# SOLVEDSequence Proof - Log Question?

Hi I'm working through the proof:

lim(n→∞) r^n = 0 (if |r| < 1)

lim(n→∞) r^n = ∞ (if |r| > 1)

using the ε-δ technique

when I'm focusing on 0 < |r| < 1 something weird happens

|r^n - 0| < ε ⇒ n > N
|r^n| < ε ⇒ n > N
n ln |r| < ln ε ⇒ n > N

n > (ln ε)/(ln|r|)

See that! The sign changes from < to > in my book and it says this is because ln|r| is negative as it's less than 1.

But isn't (ln ε) less than 1 as well, like stupendously close to zero actually!
Shouldn't the two cancel each other out the second you take the logarithm of
both sides?

Also, for 0 (the trivial case);

|r^n - 0| < ε
|r^n| < ε
n ln |r| < ln ε
------------------------
n > (ln ε) / (ln|r|)
-------------------------
n > (ln ε) / (ln|0|)
n > (ln ε) / (1)
n > ln ε

Is that the way to do it or is this:

|r^n - 0| < ε
|r^n| < ε
|0^n| < ε
0 < ε

The idea is to set N = (ln ε) / (ln|r|) and as long as n > N the proof should hold, I'm just a bit confused by this minus thing and how we're supposed to arrive at the answer i.e. because of the way everything is set up it shows that n > n, I don't see it.

MHF Hall of Honor
Hi I'm working through the proof:

lim(n→∞) r^n = 0 (if |r| < 1)

lim(n→∞) r^n = ∞ (if |r| > 1)

using the ε-δ technique

when I'm focusing on 0 < |r| < 1 something weird happens

|r^n - 0| < ε ⇒ n > N
|r^n| < ε ⇒ n > N
n ln |r| < ln ε ⇒ n > N

n > (ln ε)/(ln|r|)

See that! The sign changes from < to > in my book and it says this is because ln|r| is negative as it's less than 1.

But isn't (ln ε) less than 1 as well, like stupendously close to zero actually!
Shouldn't the two cancel each other out the second you take the logarithm of
both sides?
I think you mean
isn't (ln ε) less than 0 as well
And, yes, it is. So you have, in effect:
$$\displaystyle -An < -B$$, where $$\displaystyle A>0$$ and $$\displaystyle B>0$$
Agreed?

Well, when you change the signs of both sides of the inequality, you are, in effect, multiplying both sides by $$\displaystyle -1$$. And when you multiply (or divide) both sides of an inequality by a negative number, you reverse the direction of the inequality sign.

Thus:
$$\displaystyle -An < -B$$

$$\displaystyle \Rightarrow An > B$$

$$\displaystyle \Rightarrow n > \frac BA$$, since $$\displaystyle A>0$$
Do you see this now?

Ah yes, that's exactly what I thought. I guess I was just really angry that I didn't immediately notice the minus was implicitly there due to the logarithm being less than 1 (Angry)

But now it's all cleared up (Rofl)

Thanks, and yes I did mean ln ε < 0, well spotted (Clapping)

Thanks, have a good day (Cool)