# Uniform convergence of a series of functions

1. Aug 5, 2008

### linearfish

1. The problem statement, all variables and given/known data
Prove:
(1) the series
$$\sum_{n=0}^\infty (-1)^n x^n (1-x)$$
converges absolutely and uniformly on the interval [0,1]

(2) the series
$$\sum_{n=0}^\infty x^n (1-x)$$
converges absolutely and uniformly on the interval [0,1]

3. The attempt at a solution
I have shown, by induction, that the limiting function of the second series is 1 - xn+1, which goes to 1. Thus the series of functions converges (absolutely, since all values are positive) but is 0 at x = 1, so thus not continuous. Therefore, the convergence of the second series is not uniform. However, this also shows that the first series converges absolutely.

Where I am stuck is with uniform convergence of the first series. Using partial sums I was able to show that the series converges to (1-x)/(1+x), but how do I show this is uniform? I don't think it's enough to say that the limiting function is continuous in the given interval.

Can anyone tell me if I'm on the right track or what I can use to prove uniform convergence? Thanks.

2. Aug 5, 2008

### foxjwill

Let $$\epsilon \in \mathbb{R}$$. Show that for all $$\epsilon > 1$$, $$\sum_n \left ( {-1 \over \epsilon} \right ) (\epsilon x)^n (1-x)$$ satisfies Abel's Uniform Convergence Test.

3. Aug 6, 2008

### linearfish

I'm afraid I don't quite understand your restatement of series. Is it missing a power of n on the first term?

If that's the case, then I believe I need to show the following:
1. $${( \epsilon x)^n }$$ is uniformly bounded and eventually monotonic.
2. $$\sum_{n=0}^\infty {\left(\frac{1}{\epsilon}\right)}^n (1-x)$$ converges uniformly

Is this correct? Thanks.

4. Aug 8, 2008

### linearfish

Let me try this:

I claim:
$$\sum_{n=0}^N (-1)^n x^n (1-x) = (1-x) \frac{1 - (-x)^{N+1}}{1+x}$$
$$\sum_{n=0}^{\infty} (-1)^n x^n (1-x) = \frac{1-x}{1+x}$$

If the first converges uniformly to the second, then as n goes to infinity:
$$\left| (1-x) \left(\frac{1 - (-x)^{N+1}}{1+x}-\frac{1}{1+x}\right)\right| \rightarrow 0$$

This is true since:
$$\left| (1-x) \left(\frac{1 - (-x)^{N+1}}{1+x}-\frac{1}{1+x}\right)\right| = \left| \frac{(1-x)}{1+x} (-1) (-x)^{N+1}\right| = (x)^{N+1} \left| \frac{(1-x)}{1+x}\right| \rightarrow 0 (1) = 0$$

Hence the series converges uniformly. Can anyone validate this or pick out a problem with it? Thanks.

5. Aug 13, 2008

### linearfish

Not to bump this thread but I think I have a better solution now:
$$\left| (1-x) \left(\frac{1 - (-x)^{N+1}}{1+x}-\frac{1}{1+x}\right)\right| = \frac{1 - x}{1 + x} \left| -(-x)^{N+1} \right| = \frac{1 - x}{1 + x}(x)^{N+1} \leq (1-x)x^{N+1}$$

Keeping in mind that we are on the interval [0,1], we can take the derivative of the last function and, setting it equal to 0, find the the function has a maximum value at x=(N+1)/(N+2). Thus,

$$\left| (1-x) \left(\frac{1 - (-x)^{N+1}}{1+x}-\frac{1}{1+x}\right)\right| \leq (1-x)x^{N+1} \leq (1- \frac{N+1}{N+2})(\frac{N+1}{N+2})^{N+1} \leq \frac{1}{N+2}$$

Then this is less than any arbitrary epsilon if
$$N > \frac{1}{\epsilon} - 2$$

Since N does not depend on the choice of x, this convergence is uniform.

I think this is the right idea but I'd appreciate any comments or validations. Thanks.