Science:Math Exam Resources/Courses/MATH220/April 2011/Question 03 (b)

MATH220 April 2011

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Question 03 (b)
Use the definition of convergence for sequences, prove that Failed to parse (syntax error): {\displaystyle \lim_{n \rightarrow \infty} \frac{1-2\cos(n)}{n} = 0. }

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
As often with trigonometric function, make good use of the fact that $-1\leq \cos(x)\leq 1$ for any value of x.

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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. Here, the limit being 0 means that we simply need to show that given any Failed to parse (syntax error): {\displaystyle \epsilon > 0} there exists a number $N_{\epsilon }$ such that $\left|{\frac {1-2\cos(n)}{n}}\right|<\epsilon$ for any value of $n$ larger than $N_{\epsilon }$. Since we have that $-1\leq \cos(n)\leq 1$ we can multiply everywhere by -2 and obtain that $2\geq -2\cos(n)\geq -2$ and add 1 everywhere to obtain $3\geq 1-2\cos(n)\geq -1$ which now allows us to conclude about the size of $1-2\cos(n)$ in absolute value: Failed to parse (syntax error): {\displaystyle | 1-2 \cos (n) | \leq 3 } So we can conclude that ${\begin{aligned}\left|{\frac {1-2\cos(n)}{n}}\right|&={\frac {|1-2\cos(n)|}{n}}\\&\leq {\frac {3}{n}}\\&<\epsilon \end{aligned}}$ for any value of $n$ that is larger than $N_{\epsilon }=3/\epsilon$; this concludes our proof.

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