# Epsilon-N Proof of a Limit of a Sequence: Lim[2n/(3n+2)] = 2/3 – Article by G. Stolyarov II

**G. Stolyarov II**

**Note from the Author:**

*This proof was originally published on Associated Content (subsequently, Yahoo! Voices) in 2007. The article earned over 17,250 page views on Associated Content/Yahoo! Voices, and I seek to preserve it as a valuable resource for readers, subsequent to the imminent closure of Yahoo! Voices. Therefore, this essay is being published directly on The Rational Argumentator for the first time. ****

*~ G. Stolyarov II, July 12, 2014*

This is the first in a series of formal mathematical proofs I intend to present in order to assist anybody who has ventured into the challenging but fascinating world of advanced calculus and real analysis. I start with a fairly basic proof: the limit of the nth term of a sequence as n becomes increasingly large. This is an epsilon-N proof, which uses the following definition: lim(n approaches ∞)x_{n}= L iff for each real number ε>0, there exists a positive integer N(ε) such that if n ≥ N(ε), then │x_{n}-L│< ε, i.e., L- ε < x_{n}< L+ ε.

The epsilon-N proof has two component steps; first, we assume that our ε>0 is given and work backward to transform the inequality │x_{n}-L│< ε in order to find an appropriate value of N(ε) that corresponds to a given value of ε. Then, using the value of N(ε) we found, we work forward to show that if n ≥ N(ε), then │x_{n}-L│< ε.

The proof also uses the Archimedean Property, which states that the set of positive integers is not bounded above. There are actually four equivalent conditions that are known as the Archimedean Property:

1. If a,b are in R, a>0, b>0, then there is a positive integer n such that na>b

2. The set of positive integers is not bounded above.

3. For each real number x, there exists an integer n such that n ≤x < n+1

4. For each positive real number x, there exists a positive integer n such that 1/n ≤ x

**Prove: lim(n** **approaches ∞)[2n/(3n+2)] = 2/3**

**Proof:** Let ε>0 be given. Find N(ε)є Z^{+} such that if N(ε) < n, then │2n/(3n+2)-2/3│< ε.

Working backward to transform the inequality: │2n/(3n+2)-2/3│< ε

│6n/[3(3n+2)]-2(3n+2)/[3(3n+2)]│< ε

│[6n-2(3n+2)]/[3(3n+2)]│< ε

│[6n-6n-4]/[3(3n+2)]│< ε

│-4/[3(3n+2)]│< ε

Since ε>0, (3n+2)>0, the above inequality is the same as

4/[3(3n+2)] < ε

4/(3ε) < (3n+2)

4/(3ε)- 2 < 3n

4/(9ε)- 2/3 < n

Now I work forward to prove the original result:

Let N(ε)є Z^{+} э 4/(9ε)- 2/3 < N(ε).

Since 4/(9ε)- 2/3 is a real number, by the Archimedean Property it must be the case that some integer exists which is greater than that real number-since the set of positive integers is not bounded above. Here, we call that integer N(ε).

For all n>N, if 4/(9ε)- 2/3 < N < n, then:

4/(3ε)- 2 < 3n

4/(3ε) < (3n+2)

4/[3(3n+2)] < ε

│-4/[3(3n+2)]│< ε

│6n/[3(3n+2)]-2(3n+2)/[3(3n+2)]│< ε

**│2n/(3n+2)-2/3│< ε**

I have demonstrated the above inequality, which is sufficient to demonstrate that **lim(n** **approaches ∞)[2n/(3n+2)] = 2/3.**

I have hence proved what was desired. Another way to express that the desired objective has been obtained (which I shall use in future proofs of this sort) is the abbreviation

“Q. E. D.” of the Latin “Quod Erat Demonstraterum,” which means “that which was to be demonstrated.”

## One thought on “Epsilon-N Proof of a Limit of a Sequence: Lim[2n/(3n+2)] = 2/3 – Article by G. Stolyarov II”

Thanks a lot for your insight, can you help me work out this example Lim(as n approaches infinity) n+3/((n^2)-5)