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For $a_n = \frac{(-1){^n}(n+4)}{3n{^2}-7}$ and $\varepsilon = 0.001$

I know that $L = 0$, but how do I do the math to find $N$? thanks

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  • $\begingroup$ Hint: when applying the absolute value, the $(-1)^n$ disappears, so can you rewrite what the condition $|a_n - 0| < 0.001$ becomes? $\endgroup$
    – Basj
    Feb 9, 2019 at 11:45
  • $\begingroup$ @Basj $|\frac{(n+4)}{3n{^2}-7} - 0| < 0.001$ ? $\endgroup$
    – trizz
    Feb 9, 2019 at 11:47
  • $\begingroup$ @Basj I got many of these to practice, I'd like to get a full insight on this one so I could work on the rest. $\endgroup$
    – trizz
    Feb 9, 2019 at 11:54
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    $\begingroup$ @Basj I did find the roots, I get $n=337.287$ and $n=-3.95$ So obviously the negative root isnt valid, but since $n\in\mathbb{N}$ do I put N as 337? or 338? $\endgroup$
    – trizz
    Feb 9, 2019 at 12:14
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    $\begingroup$ PS: Just replace n by 337 and 338 in (n+4)/(3*n^2-7) and you'll see which one is good ;) $\endgroup$
    – Basj
    Feb 9, 2019 at 12:35

1 Answer 1

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If you are not demanding least such $N $ then you can use estimate $ \frac{n+4}{3 n^{2} -7} \leq \frac{n+4}{3 n^{2} -48 } = \frac{1}{3n -4} $ (you can assume n is sufficiently large) now apply $\epsilon $ on last expression to get required $N$.

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