I have two separate sequences that I using this approach on. $a_n=\frac{n^3-2n^2+3}{2n^3+7n}$ & $a_n=\frac{n^3}{2^n}$

Proof 1

$a_n=\frac{n^3-2n^2+3}{2n^3+7n}\to \frac{1}{2}$

$$\text{Let } \epsilon \gt 0 \,\, \exists N \;\;\forall n\geq N \; \left\lvert {a_n-\frac{1}{2}}\right\rvert \lt \epsilon$$ $$\left\lvert {\frac{n^3-2n^2+3}{2n^3+7n}-\frac{1}{2}}\right\rvert=\left\lvert {\frac{(2n^3-4n^2+6)-(2n^3+7n)}{4n^3+14n}}\right\rvert=\left\lvert {\frac{-4n^2-7n+6}{4n^3+14n}}\right\rvert \lt \frac{1}{2n} \lt \epsilon$$ $$\frac{1}{2\epsilon} \lt n$$ I am having difficulty in showing exactly why, $\left\lvert {\frac{-4n^2-7n+6}{4n^3+14n}}\right\rvert \lt \frac{1}{2n}$

Proof 2

$a_n=\frac{n^3}{2^n} \to 0$ $$\text{Let } \epsilon \gt 0 \,\, \exists N \;\;\forall n\geq N \; \left\lvert {a_n-0}\right\rvert \lt \epsilon$$ $$\left\lvert {\frac{n^3}{2^n}}\right\rvert=\frac{n^3}{2^n} \lt \frac{?}{?} \lt \epsilon$$ I am lost as to where to proceed on this one. Any input would be greatly appreciated . Thanks

  • $\begingroup$ The numerator in part 1 should be $-4n^2-1$. $\endgroup$ Mar 29 '19 at 23:27
  • $\begingroup$ Just edited the sequence. I had copied it wrong here Thank you for catching that $\endgroup$
    – MAC
    Mar 29 '19 at 23:34
  • $\begingroup$ I wasn't talking about the sequence, but about the algebra mistake you made in the numerator in the first exercise. $\endgroup$ Mar 30 '19 at 0:07
  • $\begingroup$ @JohnWaylandBales the algebra mistake is result of the sequence mistake. $\endgroup$
    – MAC
    Mar 30 '19 at 2:21

On the first proof:

It would be simplest to show that $\left|\dfrac{4n^2+1}{4n^3+14n}\right|<\dfrac{1}{n}$ since, clearly $4n^3+n<4n^3+14n$.

On the second proof:

For $n\ge16$ we have that $n^4\le2^n$ from which it follows that $\dfrac{n^3}{2^n}\le\dfrac{1}{n}$.

  • $\begingroup$ Why would we be able to show that (4n^2+1)/(4n^3+14n) <1/n? $\endgroup$
    – MAC
    Mar 30 '19 at 2:24
  • $\begingroup$ Divide both side of the inequality $4n^3+n<4n^3+14n$ by $n(4n^3+14n)$. $\endgroup$ Mar 30 '19 at 5:01

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