Limits of sequences involving factorial statements

[gsmtiger18]

Homework Statement

I have to determine whether or not the following sequence is convergent, and if it is convergent, I have to find the limit.

a_n = (-2)ⁿ / (n!)

In solving this problem, I am not allowed to use any form or variation of the Ratio Test.
2. The attempt at a solution
I was thinking that I could reasonably compare the above sequence to b_n = (1/2)(-2)ⁿ/(3^n-2), like so:

(1/2)(-2)ⁿ/(3^n-2) <= (-2)ⁿ / (n!) <= (1/2)(2)ⁿ/(3^n-2)

By manipulating the b_n statement, I can make a geometric sequence:

b_n = (1/2)(-2)²(-2/3)^n-2

I know this converges since r = -2/3, and the absolute value of r is less than 1. Taking the limit of the sequence (b_n) as n goes to infinity, I find that the value of the sequence approaches zero. Then, since both b_n and its absolute value converge to zero, I can say that a_n converges and its limit is zero as well, by the Squeeze Theorem.

I honestly have no idea if this is right or not. My professor gave the class this problem as part of a practice test, and we weren't given an answer key. The closest problems I can find in my textbook all use the Ratio Test in their solutions, so they're no help. The corresponding test is on Monday, and we will not be going over any solutions to the practice test beforehand, so I'd really like to make sure my thought process is correct.

 

[LCKurtz]

Homework Statement

I have to determine whether or not the following sequence is convergent, and if it is convergent, I have to find the limit.

a_n = (-2)ⁿ / (n!)

In solving this problem, I am not allowed to use any form or variation of the Ratio Test.

Ratio test?? There is no ratio test for sequences. Or is this the nth term of a series? It matters and students often confuse the two. Are you testing a sequence or a series for convergence?

2. The attempt at a solution
I was thinking that I could reasonably compare the above sequence to b_n = (1/2)(-2)ⁿ/(3^n-2), like so:

(1/2)(-2)ⁿ/(3^n-2) <= (-2)ⁿ / (n!) <= (1/2)(2)ⁿ/(3^n-2)

Inequalities like that with alternating signs don't work. And factorials grow faster then exponentials, so even with absolute values, that first inequality wouldn't work for large n.

 

[gsmtiger18]

Ratio test?? There is no ratio test for sequences. Or is this the nth term of a series? It matters and students often confuse the two. Are you testing a sequence or a series for convergence?
Inequalities like that with alternating signs don't work. And factorials grow faster then exponentials, so even with absolute values, that first inequality wouldn't work for large n.

I'm testing a sequence for convergence. I know that factorials grow faster than exponentials but simply stating that isn't enough.

 

[LCKurtz]

Try looking at the absolute value and write out the factors of numerator and denominator and see if that gives you any ideas.

 

[gsmtiger18]

I've already tried that. It's just 2 * 2 * 2... to infinity in the numerator and an infinitely long factorial in the denominator. Doesn't tell me anything.

 

[LCKurtz]

Do it for $a_n$, not $a_\infty$, whatever that might mean.

 

[gsmtiger18]

Same thing, only the factorial ends in (n-1) * n and 2 is multiplied by itself n times. I still don't have a clue what that tells me.

 

[LCKurtz]

Does it look like this:$$
\frac{2\cdot 2 \cdot 2 \cdots 2\cdot 2}{1\cdot 2 \cdot 3 \cdots n-1\cdot n}$$They don't quite line up on screen, but you have a bunch of fractions there you can think about.

 

[gsmtiger18]

Yes, that's how it should look. Again, I'm not sure what it is I'm supposed to be seeing here.

 

[LCKurtz]

You are trying to show it goes to zero. What if you group it like this:$$
\frac{2\cdot 2}{1\cdot 2}\left(\frac{2}{3}\frac 2 4 \cdots \frac 2{n-1}\right)\frac 2 n$$Can you see any useful overestimates that would still go to zero?

 

[gsmtiger18]

I know that 2³/2n goes to zero. Would that be a valid overestimate?

 

[LCKurtz]

I know that 2³/2n goes to zero. Would that be a valid overestimate?

Not sure why you would write $\frac{2^3}{2n}$ instead of $\frac 4 n$, but as to whether it is a valid overestimate, you tell me. Can you give cohesive steps or argument why $|a_n|\le \frac 4 n$? That is what your teacher would want on an exam. You have to say why...

 

[gsmtiger18]

That's the problem. I don't have any idea how to do that. At all. All we were told was that we had to know how to do these kinds of problems. It was never taught in class.

