- #1
Samuelb88
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Homework Statement
What is the limit of [tex] { \frac{n+1}{2n} }[/tex] 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 [tex] { \frac{n+1}{2n} } [/tex] behaves like [tex] { \frac{n}{2n} } [/tex], so we expect the limit to be 1/2 as n --> oo. To prove this, we observe that
[tex] | S_n - L | = | { \frac{n+1}{2n} } - { \frac{1}{2} } | = | { \frac{1}{2n} } | = { \frac{1}{2n} } < E[/tex]
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 [tex] |S_n - L | < E[/tex].
My question is, why didn't they choose N to be any integer > 1/(2ε) ? Since:
[tex] { \frac{1}{2n} } < E[/tex] implies [tex] { \frac{1}{2E} } < n [/tex]
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)