Need clarification on limit of sequence.

[Samuelb88]

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 | &lt; E.

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

{ \frac{1}{2n} } &lt; E implies { \frac{1}{2E} } &lt; 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)

 

[praharmitra]

I suspect it's a printing error.

 

[╔(σ_σ)╝]

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

{ \frac{1}{2n} } &lt; E implies { \frac{1}{2E} } &lt; 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.

 

[Samuelb88]

Thanks, both of you!