Uniform Convergence of Series on (0,1)

AI Thread Summary
The series \(\sum_{n=0}^{\infty} x^n(1-x)\) converges pointwise to 1 on the interval (0,1). However, uniform convergence is determined by the supremum of the difference between the series and its limit, which does not tend to 0 as n approaches infinity. The Weierstrass M-test is applicable here, as a suitable sequence of constants \(M_n\) must be found. Using \(M_n = 1\) for all n satisfies the conditions of the M-test, confirming uniform convergence. Thus, the series converges uniformly on (0,1).
latentcorpse
Messages
1,411
Reaction score
0
Does \sum_{n=0}^{\infty} x^n(1-x) converge uniformly on (0,1)?

S_n(x)=\sum_{k=0}^{\infty} x^k(1-x)=\frac{1-x^{n+1}}{1-x}(1-x)=1-x^{n+1}} \rightarrow 1 as n \rightarrow \infty
so we get pointwise convergence to 1

now we test for uniform convergence

d_{\infty}(S_n,1) = sup |S_n(x)-1|= sup x^{n+1} now remembering that x is in (0,1), i said that this should tend to 0 as n goes to infinity giving uniform convergence but the answers say that it doesn't go to 0 and so convergence is not uniform, merely pointwise - why is this?
 
Physics news on Phys.org


Hint: this problem is screaming Weierstrass M-test.
 


we can find a sequence of positive integers M_n such that |f_n(x) \leq M_n \forall x \in (0,1) so let M_n=x^n then since \sum M_n converges, \sum f_n(x) converges by Weierstrass M Test?

why is it screaming M test?
 


It is screaming M test because that is what you have to use to answer the question.

You can't use xn for Mn because Mn is independent of x. Try again.
 


just M_n=1 \forall n then? this obviously converges and so we would get uniform convergence...?
 


Yes, Mn = 1 works and so by the M-test, you have uniform convergence on (0,1).
 

Similar threads

Replies
5
Views
1K
Replies
10
Views
3K
Replies
3
Views
3K
Replies
4
Views
2K
Replies
3
Views
2K
Replies
16
Views
3K
Back
Top