# Does this cubic sequence converge or diverge?

Prove that the sequence $$a_n = 8n^3 + n^2 - 2$$ either converges or diverges. If it converges, find the value it converges to.

What I have so far:

Since the $$\lim_{ n\rightarrow \infty} a_n= \infty$$ the sequence diverges to infinity. However, I am having trouble proving this with the formal definition. Any guidance/helpful tips?

• – dxiv Mar 28 '18 at 17:08
• "What I have so far"? You have a sentence saying what your conclusion ought to be. That is not a starting point and it assumes the consequence of what youre trying to prove. – CogitoErgoCogitoSum Mar 28 '18 at 17:19

For any $M\in\mathbb{R}$, let us choose $N\in\mathbb{N}$ such that $N > \max\{2,M\}$, which you can find because of the Archimedean principle.

For $a_n = 8n^3 + n^2 - 2$, note that $\forall n\in\mathbb{N}$ where $n\ge N$ we have $a_n = 8n^3 + n^2 - 2 > 8n^3 > n \ge N > M$.

Thus, for any $M$ we choose, we can always find an $N$ such that $\forall n\ge N$, we observe $a_n> M$. This precisely defines the statement $\lim\limits_{n\to\infty}a_n=+\infty$.

QED

Note that since $n^2-2>0$ for $n>1$ we have $$8n^3<8n^3+n^2-2$$ for any $n\in \mathbb{N}$. Fix $M\in \mathbb{N}$, then taking $$n\geq \left\lceil\left(\lceil M/8\rceil\right)^{1/3}\right\rceil$$ insures $$8n^3\geq M\implies 8n^3+n^2-2\geq M$$ since our choice of $M$ was arbitrary, we can make our sequence as large as we want, and $$\lim_{n\to \infty} 8n^3+n^2-2=+\infty$$

• A nit in the first line $1^2-2 \not \gt 0$, but we are interested in large $n$ – Ross Millikan Mar 28 '18 at 20:25
• @RossMillikan you are right of course, amending – qbert Mar 28 '18 at 20:35

$$a_n=8n^3+n^2-2\\a_m=8m^3+m^2-2\\ a_n-a_m=8(n^3-m^3)+n^2-m^2$$ Now let's choose $\varepsilon = 1$, for all natural number $M$ I can choose $n=M+2,m=M+1$ so we get $a_n-a_m=24M^2+78M+59>M\ge 1$, hence it is not Cauchy hence does not converges. To show it disverges to infinity you just left to show that $a_n>a_{n-1}$