1. ## convergent sequences

Hi,I'm having problem determining whether this sequence is convergent or divergent:
an= 3n+2n!/1+n^3

Could someone please show me how to solve this problem. Thankyou very much xx

2. $\displaystyle a_n=\frac{3n+2n!}{1+n^3}$

First the intuition: the sequence $\displaystyle n!,\, n\geq 0$ grows way quicker than any polynomial. Hence the sequence diverges to $\displaystyle +\infty$.

Now for the proof. Because of the intuition, we try to bound $\displaystyle a_n$ from below, keeping the $\displaystyle n!$: for instance, for $\displaystyle n\geq 1$, $\displaystyle a_n\geq\frac{2n!}{2n^3}=\frac{n!}{n^3}$ (since $\displaystyle 1+n^3 \leq 2n^3$).

In a short while, you'll just have to say "$\displaystyle n^3$ is negligible compared to $\displaystyle n!$, hence $\displaystyle \frac{n!}{n^3}\to_n+\infty$." But at that time, you may prefer a proof of this fact (it depends of what you know already).

So let $\displaystyle b_n=\frac{n!}{n^3}$. We prove that $\displaystyle (b_n)_n$ is larger than a divergent geometric sequence. For $\displaystyle n\geq 1$, $\displaystyle \frac{b_{n+1}}{b_n}=\frac{(n+1)!}{(n+1)^3}\frac{n^ 3}{n!}=(n+1)\left(\frac{1}{1+\frac{1}{n}}\right)^3$. Because $\displaystyle \frac{1}{n}\to_n 0$, we have thus $\displaystyle \frac{b_{n+1}}{b_n}\to_n +\infty$. Hence there exists $\displaystyle n_0\geq 1$ such that, for all $\displaystyle n\geq n_0$, $\displaystyle \frac{b_{n+1}}{b_n}\geq 2$, i.e. $\displaystyle b_{n+1}\geq 2 b_n$. Hence, for $\displaystyle n\geq n_0$, $\displaystyle b_n\geq 2b_{n-1}\geq 2^2 b_{n-2}\geq\cdots\geq 2^{n-n_0}b_{n_0}$. Now you know that $\displaystyle 2^n\to_n+\infty$ and $\displaystyle b_n\geq 2^n 2^{-n_0}b_{n_0}$ (and $\displaystyle b_{n_0}\neq 0$), so that $\displaystyle b_n\to_n+\infty$.

3. $\displaystyle a_n=\frac{3n+2n!}{1+n^3}$
For $\displaystyle n\geq 3$,
$\displaystyle \frac{3n+2n!}{n^3+1} \geq \frac{n!}{n^3} \geq \frac{n!}{n(n-1)(n-2)} = (n-3)(n-4)...(2)(1)$

Of course this is divergent.

4. Originally Posted by ThePerfectHacker
For $\displaystyle n\geq 3$,
$\displaystyle \frac{3n+2n!}{n^3+1} \geq \frac{n!}{n^3} \geq \frac{n!}{n(n-1)(n-2)} = (n-3)(n-4)...(2)(1)$
Yes, thank you for this short solution (what I wrote could have aimed at proving $\displaystyle n^k=o_n(n!)$ for general $\displaystyle k$). There is a little mistake though, that I'd correct into:
$\displaystyle \frac{n!}{n^3}=\frac{n(n-1)(n-2)}{n^3}(n-3)!=\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)(n-3)!\geq\frac{2}{3}\frac{1}{3}(n-3)!=\frac{2}{9}(n-3)!$
for $\displaystyle n\geq 3$.