# Converge Test on the series $\sum \limits_{n=0}^{\infty} \left(\frac{2n+n^3}{3-4n}\right)^n$

I want to show, that $$a:=\sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n$$ is not converging, because $$\lim \limits_{n \to \infty}(a)\neq 0 \; (*)$$. Therefore, the series can't be absolute converge too.

Firstly, I try to simplify the term. After that I want to find the limit.

Unfortunately, I can't seem to find any good equation with that I can clearly show $$(*)$$. \begin{align} \sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n&=\left (\dfrac{\not{n}\cdot (2+n^2)}{\not n \cdot (\frac{3}{n}-4)}\right)^n\\ &=\left(\dfrac{2+n^2}{\frac 3n-4}\right)^n\\ &= \cdots \end{align}

How to go on?

• Do you know the root test? – Clarinetist Oct 29 '18 at 20:03
• You mean that we need to show, that $\lim \limits_{n \to \infty} \sqrt[n]{\mid a_n \mid}<1$? – Doesbaddel Oct 29 '18 at 20:06

Recall that

$$\sum_0^\infty a_n <\infty \implies a_n \to 0$$

therefore if $$a_n \not \to 0$$ the series can’t converge.

In that case for $$n\ge 3$$ we have

$$\left|\dfrac{2n+n^3}{3-4n}\right|=\dfrac{2n+n^3}{4n-3}>\dfrac{n^3}{4n}=\frac{n^2}4\ge2$$

and then

$$|a_n|=\left|\dfrac{2n+n^3}{3-4n}\right|^n\ge 2^n$$

Refer also to the related:

• Therfore it is $\neq 0$ and divergent. Is that enough for a formal proof? – Doesbaddel Oct 29 '18 at 20:46
• @Doesbaddel Absolutely, it is the first test to do for any series! – gimusi Oct 29 '18 at 20:47
• Alright, I was a bit confused, because it looked so simple. – Doesbaddel Oct 29 '18 at 20:49
• @Doesbaddel Here is the formal proof. Recall that it is a necessary condition but not sufficient, that is if $a_n \to 0$ we cannot conlcude that $\sum a_n$ converges but if $a_n \not \to 0$ we can conclude that the series doesn't converges. – gimusi Oct 29 '18 at 20:50
• @Doesbaddel You are welcome! Bye – gimusi Oct 29 '18 at 21:08

I'm a bit confused by your choice of notation for it seems like you write that a series diverges if its limit is not $$0$$ what is not true. So i think you mean that we want to show that

$$\frac{(2n+n^3)^n}{(3-4n)^n}\not\rightarrow 0$$

as $$n\rightarrow \infty$$ for this implies that the series does not converge. Note also that

$$\left(\frac{2+n^2}{3/n-4}\right)^n\neq \frac{2^n+n^{2n}}{3^n\cdot (1/n)^n-4^n}$$

$$\frac{2n+n^3}{3-4n}\rightarrow ?$$

as $$n\rightarrow \infty$$?

• Well $\left|\frac{2n+n^3}{3-4n}\right| = \frac{2n+n^3}{4n-3}$ if $n\geq 1$. Now if $p>1$ then which of the following holds: $p<p^n$ or $p\geq p^n$ – Olof Rubin Oct 29 '18 at 20:15
• If the series converges to $0$ then it converges and hence does not diverge – Olof Rubin Oct 29 '18 at 20:18
• Therefore $\left|\frac{2n+n^3}{3-4n}\right|^n$ satisfies...? – Olof Rubin Oct 29 '18 at 20:25
• I was thinking $\left|\frac{2n+n^3}{3-4n}\right|^3>\left|\frac{2n+n^3}{3-4m}\right|$ – Olof Rubin Oct 29 '18 at 20:42
• yes and since the left-hand side is unbounded so must the right-hand side be – Olof Rubin Oct 29 '18 at 20:44

$$\sum \limits_{n=0}^{\infty} \left(\underbrace{\dfrac{2n+n^3}{3-4n}}_{a_n}\right)^n$$

Root test: Prove:$$\lim \limits_{n \to \infty} \sqrt[n]{\mid a_n \mid}<1 \implies a_n \text{ is absolute converging.}$$

\begin{align} &\lim \limits_{n \to \infty} \sqrt[n]{\left( \dfrac{2n+n^3}{3-4n}\right)^n} \\ =&\lim \limits_{n \to \infty}=\frac{2n+n^3}{3-4n}\\ =&\lim \limits_{n \to \infty}=\dfrac{n(n^2+2)}{n(\frac 3n-4)}\\ =&\lim \limits_{n \to \infty}=\dfrac{\overbrace{n^2+2}^{\infty}}{\underbrace{\frac 3n-4}_{0}}\\ =&\lim \limits_{n \to \infty}(a_n)=\infty\\& \implies a_n \text{ and } \sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n \text{ are not absolute convergent.} \end{align}

I just noticed, that this would only prove absolute convergence and not convergence.

Let $$n \ge 4$$.

$$|a_n|:=\left |\dfrac{n^3+2n}{4n-3}\right |^n \gt$$

$$\left (\dfrac{n^3}{4n}\right )^n \ge n^n.$$

$$\lim_{n \rightarrow \infty} |a_n| \not = 0$$.

Hence the series does not converge.