# Homework Help: Convergence of sequences

1. Mar 15, 2013

### Zondrina

1. The problem statement, all variables and given/known data

Determine the convergence, both pointwise and uniform on [0,1] for the following sequences :

(i) $s_n(x) = n^2x^2(1 - cos(\frac{1}{nx})), x≠0; s_n(0) = 0$
(ii) $s_n(x) = \frac{nx}{x+n}$
(iii) $s_n(x) = nsin(\frac{x}{n})$

2. Relevant equations

$s_n(x) → s(x)$ as $n→∞$

3. The attempt at a solution

(i) So for this one, I can split it into a piecewise function since $s_n(x)$ has been defined at the origin.

Taking the limit as n → ∞, I can observe two limiting functions occuring. s(x) = 0 and s(x) = 1/2.

Therefore $s_n(x)$ converges pointwise, but not uniformly on [0,1].

(ii) As n → ∞ $s_n(x) → x = s(x)$ for all x in [0,1]. Hence $s_n(x)$ converges pointwise AND uniformly on [0,1].

(iii) Exact same limit as (ii) so the same answer will follow.

Do these look okay? It seems way too straightforward so it has me a bit worried.

If anyone could confirm it would be great :). Thanks.

2. Mar 15, 2013

### Dick

Yes, the conclusions about convergence are correct. But I don't see any reason why you think the convergence is uniform. A sequence of continuous functions can converge to a continuous function in a way that is not uniform. You need a better reason than just that the limit is continuous.

Last edited: Mar 15, 2013
3. Mar 15, 2013

### Zondrina

As in for the second and third ones? I think I see what you mean.

(ii)

$s_n(x) = 0 → 0$ as $n→∞$ for $x = 0$.
$s_n(x) = n/(n+1) → 1$ as $n→∞$ for $x = 1$.

So we get only pointwise convergence for (ii) and not uniform for all x in [0,1].

(iii)

$s_n(x) = 0 → 0$ as $n→∞$ for $x = 0$.
$s_n(x) = nsin(1/n) → 1$ as $n→∞$ for $x = 1$.

So as is in (ii) we only get pointwise and not uniform for all x in the interval [0,1].

4. Mar 15, 2013

### Dick

No, that's not what I meant at all. You were correct that they are uniformly convergent. You just didn't give a good reason (or any) reason why.

5. Mar 15, 2013

### Zondrina

Are you implying I need to go all the way back to the epsilon definition for this?

6. Mar 15, 2013

### Dick

No, I wouldn't go back that far. Don't you have any other theorems that will show that convergence is uniform? Here's a hint. In both of those cases $s_n(x)$ is differentiable and the derivative is bounded on [0,1] independent of n. Could that be useful?

7. Mar 15, 2013

### Zondrina

Uhm I think I have this theorem I just found :

Suppose $\{s_n'(x)\}$ converges uniformly on an interval I.

If each $s_n'(x)$ is continuous and $\{s_n(x_0)\}$ converges for some $x_0 \in I$, then $\{s_n(x)\}$
converges uniformly on I to s(x) and :

$lim_{n→∞} s_n'(x) = s'(x)$

EDIT : So $s_n'(x) = \frac{n^2}{(n+x)^2}$

So $s_n'(x)$ is continuous for all x in the interval.

If we choose x = 0, we see the sequence will converge to 1.

My only question is the "Suppose {s_n'(x)} converges uniformly on an interval I". Would I say : As n goes to infinity, {s_n'(x)} goes to 1. Which also happens to be the derivative of x just like the theorem is describing.

Last edited: Mar 15, 2013
8. Mar 15, 2013

### Dick

Well that just displaces the problem from showing $s(x)$ converges uniformly to showing $s_n'(x)$ converges uniformly. If you can do that then go ahead. I was thinking of something more like if $|s'_n(x)|<M$ for some constant M then if $s_n$ converges pointwise then it converges uniformly. If you can't find a theorem like that then think how you might prove it yourself. [0,1] is compact.

9. Mar 15, 2013

### Zondrina

Wait. So you're saying that if I can bound all the positive terms of my derivative by some constant M_n, then if I have just pointwise convergence of s_n, then it converges uniformly.

$|s_n'(x)| = \frac{n^2}{(n+x)^2} ≤ \frac{n^2}{n^2} = 1 = M_n$

Since we already have shown s_n(x) is pointwise convergent, then because its derivative is bounded we can conclude that s_n(x) is uniformly convergent.

10. Mar 15, 2013

### Dick

No, no subscript n. You've shown $|s_n'(x)|$ is bounded by 1. Independent of n. That's important. Showing each $|s_n'(x)|$ is bounded doesn't help. They all have to be bounded by the same constant. And they are, as you've shown. Can you say how this would show pointwise convergence of $s_n(x)$ would show $s_n(x)$ converges uniformly? Think compactness of [0,1] and epsilon. Wish I could think of the name of a theorem to refer to.

11. Mar 15, 2013

### Zondrina

Hmm I just read about what a compact space is, but sadly I don't think I'm allowed to use any of this ( the thing about the limit points made sense ).

