that the Weierstrass M-Test gives me "Absolutely Uniformly Convergence" (that is, the absolute value of the function is also uniformly convergent) for free. Am I right? Is that always going to hold true, or is this a property of power series?