- 26 Oct '04 02:24I 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?
- 26 Oct '04 04:50

Well let's see. Just so that I'm sure. Is it [(n+1)/n]^n or is it [n+(1/n)]^n?*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?**

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

[1+(1/n)]^n which as n-> 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 -> 0, and so the sequence -> 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. - 26 Oct '04 08:41This looks like it might help you.

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. - 26 Oct '04 11:24 / 1 editIf 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. - 26 Oct '04 16:41I 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!
- 26 Oct '04 17:08 / 1 edit

Oh yeah I remember those. Nice little application of a geometric series. I think I had that exact problem back win I took calculus.*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!**

Thank you all for the correction on e.

- 26 Oct '04 17:19

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....*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.

- 27 Oct '04 10:57

e = 2.7 1828 1828 459045 2353 6028 7549 ......*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. - 16 Nov '04 20:12Here'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-->inf) 1/n = 0, we can take the lim(x-->0) instead. Using L'Hopital's Rule, we get:

lim(x-->0) ln(1+x)/x = lim(x-->0) ln(1+x)'/x' = lim(x-->0) (1/(1+x))/1

lim(x-->0) (1/(1+x))/1 = (1/1)/1 = 1

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

lim(x-->0) f(n) = e

Therefore:

lim(n-->inf) f(n) = e

- 16 Nov '04 23:48 / 1 edit

42 feet*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!**

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