I still think I was correct in an earlier post in saying this ( Slightly modified now ) :

For $x = 0, \space s_n(x) = 0 → 0$ as $n→∞$.
For $x ≠ 0, \space s_n(x) = nx/(n+x) → x$ as $n→∞$.

So we have two values for the limiting function s(x) depending on x.

I don't think this is incorrect at all and would result in pointwise, but not uniform convergence.

Differentiating, I would get :

For x = 0. $s_n'(x) = 0 → 0$ as n→∞.
For x ≠ 0. $s_n'(x) = \frac{n^2}{(n+x)^2} → 1$ as n→∞.

So the convergence is uniform.

Since $\{s_n(0)\}$ converges, we have one point where $\{s_n(x)\}$ converges and we know that $\{s_n'(x)\}$ converges uniformly so it follows that $\{s_n(x)\}$ converges uniformly.

Last edited: Mar 15, 2013
12. Mar 15, 2013

### Dick

That. Is. Completely. Incorrect. The convergence is uniform. It's late and I'm running out of ways to convince you. But nx/(n+x) does not converge to 1. It converges to x. You had that part right to begin with. What changed your mind?

13. Mar 15, 2013

### Zondrina

I had a moment of doubting myself for some reason. I added a bunch to the prior post that makes more sense than what I had before.

14. Mar 16, 2013

### Dick

You just don't seem to be grappling with what uniform means. I just had a great idea. Why don't you graph s_n(x)=nx/(x+n) for some values of n, like 1, 10, 100. Does that give you some idea how you could directly prove the convergence is uniform (by epsilon ideas - not theorems)?

15. Mar 16, 2013

### Zondrina

Going back to epsilons would probably make me feel more comfortable anyway.

$\forall ε > 0, \exists N \space | \space n>N \Rightarrow |s_n(x) - s(x)| < ε, \forall x \in [0,1]$

$|s_n(x) - s(x)| = |\frac{nx}{n+x} - x| = \frac{x^2}{n+x} ≤ \frac{1}{n+1}$

So choosing $n > \frac{1}{ε} - 1 \Rightarrow |s_n(x) - s(x)| < ε, \forall x \in [0,1]$

Hence (ii) converges both pointwise and uniformly.

16. Mar 16, 2013

### Dick

That's the idea.

17. Mar 16, 2013

### Zondrina

See, I understand when I go back to the epsilon definition for some reason, but the theorems he's provided for us are not sufficient for this problem I think.

(iii)

$\forall ε>0, \exists N \space | \space n>N \Rightarrow |s_n(x) - s(x)| < ε, \forall x \in [0,1]$

$|s_n(x) - s(x)| = |nsin(x/n) - x| ≤ nsin(1/n) - 1$

Having a small nitch with the arithmetic here. I can't isolate for n in terms of epsilon.

EDIT : Wait, what if :

$|s_n(x) - s(x)| = |nsin(x/n) - x| ≤ |n||sin(x/n)| + |x| ≤ n + 1$

So choosing $n > ε -1 \Rightarrow |s_n(x) - s(x)| < ε, \forall x \in [0,1]$

18. Mar 16, 2013

### Dick

|x| isn't very small on [0,1]. That's not going to work. Look at the function x-n*sin(x/n). Is it increasing or decreasing on [0,1]?

19. Mar 16, 2013

### Zondrina

Wouldn't it be increasing? It's derivative is 1 - cos(x/n).

$2nπ$ seems to be relevant here as it is the root of the derivative.

Last edited: Mar 16, 2013
20. Mar 16, 2013

### Dick

Sure. It's increasing. It's zero at x=0 and increases to 1-n*sin(1/n) at x=1. Does that give you any ideas about how to pick an N corresponding to an epsilon?

Last edited: Mar 16, 2013
21. Mar 16, 2013

### Zondrina

No, I've looked at this for awhile and it doesn't seem obvious.

My question is why consider x-n*sin(x/n)? All that does is consider s(x) - sn(x) instead of the other way around.

What did I do wrong in the edit portion of my post a few posts ago? I applied the triangle inequality and then saw n+1 as an upper bound since |sin(x)| ≤ 1 and |x| = x since x is positive on [0,1] as is n.

22. Mar 16, 2013

### Dick

The triangle inequality is true, but it's not useful. |x-n*sin(x/n)| gets small as n increases. |x|+|n*sin(x/n)| doesn't. It approaches 2*|x| which you are not going to be able to make less than epsilon. By showing it's increasing you've shown |x-n*sin(x/n)|<1-n*sin(1/n) for all x in [0,1]. What's the limit of the right side of that inequality?

23. Mar 16, 2013

### Zondrina

$|x-nsin(x/n)| ≤ 1-nsin(1/n), \forall x \in [0,1]$

The limit of the right side is 0 as n goes to infinity.

24. Mar 16, 2013

### Dick

So? Conclusion?

25. Mar 16, 2013

### Zondrina

Er..

I've shown $|s(x) - s_n(x)|$ is increasing and that it's bounded by $1-nsin(1/n)$ for all x in [0,1].

Wait, that means it's uniformly convergent because I've shown that if I take the limiting function s(x) and take away sn(x) I get zero as n goes off to infinity so that lim n→∞ sn(x) = s(x) for all x in [0,1].