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

Question

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?

Answer 1

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

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

see https://en.wikipedia.org/wiki/Freshman%27s_dream

Instead what can you say about

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

as $n\rightarrow \infty$?

Answer 2

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:

• Does the series $\sum_{n=1}^\infty (-1)^n\ln(n)$ converge or diverge?

Answer 3

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

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.

Answer 4

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.