Question

\[a_n=\frac{(-3)^n}{n!}\]
\[= \frac{(-3)\times(-3)\times(-3)\times(-3)\times...\times (-3)}{1\times2\times3\times4...\times n}\]
\[\ge \frac{-3}{1} \times \frac{(-3)^{n-2}}{3^{n-2}} = \frac{(-3)}{1}\left(\frac{(-3)}{3}\right)^{n-2}\]
\[-1 <1\]
converges to -1?

Answer 1

0

Answer 2

say what?

Answer 3

maple gave me that lol took limit as n->infinity

Answer 4

what's wrong with my thought process @KingGeorge?

Answer 5

He's using the (not so obvious) fact that \(f(x)=x!\) trumps \(f(x)=c^x\) as \(x\to\infty\).

Answer 6

I'm not actually sure what you've done wrong. However, I can't tell why it would prove convergence to anything either.

Also, @timo86m don't just give out answers like that.

Answer 7

If you wanted to prove convergence, the alternating series test would be a good bet.

Answer 8

here is the question from my book. "Determine whether the sequence converges or diverges. If it converges, find the limit."

Answer 9

First, use the alternating series test to determine if it converges. When you use that, you'll easily see what the limit is.

Answer 10

but it's sequences

Answer 11

In this case, it's basically the same thing. Look at \[\lim_{n\to\infty}(-1)^{n}\frac{3^n}{n!}\]

Answer 12

Using the same concept of the alternating series test, you can just look at \[\lim_{n\to\infty}\frac{3^n}{n!}\]

Answer 13

here is what I did...I took Satellites train of thought and applied it to this question. I guess it's not that simple?
http://openstudy.com/users/mathsofiya#/updates/4ff89c0be4b058f8b7631d1b

Answer 14

If that limit I just posted above is anything but 0, it diverges. Otherwise, it converges to 0.

Answer 15

Here's a good intuition for why \(x!\) increases faster than \(c^x\). Think about a large value of \(x\), and think about what increasing it by one does to each expression. For the \(x!\), you multiply the old value by \(x+1\), whereas for \(c^x\) you multiply by \(c\). Since \(c\) is fixed, and \(x\) tends to infinity, \(x!\) will always eventually increase faster than \(c^x\) as \(x\to\infty\).

Answer 16

makes sense @nbouscal

Answer 17

its zero cuzz n! grows larger than even abs(3^n)

Answer 18

and \[\frac{3n}{\infty}=0\]

Answer 19

silly question...so this sequence is finite?

Answer 20

don't shoot me

Answer 21

No no no, the sequence is still infinite. We're just looking at how these things act as they tend towards infinity by looking at examples of them increasing at specific points.

Answer 22

ok :P

Answer 23

n! grows a heck of a lot faster than some integer^n

Answer 24

So basically the key point to look at is where the factorial starts growing faster than the exponential. When n=c, the factorial starts to grow faster than the exponential does, and it's just a matter of time for the factorial to catch up and pass the exponential in value. Then it will continue to grow faster as you approach infinity, which means the sequence converges to zero.

Answer 25

ok...let's see if I understand this. The sequence is infinite, but eventually converges to zero.

Answer 26

right.

Answer 27

yes

Answer 28

Hmm I don't like the use of the word eventually there.

Answer 29

That implies that there is a point at which the terms of the sequence are zero, and that is not the case.

Answer 30

We say that the sequence converges to zero, but that does not mean that it actually gets there.

Answer 31

how would you rephrase this sentence?
"The sequence is infinite, but eventually converges to zero."
the sequence is infinite, and converges to zero?

Answer 32

Yes. The word eventually implies that there is a point at which the sequence converges to zero, but that's not what convergence means. No matter how big an \(n\) you choose, \(\dfrac{(-3)^n}{n!}\) will always be nonzero.

Answer 33

I think this is an important distinction to understand when talking about sequence convergence. Sequence convergence simply means that you can get as close as you like to zero by going far enough up the sequence, it doesn't mean the sequence ever actually reaches zero.

Answer 34

I don't see that the term "eventually" is incorrect. Converging to 0 does not mean that it eventually equals 0.

Answer 35

Eventually implies temporality to the convergence, and convergence does not have temporality.

Answer 36

In this case then, eventually means "immediately" since it immediately starts to converge to 0. The two aren't mutually distinct.

Answer 37

"Starts to converge" is still attributing temporality to the convergence. Convergence is not temporal. A sequence either converges or it does not. There is not a point at which it starts to converge.

Answer 38

Well it starts at \(n=0\) (or 1 in some cases).

Answer 39

What are you going to do, define a value of epsilon where we can say the convergence has begun? Convergence is on-off, there is no notion of time involved.

Answer 40

Either there's a delta for every epsilon or there is not.

Answer 41

I could also argue that for the first few terms for \(n=0,1,2\) the absolute value of the sequence is increasing, and is thus not converging to 0.

Answer 42

If you were to argue as such, you would be misunderstanding what convergence means.

Answer 43

Convergence isn't an action, it is an attribute.

Answer 44

Maybe that's why I don't like analysis very much :P

Answer 45

It can be convenient to think about convergence as an action that the sequence takes, but it is not really proper. A sequence converges if there's a delta for every epsilon. It doesn't converge if there isn't.

Answer 46

Haha, I don't blame you for that. I think most people don't like analysis very much.

Answer 47

And now we've probably confused MathSofiya much more than we have helped :P

Answer 48

no I like the discussion:P

Answer 49

thanks guys!!!

Answer 50

you're welcome.

Answer 51

While the results alternate between negative and positive, they are still ALWAYS getting closer and closer to 0

That's a bad picture, but it's a little something like that.