# 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

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?

• Since I am unfamiliar with the topic, the following idea might well fail to solve the problem: Anyway, my first try would be to let $~r~$ be any fixed (but unspecified) element in $~\Bbb{R^+},~$ and try to determine $~\displaystyle \lim_{n \to \infty} \left\{ ~n \times ~\left[ ~log(n) - \log(n-r) ~\right] ~\right\}.$ Commented May 26 at 2:13
• Have you tried the ratio test? Commented May 26 at 6:04

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 $$n \in \mathbb{N}$$.

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

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

what do you think ? please let me know!

I tried to use least machinery possible, e and algebra