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Prove (with the $\epsilon$-$N$ method) that the sequence $(3n^3 +2)/n^3$ converges. Determine the limit first. Can someone help me with this? I have seen many examples but it I still don't get it. If someone can explain step by step (especially why we need to use the $N$?) I would be very very thankful.

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3 Answers

As this is likely to be a homework, just few hints: $$ a_n := \frac{3n^3 +2}{n^3} = 3 + \frac{2}{n^3} $$ As $n$ increases, the term $2/n^3$ clearly becomes smaller and smaller. Now for any $\epsilon > 0$, find $N$, such that for all $n > N$, $| a_n - 3 | < \epsilon $.

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Well, you can rewrite $\frac{3n^3 + 2}{n^3}$ as $3 + \frac{2}{n^3}$. Now observe what happens if you put in larger and larger $n$. What can you say about the term $\frac{2}{n^3}$ then? And what does that tell you about the limit?

To formally prove that some $c$ that you have determined previously is actually the limit, you will have to show that for every $\epsilon > 0$ there is an $N$ such that for every $n \geq N$ you have $\left|\frac{3n^3 + 2}{n^3} - c\right| < \epsilon$. In other words, show that you can make all values of the sequence after a certain starting point lie arbitrarily close to the limit, provided that you pick a suitable starting point. Hint: This is the same as showing that all values of $\frac{2}{n^3}$, $n \geq N$ are $< \epsilon$, for an arbitrary epsilon, if you pick $N$ large enough.

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I think I am starting to understand. I only got 2 examples in the book where I stared for 2 days and still I didn't get it. Thanks both of you for your effort to help people! –  Jeroen Oct 7 '12 at 14:29
@Jeoren Once you've solved it, you could post your solution as an answer. That way, you'll get feedback, and also earn some reputation points. –  fgp Oct 7 '12 at 18:30
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Proof: Let $\epsilon > 0$ be given. You want to find a $N$ such that if $n\geq N (>0)$, then $\lvert a_n - 3\lvert < \epsilon$. Consider what you want: $$ \lvert a_n - 3 \lvert = \left\lvert \frac{2}{n^3}\right\lvert = 2 \frac{1}{n^3}. $$

Hence you just need to find a $N$ such that if $n> N$, then $2\frac{1}{n^3} < \epsilon$. Note that this $N$ will be dependent on $\epsilon$ in some way.

Just to show an example. Say that $\epsilon = 0.1$. Then what would $N$ need to be for $n> N$ to imply that $\frac{2}{n^3} < 0.1$? Well, we would need that $$ \begin{align} 2 &< 0.1\cdot n^3 \Rightarrow \\ \frac{2}{0.1} = 20 &< n^3 \Rightarrow \\ \sqrt[3]{20} &< n. \end{align} $$ So for any $n> N = \sqrt[3]{20}$ you have the desired conclusion.



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