Science:Math Exam Resources/Courses/MATH220/December 2009/Question 08 (b)

MATH220 December 2009

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Question 08 (b)
Let $\displaystyle {}s_{n}$ be a convergent sequence such that $\displaystyle {}s_{n}<q$ for all $n\in \mathbb {N}$. Prove that $\lim _{n\to \infty }s_{n}\leq {}q.$

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Use a proof by contradiction. If the limit tends to a number bigger than q, justify why eventually some terms must exceed q.

Hint 2
Start your proof by setting $\displaystyle \epsilon =(c-q)/2$.

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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. Assume towards a contradiction that $\lim _{n\to \infty }s_{n}=c>q$ for some real value c. Let $\displaystyle \epsilon =(c-q)/2$ which is positive by the above assumption. By the definition of a limit, we see that there is some index N such that for all natural numbers n bigger than N, we have that $|s_{n}-c|<\epsilon ={\frac {c-q}{2}}.$ If $\displaystyle s_{n}-c>0$ this is a contradiction, because $s_{n}>c$ contradicts the given fact that $s_{n}<q$, since we assumed $q<c$. So it must hold that $\displaystyle s_{n}-c<0$, that is, $\displaystyle c-s_{n}>0$. Hence, we can remove the absolute values and get that $c-s_{n}<\epsilon ={\frac {c-q}{2}}$ Rearranging gives $s_{n}>{\frac {c+q}{2}}>{\frac {q+q}{2}}=q.$ This is a contradiction.

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