# Find $N\in\mathbb{N}$ for which $n>N$ makes $\left | a_n-L \right |<\varepsilon$

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

• Hint: when applying the absolute value, the $(-1)^n$ disappears, so can you rewrite what the condition $|a_n - 0| < 0.001$ becomes?
– Basj
Feb 9, 2019 at 11:45
• @Basj $|\frac{(n+4)}{3n{^2}-7} - 0| < 0.001$ ? Feb 9, 2019 at 11:47
• @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. Feb 9, 2019 at 11:54
• @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? Feb 9, 2019 at 12:14
• PS: Just replace n by 337 and 338 in (n+4)/(3*n^2-7) and you'll see which one is good ;)
– Basj
Feb 9, 2019 at 12:35

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 $$N$$.