 

[LCKurtz]

Well, you asked me whether $\frac 4 n$ was a valid overestimate. You must have had some reason to think it was. How did you get that and why did you think it might be an overestimate? You had some reason. Just tell me.

 

[gsmtiger18]

Well, my thinking is that n! becomes infinitely large. That said, I think that since n! is in the numerator, I should look for a value in the numerator which is comparatively less than n! and also becomes infinitely large. However, I'm unsure how the numerator is established. Also, why did you divide up the fractions the way you did?

 

[LCKurtz]

Well, my thinking is that n! becomes infinitely large. That said, I think that since n! is in the numerator, I should look for a value in the numerator which is comparatively less than n! and also becomes infinitely large. However, I'm unsure how the numerator is established. Also, why did you divide up the fractions the way you did?

I don't think that tells me how you came up with $\frac 4 n$.

That last question is the key. What do you notice about the fractions in the parentheses? What would be a nice overestimate for them?

 

[gsmtiger18]

The numerator is constant, and the denominator is continuously increasing. I'm grasping at straws at this point, but could an overestimate be 2/n?

 

[LCKurtz]

You need to quit thinking about things increasing or n changing. We are looking at$$
|a_n|=\frac{2\cdot 2}{1\cdot 2}\left(\frac{2}{3}\frac 2 4 \cdots \frac 2{n-1}\right)\frac 2 n$$for fixed n. We might wonder later about what happens to it if $n$ is large, but right now we are looking at this expression. I asked you if you notice anything about the fractions in the parentheses. You haven't answered that.

Earlier in this thread, you somehow came up with $\frac 4 n$ as a possible overestimate for this expression. There is a good reason to do that but you haven't told me how you got it.

 

[gsmtiger18]

Forgive me if I sound irritated, but the whole reason I asked this question is I HAVE NO CLUE WHAT I'M DOING. I chose 4/n because it was what resulted from the terms on either side of the parentheses. I don't notice anything about the fractions in parentheses.

 

[LCKurtz]

Forgive me if I sound irritated, but the whole reason I asked this question is I HAVE NO CLUE WHAT I'M DOING. I chose 4/n because it was what resulted from the terms on either side of the parentheses. I don't notice anything about the fractions in parentheses.

It's not a good thing to have "no clue" with a test coming Monday, is it. I have led you to a critical step in this problem and I'm trying to get you to think about it. I'm going to ask again for you to look at those fractions. You can't not notice anything about them. Are they complex numbers (no), are they negative (no), are they irrational numbers (no), are they positive (yes) etc.

So instead of telling me you have no clue, I want you to tell me what you do know about those fractions in addition to the fact that they are positive. Write several of them down if you have to. Look at them.

 

[gsmtiger18]

It's not a good thing to have "no clue" with a test coming Monday, is it. I have led you to a critical step in this problem and I'm trying to get you to think about it. I'm going to ask again for you to look at those fractions. You can't not notice anything about them. Are they complex numbers (no), are they negative (no), are they irrational numbers (no), are they positive (yes) etc.

So instead of telling me you have no clue, I want you to tell me what you do know about those fractions in addition to the fact that they are positive. Write several of them down if you have to. Look at them.

They're continuously decreasing, so each successive fraction is smaller than the last. That would suggest that the sequence converges. Like you said, they are not complex or irrational numbers. Every fraction within the parentheses is less than 1. Is that the important part?

 

[LCKurtz]

They're continuously decreasing, so each successive fraction is smaller than the last. That would suggest that the sequence converges. Like you said, they are not complex or irrational numbers. Every fraction within the parentheses is less than 1. Is that the important part?

Bingo. That wasn't so difficult, was it? So, because of that what could you put on the right side of this:$$
|a_n| = \frac{2\cdot 2}{1\cdot 2}\left(\frac{2}{3}\frac 2 4 \cdots \frac 2{n-1}\right)\frac 2 n \le ~?$$

 

[gsmtiger18]

4/n?

 

[LCKurtz]

4/n?

Are you asking me or telling me? Do you see why?

 

[gsmtiger18]

Are you asking me or telling me? Do you see why?

The product of the fractions outside the parentheses is 4/n. Since the value of the fractions within the parentheses is less than one, 4/n multiplied by that value is obviously going to be less than or equal to 4/n.

 

[LCKurtz]

Good. Now if you put that on the right side of that inequality do you see that you have, by the squeeze principle, a very short, simple argument that $a_n\to 0$ as $n\to \infty$? I assume you know $|a_n|\to 0$ implies $a_n\to 0$.

 

[gsmtiger18]

Yes, I know that. Thank you so much.