# Sequence question:

!~TONY~!
Posers and Puzzles 26 Oct '04 02:24
1. !~TONY~!
1...c5!
26 Oct '04 02:24
I am studying for a quiz tomorrow in Calc II and ran across I problem that I know we did in class but I don't know how to do it and didn't copy it down. (idiot, I know). The question is : Find whether the sequence (N + 1/ N) ^ N from N= 1 to infinity converges or diverges, and if it converges, where? The answer is e, but I didn't some mumbo jumbo that didn't work, and I don't even know if it's mathematically legal. Anyone know how to do this?
2. telerion
True X X Xian
26 Oct '04 04:50
Originally posted by !~TONY~!
I am studying for a quiz tomorrow in Calc II and ran across I problem that I know we did in class but I don't know how to do it and didn't copy it down. (idiot, I know). The question is : Find whether the sequence (N + 1/ N) ^ N from N= 1 to infinity converges or diverges, and if it converges, where? The answer is e, but I didn't some mumbo jumbo that didn't work, and I don't even know if it's mathematically legal. Anyone know how to do this?
Well let's see. Just so that I'm sure. Is it [(n+1)/n]^n or is it [n+(1/n)]^n?

The first one just goes to 1. Since it can be rewritten

[1+(1/n)]^n which as n-&gt; infinite goes to 1^n = 1.

So I don't suppose that's the version of it.

The second one clearly diverges. Again in the limit, 1/n -&gt; 0, and so the sequence -&gt; infinite^infinite .

You're sure the answer is e?

Maybe the question was a little different?

Well, that's my two cents. I'll differ to one of the mathematicians on the site if I'm missing something.
3. AThousandYoung
All My Soldiers...
26 Oct '04 08:141 edit
http://mathworld.wolfram.com/e.html

e is defined as lim(n=&gt;infinity) (1+(1/n))^n = ((n+1)/n)^n

There's all kinds of stuff about e there.

4. AThousandYoung
All My Soldiers...
26 Oct '04 08:29
[1+(1/n)]^n which as n-&gt; infinite goes to 1^n = 1.

This is flawed, because even as 1/n gets very very small, the power of the sum of 1+(1/n) gets really, really huge.
5. AThousandYoung
All My Soldiers...
26 Oct '04 08:41

http://mathforum.org/library/drmath/view/51954.html

Use the Binomial Theorem to expand the limit.

(1 + 1/n)^n = 1 + 1 + (1 - 1/n)/2! + (1 - 1/n)(1 - 2/n)/3! + ... + (1 - 1/n)(1 - 2/n)(...)(1 - (n-1)/n)/n!.

The last term clearly approaches zero, since n! approaches infinity and none of the terms in the numerator approach infinity. In fact, (1 - (n-1)/n) approaches zero by itself (1-1). So, the series converges.

I don't know how to prove the answer is e though. Maybe you can wade through those websites and figure it out.
6. royalchicken
CHAOS GHOST!!!
26 Oct '04 11:241 edit
If it's (n+1/n)^n then it diverges. If it's (1+1/n)^n, then it converges to e as follows:

(1+1/n)^n = 1+(n/n)+n(n-1)*n^-2/2 + n(n-1)(n-2)*n^-3/6... +n^-n

Letting n --&gt; infinity and canceling ns:

= 1 + 1 + 1/2 + 1/6 + ...

= e.

EDIT The last bit is just the definition of e.
7. !~TONY~!
1...c5!
26 Oct '04 16:41
I figured it out today before the quiz. Good thing there was nothing of the sort on the test. Instead he unleashed this one. A ball is dropped from 6 feet above the ground, with each subsequent bounce being (3/4) that of the next. What is the total vertical distance traveled by the ball. I got it, but it took me about 15 minutes!
8. telerion
True X X Xian
26 Oct '04 17:081 edit
Originally posted by !~TONY~!
I figured it out today before the quiz. Good thing there was nothing of the sort on the test. Instead he unleashed this one. A ball is dropped from 6 feet above the ground, with each subsequent bounce being (3/4) that of the next. What is ...[text shortened]... traveled by the ball. I got it, but it took me about 15 minutes!
Oh yeah I remember those. Nice little application of a geometric series. I think I had that exact problem back win I took calculus.

Thank you all for the correction on e.

9. !~TONY~!
1...c5!
26 Oct '04 17:19
Originally posted by telerion
Oh yeah I remember those. Nice little application of a geometric series. I think I had that exact problem back win I took calculus.

Thank you all for the correction on e.

Yeah, it took me quite a bit to figure out it was a geometric series. I was messing around with telescopic series' for a about 5 minutes before that...hehehehehe....
10. 27 Oct '04 10:57
Originally posted by royalchicken
If it's (n+1/n)^n then it diverges. If it's (1+1/n)^n, then it converges to e as follows:

(1+1/n)^n = 1+(n/n)+n(n-1)*n^-2/2 + n(n-1)(n-2)*n^-3/6... +n^-n

Letting n --> infinity and canceling ns:

= 1 + 1 + 1/2 + 1/6 + ...

= e.

EDIT The last bit is just the definition of e.
e = 2.7 1828 1828 459045 2353 6028 7549 ......
11. royalchicken
CHAOS GHOST!!!
27 Oct '04 14:05
Right, but that wouldn't make a very useful definition, because it doesn't recur or stop, and without a definition we'd have no way of calculating it anyway. So we say:

e = exp(1) = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + ...
12. PBE6
Bananarama
16 Nov '04 20:12
Here's an easier way to get the answer:

Let f(n) = (1+1/n)^n. Taking the natural logarithm of both sides we get:

ln(f(n)) = ln[(1+1/n)^n] = n*ln(1+1/n)

Now let x = 1/n:

n*ln(1+1/n) = ln(1+x)/x

Since lim(n--&gt;inf) 1/n = 0, we can take the lim(x--&gt;0) instead. Using L'Hopital's Rule, we get:

lim(x--&gt;0) ln(1+x)/x = lim(x--&gt;0) ln(1+x)'/x' = lim(x--&gt;0) (1/(1+x))/1

lim(x--&gt;0) (1/(1+x))/1 = (1/1)/1 = 1

Therefore lim(x--&gt;0) ln(f(n)) = 1. Taking the antilogarithm of both sides we get:

lim(x--&gt;0) f(n) = e

Therefore:

lim(n--&gt;inf) f(n) = e

13. The Plumber
Leak-Proof
16 Nov '04 23:481 edit
Originally posted by !~TONY~!
I figured it out today before the quiz. Good thing there was nothing of the sort on the test. Instead he unleashed this one. A ball is dropped from 6 feet above the ground, with each subsequent bounce being (3/4) that of the next. What is ...[text shortened]... traveled by the ball. I got it, but it took me about 15 minutes!
42 feet

(or 6 feet if you want to be a wise guy)ðŸ˜‰