# Need clarification on limit of sequence.

## Homework Statement

What is the limit of $${ \frac{n+1}{2n} }$$ as n --> oo. Prove your answer.

## The Attempt at a Solution

This is example from my book. Here is the problem:
*I am using the capital letter "E" instead of "ε" in latex-code.

Intuitively, 1 is small relative to n as n gets large, so $${ \frac{n+1}{2n} }$$ behaves like $${ \frac{n}{2n} }$$, so we expect the limit to be 1/2 as n --> oo. To prove this, we observe that

$$| S_n - L | = | { \frac{n+1}{2n} } - { \frac{1}{2} } | = | { \frac{1}{2n} } | = { \frac{1}{2n} } < E$$

We observe that if n > 2/ε, then 0 < 1/2n < ε. Thus taking N to be any integer > 2/ε, we have as required that n>N, then $$|S_n - L | < E$$.

My question is, why didn't they choose N to be any integer > 1/(2ε) ? Since:

$${ \frac{1}{2n} } < E$$ implies $${ \frac{1}{2E} } < n$$

I see that 1/(2ε) < 1/ε < 2/ε < N. Is there a reason why N > 2/ε was chosen rather than N > 1/(2ε)

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I suspect it's a printing error.

My question is, why didn't they choose N to be any integer > 1/(2ε) ? Since:

$${ \frac{1}{2n} } < E$$ implies $${ \frac{1}{2E} } < n$$

I see that 1/(2ε) < 1/ε < 2/ε < N. Is there a reason why N > 2/ε was chosen rather than N > 1/(2ε)
I am not sure if it is a printing error or not since it works.
Their bound is not at "tight" but it works.

Thanks, both of you!