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Need clarification on limit of sequence.

  • Thread starter Samuelb88
  • Start date
  • #1
162
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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)
 

Answers and Replies

  • #2
311
1
I suspect it's a printing error.
 
  • #3
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)
I am not sure if it is a printing error or not since it works.
Their bound is not at "tight" but it works.
 
  • #4
162
0
Thanks, both of you!
 

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