how do I find all $k \in \mathbb{R} $ such that $\lim_{n \to \infty} a_n = +\infty$ for a given sequence without using approximations

Question

I have a sequence $(a_n)_{n \in \mathbb{N}}$ of positive terms defined by the recurrence relation

$ \frac{a_{n+1}}{a_n} = \left(\frac{2n}{2n + k + 4}\right)^{2n}, $

where $k \in \mathbb{R}$ and $k \neq -4$. I need to determine all values of $k$ for which $\lim_{n \to \infty} a_n = +\infty$.

I would like to avoid using Taylor series approximations, binomial approximations.

Ratio-test, cauchy's root-test aswell as euler limit and subsequences are the tools that have been taught, how should I proceed with this problem?

Answer 1

The sequence $(1 +\frac{1}{n})^n$ is strictly increasing and tends to $e$. The sequence $(1 -\frac{1}{n})^n$ is strictly decreasing and tends to $e^{-1}$. With these two facts we can answer the question. I'll develop the argument for $k < -4$ only. The other case is similar.

If $k < -4$, setting $a = -k -4 > 0$, we have $$ \left(\frac{2n}{2n +k +4}\right)^{2n} = \left(\frac{2n +k +4 -(k +4)}{2n +k +4}\right)^{2n} = \left(1 +\frac{a}{2n -a}\right)^{2n} = \left( \left(1 +\frac{a}{2n -a}\right)^{\frac{2n -a}{a}} \right)^{\frac{2an}{2n -a}} \longrightarrow e^a. $$ Hence there exists an integer $N$ such that for every $n \geq N$ we have $$ \frac{a_{n +1}}{a_n} = \left(\frac{2n}{2n +k +4}\right)^{2n} > e^a -\epsilon > 1. $$ We conclude that $a_n$ goes to infinity with $n$.

Remark. I assume that $a_0 > 0$ and that $k$ is such that $2n +k +4$ is never zero when $n \in \mathbb{N}$.

Answer 2

$\frac{a_{n+1}}{a_n} = \left( \frac{2n}{2n+ k + 4}\right)^{2n} \\ = \left( 1 + \frac{2n}{2n+ k + 4 }- 1 \right)^{2n} \\ = \left(1 + \frac{2n - \left( 2n + k + 4\right)}{ 2n + k + 4} \right)^{2n} \\ =\left(1 + \frac{-k -4}{ 2n + k + 4} \right)^{2n} = \left(1 + \frac{1}{\frac{ 2n + k + 4}{-k -4}} \right)^{2n} \\ = \left(1 + \frac{1}{\frac{ 2n + k + 4}{-k -4}} \right)^{\frac{ 2n + k + 4}{-k -4} \cdot \frac{-k -4}{ 2n + k + 4} \cdot 2n} \\ \\ \text{check this out:}\\ \\ \frac{-k-4}{2n + k + 4} \cdot 2n = \frac{2n\left(-k-4\right)}{2n\left(1 + \frac{k}{2n} + \frac{2}{n}\right)} \\ \lim_{n \to \infty} \frac{-\left(k+4\right)}{1 + \frac{k}{2n} + \frac{2}{n}} = -k-4 \\ \lim_{n \to \infty} \left(1 + \frac{1}{\frac{ 2n + k + 4}{-k -4}} \right)^{\frac{ 2n + k + 4}{-k -4} \cdot \frac{-k -4}{ 2n + k + 4} \cdot 2n} = e^{-k-4}$

$\implies e^{-k-4} > 1 \\ \implies e^{-k-4} > e^0 \\ \implies -k-4 > 0 \\ \implies -k > 4 \\ \implies k < -4 \\ k \in (-\infty, -4)$

what do you think ? please let me know!

I tried to use least machinery possible, e and algebra