# Uniform convergence of series

1. Apr 8, 2009

### latentcorpse

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?

2. Apr 8, 2009

### e(ho0n3

Re: Convergence

Hint: this problem is screaming Weierstrass M-test.

3. Apr 9, 2009

### latentcorpse

Re: Convergence

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?

4. Apr 9, 2009

### e(ho0n3

Re: Convergence

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.

5. Apr 9, 2009

### latentcorpse

Re: Convergence

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

6. Apr 9, 2009

### e(ho0n3

Re: Convergence

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