# Homework Help: Show $\{nx^{n}(1-x)\}_{n = 0}^{\infty}$ converges uni

1. Jun 5, 2015

### Euklidian-Space

1. The problem statement, all variables and given/known data
show $\{nx^{n}(1-x)\}$ converges uni on [0,1]

2. Relevant equations

3. The attempt at a solution
Note, that $\sum nx^n(1-x)$ converges uni on [0,1] by Abels' Theorem. Therefore the series follows the cauchy criterion,
$\forall \epsilon > 0$, $\exists N > 0$ such that if n,n > N then

$$\left|(m+1)x^{m+1}(1 - x) + ... + nx^{n}(1- x)\right| < \epsilon\,\,\, \forall x \in [0,1]$$,

Now since $nx^n(1 - x) \geq 0$ for all n we can say

$$\left|nx^{n}(1 - x) - mx^{m}(1 - x)\right| < \epsilon\,\,\, \forall x \in [0,1].$$

Which implies that $\{nx^n(1 - x)\}$ is uniformly cauchy $\forall x \in [0,1]$ and therefore uniformly convergent for $x \in [0,1]$

Dont know if this is right

2. Jun 6, 2015

### wabbit

Care to elaborate ?

Note that once you've said that, you're done - if the series converges uniformly so does the sequence of course, there is no need for a Cauchy argument. But your first step is mysterious (and unnecessary).

3. Jun 6, 2015

### Euklidian-Space

Well my version of Abel's theorem states
Let $g(x) = \sum a_{n}x^{n}$ that converges at some point R > 0. Then the series converges uniformly on [0,R]. I used the cauchy argument since we never proved series uniformly convergent implies the sequence is uniformly convergent.

4. Jun 6, 2015

### micromass

That is:
1) Incorrect, if $g(x)$ converges at some point $R>0$, then you are only guaranteed convergence on $[0,r]$ for $r<R$.
2) Not what Abel's theorem says.

See http://jlms.oxfordjournals.org/content/s1-39/1/81.extract [Broken]

Last edited by a moderator: May 7, 2017
5. Jun 6, 2015

### Euklidian-Space

i am quite literally picking this up verbatim from my text
it is attached to this post

#### Attached Files:

• ###### Selection_005.png
File size:
60.3 KB
Views:
106
Last edited by a moderator: May 7, 2017
6. Jun 6, 2015

### micromass

Which book is this? The flaw in the proof seems to be assuming that $\left( \frac{x}{R} \right)^n$ is monotone decreasing, which is not the case for $x=R$.

7. Jun 6, 2015

### Euklidian-Space

understanding analysis, Abbot

8. Jun 6, 2015

### micromass

OK, never mind. The theorem and proof is correct. Your proof is correct too then.

9. Jun 6, 2015

### micromass

But, your proof is unnecessarily complicated by the way. It is much easier to find the maximum of the function on each interval, and to prove those maxima converge to $0$. There is no real reason to transform this into a power series.

10. Jun 6, 2015

### Euklidian-Space

Yeah we havent been given that theorem, this was me trying to tackle this thing with the tools available

11. Jun 7, 2015

### wabbit

Actually, your sequence $f_n(x)=nx^n(1-x)$ does not converge uniformly on $[0,1]$.
It converges pointwise to zero on that interval. However, if you set $x_n=\frac{n}{n+1}$ which is the value at which $f_n$ reaches its maximum over $[0,1]$ , you find that $f_n(x_n)=(\frac{n}{n+1})^{n+1}=(1-\frac{1}{n+1})^{n+1} \ _{\overrightarrow{n\rightarrow\infty}}\frac{1}{e}>0$ and this proves that the convergence is not uniform.

Edit : the reason your proof is incorrect, and the theorem you quote is not applicable here, is that $\sum nx^n(1-x)$ is not a power series. You can of course write it as $(1-x)\sum nx^n$ but then you will note that the power series $\sum n x^n$ does not converge for $x=1$ so you cannot use that theorem.

Last edited: Jun 7, 2015
12. Jun 7, 2015

### Euklidian-Space

ok, so if it doesnt converge uni then isnt it weird that
$$\lim_{n\rightarrow \infty}\int_{0}^{1} f_{n} \rightarrow 0?$$

Isn't uniform convergence a necessary condition for that?

#### Attached Files:

• ###### Selection_006.png
File size:
7 KB
Views:
96
13. Jun 7, 2015

### wabbit

No, uniform convergence is a sufficient condition for the convergence of the integral, but by no means a necessary one - as this example shows.

14. Jun 7, 2015

### Ray Vickson

This just implies that the series $\sum_{n=0}^N x^n$ converges uniformly on $[0, 1 - \epsilon]$ for any small $\epsilon > 0$. Thus, $\sum_{n=0}^N (1-x)x^n$ converges to 1 uniformly on $[0, 1-\epsilon]$. This does not say what happens right at $x = 1$, because then you have a series of terms that all = 0 (and, somehow, you want that series to converge to 1?)

15. Jun 7, 2015

### Euklidian-Space

hmmm, never said anything about converging to 1...I said that the theorem implies that it converges uni on [0,1]. It does say what happens at x = 1. The original series equals zero at x = 1. Therefore it converges at some point x = R = 1 > 0. Therefore it converges for [0,R] = [0,1]

16. Jun 7, 2015

### wabbit

As I mentionned in my previous post, the original series you wrote is not a power series so you cannot apply Abel's theorem to it.

17. Jun 7, 2015

### Euklidian-Space

my only issue wabbit is that in my class we did not get the theorem you applied to show it is not uni conv. i do not know how to approach it without that theorem

18. Jun 7, 2015

### Ray Vickson

You did not mention convergence to 1, but I did.

You have a series that converges uniformly to 1 for $0 \leq x \leq 1 - \epsilon$ (for any small $\epsilon > 0$), but which converges to 0 when $x = 1$ exactly. You may call that uniform convergence on [0,1], but I---for one---would not.

19. Jun 7, 2015

### Euklidian-Space

I do not see where you are getting that the series converges to 1. thats not even the point wise limit. Maybe i am misunderstanding you

20. Jun 7, 2015

### wabbit

I did not use a theorem in particular, could you be more specific as to what you find unclear in the proof I gave ?

21. Jun 7, 2015

### Ray Vickson

For $|x| < 1$ we have
$$\sum_{n=0}^N x^n = \frac{1-x^{N+1}}{1-x} \to \frac{1}{1-x}$$
as $N \to \infty$. Thus, $\sum_n (1-x) x^n \to 1$.

22. Jun 7, 2015

### Euklidian-Space

you used the lim sup right? Finding the maximum of $f_{n}$ and showing that the limit of $f_{n}(x_{max})$ does not go to zero?

23. Jun 7, 2015

### wabbit

I did not exactly use a limit-sup (and certainly no theorem about lim sup) but yes for the rest. It would be easier if you quoted the step you are having difficulty with so I could elaborate on it.

24. Jun 7, 2015

### Euklidian-Space

well, what you used to show it is not uni conv we have not been taught, therefore i must find another way to show that the sequence is not uni conv

25. Jun 7, 2015

### wabbit

No. I used only the definition of uniform convergence, and you can follow that proof. Try it.