# Find the lowest value of N

1. Jun 25, 2012

### Mosaness

1. If we accept the fact that the sequence {n/(n + 1)}n = 1 converges to the limit L = 1, then for every ε>0 there exists an integer N such that abs value of an - L = abs value of n/(n + 1) - 1< ε when n>/ N. In each part, find the smallest value of N for the given value of ε.

2.a. ε = 0.25
b. ε = 0.1
c. ε = 0.001

The way I did this was:
abs value of n/(n + 1) - 1 < ε
abs value of n/(n + 1) - 1< 0.25
abs value of n/(n + 1) - 1= 0.25
n/(n + 1) = 1.25
n = 1.25n + 1.25
-0.25n = 1.25
n = 5 because of the absolute value bars, therefore N = 4.

However, I do not think I did this correctly. Can someone show me the correct way?

Last edited by a moderator: Jun 26, 2012
2. Jun 26, 2012

### klondike

What it says is "for any ε>0, there exists N, such that if n>N, then |n/(n+1)-1|<ε."

The inequality with the abs expands to
$$-\epsilon<\dfrac{n}{n+1}-1<+\epsilon$$

The right part of the inequality is automatically satisfied for all n because the fraction is always less than 1, so you only work on the left part.

3. Jun 26, 2012

### Mosaness

Thank you for explaining :)

However, one last question, where is the 1 coming from? Isn't ε = 0.25?

4. Jun 26, 2012

### klondike

You are welcome. Limit is the most interesting and fundamental thing in calculus 101. It's hard for beginner. I remembered our prof said don't feel bad if you don't get it at first, it took the best mathematicians centuries to get it right.

Back to your question, the 1 comes from the limit of the sequence as you agree. The epsilon-N language can be said in plain English in this way "we always pick a large enough n to make n/(n+1) close to 1 as much as we please." There are always 3 important things in the limit, the sequence, the limit, and the distance between the limit (1 in this case) the sequence evaluated at n. The distance is ε(n) which is a function of n. Your task is to find the minimal n for the given ε that makes the inequality happy.

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