<s>[INST] Prove that $\frac{n^2+(-1)^nn+2}{7n^2+3}$ converges to $\frac{1}{7}$ I want to show that $\frac{n^2+(-1)^nn+2}{7n^2+3}$ converges to $\frac{1}{7}$ using the definition of convergence.
Skratch work:
I need $\mid\frac{n^2+(-1)^nn+2}{7n^2+3}-\frac{1}{7}\mid<\epsilon$.
So I take $\frac{n^2+(-1)^nn+2}{7n^2+3}-\frac{1}{7}=\frac{7(-1)^nn+11}{7(7n^2+3)}$.
For $n=1$, $\frac{7(-1)^nn+11}{7(7n^2+3)}=\frac{-4}{70}$
For $n=2$, $\frac{7(-1)^nn+11}{7(7n^2+3)}=\frac{24}{196}$
For $n=4$, $\frac{7(-1)^nn+11}{7(7n^2+3)}=\frac{39}{805}$.
So I think I can safely say that $\frac{7(-1)^nn+11}{7(7n^2+3)}<\frac{1}{n}$.
Now I can start the proof:
Let $\epsilon>0$. And choose $N>\frac{1}{\epsilon}$. Hence, $\frac{1}{N}<\epsilon$. So $n>N \implies \mid\frac{n^2+(-1)^nn+2}{7n^2+3}-\frac{1}{7}\mid < \frac{1}{n}< \frac{1}{N} < \epsilon$.
Does this proof look correct? I think I may have made a mistake in choosing $N$. Please let me know. Thanks
 [/INST] 
you can choose n in better way
you can use sup of a(n) or a bit more to find n>N ,which satisfy that relation
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