Science:Math Exam Resources/Courses/MATH220/December 2010/Question 08 (b)

MATH220 December 2010

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Question 08 (b)
Decide whether the following sequence converges or diverges. Prove your answer using the definition of convergence. $c_{n}={\frac {2n^{3}-5n+2}{3n^{3}+8n+3}}.$

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
Using basic calculus, we can see that the limit is 2/3. Now we just need to prove it. Start by picking $\epsilon >0$ and proceed from there.

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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. This sequence converges to 2/3. To see this, consider the following ${\begin{aligned}\left|c_{n}-{\frac {2}{3}}\right|&=\left|{\frac {2n^{3}-5n+2}{3n^{3}+8n+3}}-{\frac {2}{3}}\right|\\&=\left|{\frac {2n^{3}-5n+2}{3n^{3}+8n+3}}-{\frac {2(n^{3}+8n/3+1)}{3(n^{3}+8n/3+1)}}\right|\\&=\left|{\frac {2n^{3}-5n+2}{3n^{3}+8n+3}}-{\frac {2n^{3}+16n/3+2)}{3n^{3}+8n+3}}\right|\\&=\left|{\frac {-31n/3}{3n^{3}+8n+3}}\right|\\&=\left|{\frac {-31n}{9n^{3}+24n+9}}\right|\\&={\frac {31n}{9n^{3}+24n+9}}\\&<{\frac {31n}{9n^{3}}}\\&={\frac {31}{9n^{2}}}\end{aligned}}$ where the inequality above holds since we made the denominator smaller hence making the fraction larger. Now let $\epsilon >0$ and set $n_{0}={\sqrt {\frac {9}{31\epsilon }}}$. It follows that for each $n\geq n_{0}$ we obtain $\left|c_{n}-{\frac {2}{3}}\right|<{\frac {31}{9n^{2}}}\leq {\frac {31}{9n_{0}^{2}}}=\epsilon$ which proves the convergence.

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