# Sequences limit law problem

1. Jan 2, 2014

### alingy1

Hello everyone,
I'm starting to study sequences.
I'm on Stewart's Calculus textbook (single variable, 7th edition, for those who have it, on p. 693).
Now, I'm at the part where the limit laws are "transferred" to sequences.

(I'm sorry. I do not know how to code. I hope this is clear to you.)

lim as n->infinity of anp=[lim as n->infinity of an]p
IF p>0 AND an>0

Why do they add the IF p>0 AND an>0? I do not see why the formula would be wrong without that.

2. Jan 2, 2014

### PeroK

Can you think of a counterexample if p = -1?

What could happen to a_n that would prevent the formula from being true?

Also, if a_n altenates between +1 and -1, can you see a counterexample?

3. Jan 2, 2014

### alingy1

Why would a_n alternating change anything?
I tried to think about what would happen, but everything still makes sense.
Can you give me the counterexample you are thinking of? I'd like to graph it.

4. Jan 2, 2014

### PeroK

$If \ a_n = -1^n$

Then what is:

$lim_{n→∞}a_n$

5. Jan 2, 2014

### alingy1

lim of a_n as n->infinity is -1
lim of (a_n)^(-3) as n->infinity is -1
(lim of a_n as n->infinity)^(-3) is -1
The law seems to work! I think I'm doing something clearly wrong.

6. Jan 2, 2014

### PeroK

In this case, the limit does not exist. The sequence alternates between 1 and -1 doesn't get stay close to either. So:

$(lim_{n→∞}a_n)^2$ does not exist

But:

$lim_{n→∞}a_{n}^2 = 1$

7. Jan 2, 2014

### alingy1

Did you mean a_n=(-1)^n? I think if a_n=-1^n, the limit is -1.

8. Jan 2, 2014

### PeroK

Why would it be -1 in preference to 1? The sequence goes:

-1, 1, -1, 1, -1, 1, -1, 1...

Why would that converge to -1?

9. Jan 2, 2014

### alingy1

10. Jan 2, 2014

### vela

Staff Emeritus
Yes, PeroK meant the former. The expression -1^n is a bit ambiguous as some people will read it as -(1^n) and others will read it as (-1)^n.

11. Jan 2, 2014

### alingy1

Awesome! Thanks a lot. I'm starting to get this. Can I extrapolate that the a_n being >0 only adresses the issue of signs? What about the fact that p>0? I searched the web and no document explains it. Videos online, including KhanAcademy does not seem to address the issue!

12. Jan 2, 2014

### PeroK

For the other one, try a_n = n and p = -1:

$lim_{n→∞}a_{n}^{-1} = lim_{n→∞}\frac{1}{n} = 0$

But:

$lim_{n→∞}a_{n} = lim_{n→∞}n$ does not exist

so:

$(lim_{n→∞}n)^{-1}$ is not defined

13. Jan 2, 2014

### vela

Staff Emeritus
PeroK already gave you a counterexample with p=2 and $a_n = (-1)^n$. In that case, $a_n^p = 1$, so the limit on the lefthand side exists. The limit on the righthand side, however, doesn't.

What about if p=1/2 but $a_n$ wasn't necessarily positive?

What about if p=-1 and $a_n \to \infty$?

What about p=0 and $a_n=n$?

Can you see why the equality wouldn't hold in each of these cases?

14. Jan 2, 2014

### alingy1

Thanks PeroK!
Vela:
What about p=0 and a_n=n? This seems to work:
Left-hand limit=1
Right-hand limit=1

15. Jan 2, 2014

### vela

Staff Emeritus
Really? What's $\lim_{n \to \infty} n$?

16. Jan 2, 2014

### alingy1

Infinity, no?

17. Jan 2, 2014

### vela

Staff Emeritus
Better to say the limit doesn't exist.

18. Jan 2, 2014

### alingy1

True, but doesn't the answer give infinity^0=1?

19. Jan 2, 2014

### vela

Staff Emeritus
No. You can't raise something that doesn't exist to any power.

20. Jan 2, 2014

### PeroK

I'm sure the book you have will help with this, but one of the best things you can do before you go much further is:

1) Understand why ∞ is not a number. And, in particular, that no arithemtic operations are defined on it.

2) Understand precisely what we mean by n → ∞ and why this does not require ∞ to be a number.

This is very important before you go very far with real analysis.