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

MATH220 April 2005

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[hide] Question 09
Prove, directly from the $\epsilon$ -definition of the limit of a sequence, that $\lim _{n\to \infty }\left(-{\frac {2}{3}}\right)^{n}=0$

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[show] Hint
A sequence ${a_{n}}$ converges to a number L if for any $\epsilon >0$ there exists a number $N_{\epsilon }$ such that ${\text{if }}n\geq N_{\epsilon }\quad {\text{then}}\quad \|a_{n}-L\|<\epsilon$

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[show] Solution
Found a typo? Is this solution unclear? Let us know here. 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 $\left\|\left(-{\frac {2}{3}}\right)^{n}\right\|<\varepsilon \quad {\text{if }}n\geq N_{\varepsilon }$ First, we can get rid of the absolute value since $\left\|\left(-{\frac {2}{3}}\right)^{n}\right\|=\left({\frac {2}{3}}\right)^{n}$ and so we wonder if for some value of n large enough, we will have $\left({\frac {2}{3}}\right)^{n}<\varepsilon$ 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 $\underbrace {\ln \left[\left({\frac {2}{3}}\right)^{n}\right]} _{=n\ln(2/3)}<\ln(\varepsilon )$ 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 $n>{\frac {\ln(\varepsilon )}{\ln(2/3)}}$ Note that since ε is also a very small number, its logarithm is also negative and hence the fraction on the right-hand 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 $N_{\varepsilon }=\left\lceil {\frac {\ln(\varepsilon )}{\ln(2/3)}}\right\rceil$ which concludes our proof. Note It 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 $N_{0.001}=\left\lceil {\frac {\ln(0.001)}{\ln(2/3)}}\right\rceil =\left\lceil 17.0366\right\rceil =18$ So for any n ≥ 18, the value of (-2/3)ⁿ should be less than 0.001, indeed we have $\left\|\left(-{\frac {2}{3}}\right)^{17}\right\|=0.00101\quad {\text{and}}\quad \left\|\left(-{\frac {2}{3}}\right)^{18}\right\|=0.0006766$