Science:Math Exam Resources/Courses/MATH220/April 2005/Question 09
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Question 09 

Prove, directly from the definition of the limit of a sequence, that 
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Hint 

A sequence converges to a number L if for any there exists a number such that 
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Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. Let ε > 0, we want to show that for some positive integer N_{ε}, we have that First, we can get rid of the absolute value since and so we wonder if for some value of n large enough, we will have for the value of ε that we fixed (and that we think of as being a very small number). Since we are interested in the value of n, we take the logarithm on each side and obtain Now since 2/3 is a number smaller than 1, its logarithm will be a negative number, so when we divide in the above by ln(2/3) we have to flip the inequality sign, we obtain Note that since ε is also a very small number, its logarithm is also negative and hence the fraction on the righthand side is a positive number (as expected). We can thus choose N_{ε} to be the nearest integer larger than the fraction, which we can write as which concludes our proof. NoteIt is not a bad idea to see that ε proofs allow you to actually compute something about how quickly the sequence will converge to its limit. In this case for example, if we choose ε = 0.001 then the above proof gives us the value So for any n ≥ 18, the value of (2/3)^{n} should be less than 0.001, indeed we have 