Question

I'm having a difficult time understanding the definition of a limit and the proof.

The definition of a limit is defined (by my book):
Let An (from n=1 to infinity) be a sequence of real numbers. We say that the sequence An has a limit L in the set of real numbers if for every epsilon > 0, there exists a positive integer N, such that if n >= N, then abs value of An- L < epsilon.

My book proves the sequence n/n+1 has a limit of 1.

"If epsilon > 0, we are required to show that for some positive integer N, abs value of An - L < epsilon, if n >= N."

Abs value of An - L simplifies

Answer 1

to 1/(n+1) and "hence if we choose a positive integer N such that 1/(N+1) < epsilon then if n>= N, 1/n+1 <= 1/N+1 < epsilon. (I understand the algebra completely fine.) I guess I just don't understand how this is enough to conclude that the limit of the sequence is 1?

Answer 2

not sure I follow it completely myself
at least enough to conclude the limit is 1

Answer 3

what is \(a_n\)

Answer 4

Sorry, I should have included that. It's n/(n+1)

Answer 5

You have \[\left|\frac{n}{n+1}-1\right|<\epsilon\]

for any choice of \(\epsilon>0\) provided \(n
\) is large enough

Answer 6

what else could the limit be then?

Answer 7

Intuitively, I understand why the limit would be 1. I just don't understand how that applies to the definition/theorem.

Answer 8

are you saying you don't understand the definition?

Answer 9

Basically, I don't understand how they can conclude the limit of An is 1 by the proof the book gave.

Answer 10

well to should that the limit is 1 we need to show that \(\frac{n}{n+1}\) can be made to be as close to 1 as we want

so we need to show that for all \(n>N\)

we have \[\left|\frac{n}{n+1}-1\right|<\epsilon\]

...

Answer 11

do you understand that?

Answer 12

so far

Answer 13

for the most part. The "distance" between an arbitrary term of An and 1 is less than some really small, positive number epsilon?

Answer 14

although I know that the only real condition on epsilon is that it is positive.

Answer 15

any term of \(a_n\) provided \(n>N\)

the entire tail of the sequence needs to be as close to 1 as we want

Answer 16

okay, there is one part where I'm confused. What's the difference between n and N?

Answer 17

that's**

Answer 18

jas

Answer 19

there is a nice picture about half way down the page that illustrate what N is

http://math.feld.cvut.cz/mt/txta/2/txe3aa2a.htm

Answer 20

need to go for a little bit

Answer 21

okay, thanks!

Answer 22

you got this?

Answer 23

i think the crux is in the statement here
"hence if we choose a positive integer N such that \(\frac{1}{N+1} < \epsilon\)"

Answer 24

since the natural numbers are unbounded, that tells you that given any \(\epsilon\) there is an integer \(N\) with \(N>\frac{1}{\epsilon}\) i.e we can find a natural number that large
that is the same as saying we can find some
\(N\) with \[\frac{1}{N+1}<\epsilon\]