# Prove that the sequence $\left\{\frac{4n^2}{2n^3-5}\right\}$ converges to $0$.

Prove that the sequence $$\left\{\frac{4n^2}{2n^3-5}\right\}$$ converges to $0$.

The first thing I would like to do is further bound the sequence so I can get rid of the $-5$ in the bottom of the fraction. How can I do this? For negatives it's tricky since you want to make it smaller, but removing a $-5$ actually makes the denominator bigger, and not smaller.

If this cannot be done, I must do the following:

$$\left\{\frac{(4/n)}{2-5/n^3}\right\}$$ and find a bound for $2-5/n^3$, separately. If I assume $n\geq 10$, then: $$\frac{1}{n}<\frac{1}{10}\color{red}{\to} \frac{5}{n}<\frac{1}{2} \color{red}{\to} \frac{-5}{n}>\frac{-1}{2}\color{red}{\to} 2-\frac{5}{n}>\frac{3}{2}\color{red}{\to} \frac{1}{|2-5/n^3|}<\frac{2}{3}$$

Then I have that:

$$\left|\frac{(4/n)}{2-5/n^3}\right|<\frac{2}{3}\cdot\frac{4}{n}<\frac{8}{3N}<\frac{8}{3(8/3\epsilon)}=\epsilon$$

Pick $N=\max(10,8/3\epsilon)$

This excess bounding could have been avoided if I can somehow bound the fraction even more?

For $n\ge 2$, we have for all $\epsilon>0$
$$\frac{4n^2}{2n^3-5}\le \frac{4n^2}{n^3}=\frac4n<\epsilon$$
whenever $n>N=\max\left(2,\frac{4}{\epsilon}\right)$
• How did you know to pick $n\geq 2$? – K Split X Apr 17 '17 at 16:40
• $2n^3-5\ge n^3\implies n^3\ge 5$. This is true for values of $n\ge 2$. – Mark Viola Apr 17 '17 at 16:43
Two-stage bounding seems unavoidable. To simplify the works, I would do something like the following. If $n \geq 3$, then $$\frac{4n^{2}}{2n^{3}-5} < \frac{4n^{2}}{2n^{3}-n^{2}} = \frac{4}{2n-1}.$$ Given any $\varepsilon > 0$, we have $\frac{4}{2n-1} < \varepsilon$ if in addition $n > (\frac{4}{\varepsilon+1})/2$. So taking $N := \max \{ 3, (\frac{4}{\varepsilon+1})/2 \}$ suffices.