# |a| < 1 show lim |a|^n ->0

1. Nov 26, 2013

### kingstrick

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

given |a| < 1, show that the limit of |a|^n goes to 0 as n goes to infinity.

2. Relevant equations

3. The attempt at a solution

let |a|<1 and n>0 (n is a natural number, a is a real number)

then

|a^n| < 1^n

then

|a|^n < 1

then

1/n * |a|^n <= |a|^n < 1 (and we know from earlier that 1/n * |a|^n goes to zero)

so 0 <=|a|^n < 1 meaning |a|^n is bounded below.

now note that

|a| > |a|^2 > |a|^3 > ... |a|^n > |a|^(n+1) > ... since a < 1

Since a^n is progressively smaller and bounded below by zero, we know that a^n does converge by the convergent monotone sequence thm and since there are infinite items in this sequence, its limit cannot be in the set of a^n.

therefore let L be the limit,

|a^n - L| < e, since L is smaller than a^n;

a^n - L < e.

**I tried using the squeeze theorem but could not find a sequence that I new to be greater than a^n and goes to zero.** And I don't know what I am missing in order to be able to claim that L is zero.

2. Nov 26, 2013

### jbunniii

Hint: $|a|^{n+1} = |a| \cdot |a|^{n}$. What happens if you take limits of both sides of this equation?

3. Nov 26, 2013

### kingstrick

i am still missing something, cause now i am getting the lim to be e which i know is incorrect.

(n+1) * lim ln |a|= Lim (|a| * |a|^n) = lim |a| * n lim ln |a|

(n+1) = n lim |a|

(1+1/n) = lim |a|

(1+1/n)^n = (lim |a|)^n

e = lim |a|^n

4. Nov 26, 2013

### jbunniii

I'm not sure why you're taking logs, but in any case the above doesn't make any sense.

First, how did you slide the "lim" past the $n+1$ to get $(n+1) \lim \ln |a|$? Aren't you taking the limit with respect to $n$?

Second, it is not true that $\ln (|a| \cdot |a|^n) = |a| \cdot n\ln|a|$ as your second equality seems to suggest.

5. Nov 26, 2013

### jbunniii

Try taking limits directly of the left and right hand sides. Don't take logs first:
$$|a|^{n+1} = |a| \cdot |a|^n$$
Take advantage of the fact that you already concluded that $\lim |a|^n$ must exist because the sequence is bounded and monotonic.

6. Nov 26, 2013

### kingstrick

I am unsure. I remember from an earlier class that the limit of x^n is equal to n * lim ln x

so i tried extending it to calculate the limits like you suggested:

lim of a^(n+1) = (n+1) * lim ln a [1]

lim (a*a^n) = lim a * lim a^n = lim a * n * lim ln a. [2] Then i divided both sides ([1], [2]) by lim ln a

7. Nov 26, 2013

### jbunniii

It's not true. What is true is that $\ln (x^n) = n \ln (x)$ and therefore $\lim \ln (x^n) = \lim n \ln (x)$, but that isn't going to help you here.

You have established that $\lim_{n\rightarrow \infty} |a|^n$ must exist. Let's call this limit $L$. Now what is $\lim_{n\rightarrow \infty} |a|^{n+1}$?

8. Nov 26, 2013

### kingstrick

lim |a|^(n+1) = aL < L since a < 1

9. Nov 26, 2013

### jbunniii

What else is $\lim|a|^{n+1}$ equal to? Think about sequences in general. If $(x_n)$ is some sequence and $\lim x_n = L$, then what is $\lim x_{n+1}$?

10. Nov 26, 2013

### kingstrick

It should have the same limit

11. Nov 26, 2013

### jbunniii

Yes, that's right. So if we put $x_n = |a|^n$, then $x_{n+1} = |a|^{n+1}$.

If $\lim |a|^n = L$, then what is $\lim |a|^{n+1}$?

12. Nov 26, 2013

### kingstrick

lim |a|^(n+1) = L

13. Nov 26, 2013

### kingstrick

but doesn't that show

lim |a|^(n+1) = lim |a| * lim |a|^n

L = lim |a| * L
1 = lim |a| ??

14. Nov 26, 2013

### jbunniii

Yes!

Well, that would be valid if $L$ was nonzero. Notice that $\lim |a| = |a|$, so what you have just shown is that if you can divide by $L$, then $|a| = 1$. But we are given that $|a| < 1$.

We want to prove that $L$ is zero, so dividing by $L$ is not what you want to do. How can you solve $L = \lim |a| \cdot L$ for $L$?

15. Nov 26, 2013

### kingstrick

since |a| < 1 then lim |a| = |a| < 1 and L = lim |a| * L then L - L*Lim |a| = 0
which implies L(1-Lim|a|)=0 therefore L = 0

16. Nov 26, 2013

### kingstrick

got it!! thanks!

17. Nov 26, 2013

Looks good.