Proving A Rational Sequence Converges

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Problem 75: (Proving A Rational Sequence Converges)

Consider the sequence $\ds (s_n)=\left(\frac{4n^3-2n^2-7n}{5n^3-3n^2+2n-1}\right)$. In this problem, your job is to prove that $s_n\to 4/5$. You may assume that for all natural numbers, we have $5n^3-3n^2+2n-1>0$.

1. Show that $\ds(s_n-4/5)= \left(\frac{2 n^2-43 n+4}{5 \left(5 n^3-3 n^2+2 n-1\right)}\right).$
2. Find a $k_1>0$ and $M_1$ so that $|2 n^2-43 n+4|\leq k_1 n^2$ for all natural numbers $n>M_1$.
3. Find a $k_2>0$ and $M_2$ so that $|5 \left(5 n^3-3 n^2+2 n-1\right)|\geq k_2 n^3$ for all natural numbers $n>M_2$.
4. Prove that $(s_n)$ converges to $4/5$.

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