Prove that |z|^n -> 0 if |z| < 0

[Jamin2112]

Prove that |z|^n ---> 0 if |z| < 0

Homework Statement

This isn't really a homework problem; I'm preparing for an upcoming course and trying to recall how to do these basic proofs.

Homework Equations

Definition of convergence

The Attempt at a Solution

We seek a positive integer N such that for any positive number ∂,
| |z|ⁿ - 0 | < ∂ whenever n ≥ N.

Fix ∂ > 0.

| |z|ⁿ- 0| = |z|ⁿ < ∂
iff
n < log(∂) / log(|z|) (some steps omitted)


But here's the problem: I have n smaller than a positive number, since log(∂), log(|z|) < 0 when ∂, |z| < 1. What do?

 

[stringy]

I'd split it up into cases just to be a bit cleaner:

Let $\epsilon > 0$ be given and suppose $|z| < 1$.

If $\epsilon \geq 1$, then $|z|^n < 1 \leq \epsilon$ for what values of n?

If $\epsilon < 1$, put $|z| = 1/x$. Then what must be true of x?

 

[HallsofIvy]

Note that stringy is assuming that you mean |z|< 1, not |z|< 0 which is impossible anyway.

 

[Jamin2112]

If $\epsilon \geq 1$, then $|z|^n < 1 \leq \epsilon$ for what values of n?

n > log(eps) / log(|z|)

If $\epsilon < 1$, put $|z| = 1/x$. Then what must be true of x?

Then x > 1 (?)

 

[stringy]

n > log(eps) / log(|z|)

Certainly $\left|z\right|^n < 1$ for all n! Remember, we're assuming the absolute value of z is less than 1.

EDIT: This is the sort of trivial case that is dispensed with quickly at the beginning of the proof.

Then x > 1 (?)

Bingo. Then proving $\left|z\right|^n \to 0$ is the same as proving $1/x^n \to 0$ with x>1.

 

[Jamin2112]

Thanks, brah!

I have another one: |z|ⁿ / n! ----> 0


How should I go about showing this?

|z|ⁿ / n! < ∂
log( |z|ⁿ / n! ) < log(∂)
log(|z|ⁿ) - log(n!) < log(∂)
n * log(|z|) - ∑log(k) < log(∂) (where the index of the sum is k = 2, 3, ..., n)


? Seems like I'm getting nowhere with that

Maybe
|z|ⁿ / n! < ∂
[∑n!x^n-k(iy)^k/(n-k)!k!] / n! < ∂
[∑x^n-k(iy)^k/(n-k)!k!] < ∂


? Doesn't seem legit

 

[stringy]

Hmmm... I have an idea. We are to prove

$$\lim_{n\to \infty} \frac{|z|^n}{n!} = 0,$$

correct? No conditions placed on the size of |z| (it doesn't matter anyways)? We know

$$\sum_{n=0}^\infty \frac{|z|^n}{n!}$$

converges. Do you remember why and to what? What do we know about the nth term of a convergent series?

By the way, remember what n! is:

$$n! = n(n-1)(n-2)...(2)(1).$$

It's not a sum.

 

[Jamin2112]

Hmmm... I have an idea. We are to prove

$$\lim_{n\to \infty} \frac{|z|^n}{n!} = 0,$$

correct? No conditions placed on the size of |z| (it doesn't matter anyways)? We know

$$\sum_{n=0}^\infty \frac{|z|^n}{n!}$$

converges. Do you remember why and to what? What do we know about the nth term of a convergent series?

By the way, remember what n! is:

$$n! = n(n-1)(n-2)...(2)(1).$$

It's not a sum.

I know, but log(n!) = log(n*(n-1)***2*1) = log(n) + log(n-1) + ... + log(2) + log(1)

The chapter from which this practice problem was derived comes prior to the chapter on series, so I don't think I can use the fact that ∑a_n is convergent implies a_n ---> 0. We'll have to go about it another way.

 

[stringy]

I know, but log(n!) = log(n*(n-1)***2*1) = log(n) + log(n-1) + ... + log(2) + log(1)

The chapter from which this practice problem was derived comes prior to the chapter on series, so I don't think I can use the fact that ∑a_n is convergent implies a_n ---> 0. We'll have to go about it another way.

D'oh! You're right. I just saw the sigma in there and jumped to conclusions. Let me think about this for awhile. I get back.

 

[stringy]

You remember from calculus the ratio test for series? There's an analogous test for sequences. The proofs are similar.

Let $(a_n)$ be a sequence of nonzero terms such that

$$\lim_{n\to \infty} \left| \frac{a_{n+1}}{a_n}\right| =\alpha .$$

If $\alpha< 1$, then the sequence $(a_n)$ goes to zero.

Quick sketch of a proof: Suppose $| a_{n+1}/a_n| \to \alpha$ and $\alpha <1$. Then there is some N so that n>N implies $|a_{n+1}/a_n| \leq \beta$ where $\alpha < \beta < 1$. So if n>N,

$$|a_n| = \frac{|a_n|}{|a_{n-1}|} \frac{|a_{n-1}|}{|a_{n-2}|} \cdots \frac{|a_{N+1}|}{|a_{N}|} |a_{N}| \leq \beta^{n-N} |a_N| = \beta^n \frac{|a_N|}{\beta^N}.$$

Since $\beta <1$, we have $\beta^n \to0$ (we just proved this!). Therefore, by the squeeze theorem, $(|a_n|)$ and hence $(a_n)$ goes to zero.

Apply this test to your sequence and you'll have your desired result.

You can also obtain the result by using known upper and lower bounds on the factorial. You can look them up on the web. Trying to prove that the sequence converges by explicitly finding a delta is going to be tough, which is why I've avoided that method of attack.

 

[Oshada]

If I might 'hijack' this question, I have the same question, but it explicitly states to avoid logarithms and exponentials, and use the properties of convergence and the properties of a subsequence u2n to prove un = |a|^n ---> 0 when n ---> infinity if |a| < 1. The hint supplied is to let u2n = (un)^2. I'm assuming this makes use of the fact that all subsequences converge to the same limit as the sequence? And also, is application of the Bolzano-Weierstrass theorem required for this proof?

 

[stringy]

If you want to avoid logarithms, you can use Bernoulli's inequality which states

$$(1+b)^n \geq 1+nx$$

for all positive integers $n$ and real $b\geq -1$.

Since $|a| <1$, write

$$|a| = \frac{1}{1+b}$$

where $b>0$. Apply Bernoulli's inequality and you'll find a delta relatively easily.

As for this subsequence business... I'm not immediately seeing the reason why we should look at that subsequence!

We know that $(u_n)$ converges to a point u if and only if all subsequences converge to u. So it's not enough to ONLY show that $(u_{2n})$ goes to zero. We'd need to show all subsequential limits are 0.

However, we know that monotonic sequences converge if and only if they are bounded. We know $(u_n)$ is bounded (below by 0 and above by 1). So we know $(u_n)$ converges to a point u. Then if you show $(u_{2n})$ goes to zero, you know u=0.

Hmmmm... I'll think about this one for awhile and get back if I come up with anything. Meanwhile, maybe somebody else can chime in.