# Evaluate the limit $\lim_{n\to\infty}\log_a\left(\frac{4^nn!}{n^n}\right)$

Evaluate the limit: $$\lim_{n\to\infty}\log_a\left(\frac{4^nn!}{n^n}\right)\\ a>0\\ a \ne 1$$

I've started with defining another sequence. Let: $$y_n = a^{x_n} = \frac{4^nn!}{n^n}$$

Consider the fraction: $$\frac{y_{n+1}}{y_n} = \frac{4^{n+1}(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{4^nn!}\\ = \frac{4n^n}{(n+1)^n}$$

Consider the limit: \begin{align} \lim_{n\to\infty}\frac{y_{n+1}}{y_n} &= \lim_{n\to\infty}\frac{4n^n}{(n+1)^n} \\ &= \lim_{n\to\infty}4\left(\frac{n}{n+1} \right)^n \\ &= {4\over e} > 1 \end{align}

So by this $$y_n$$ is divergent. Which means: $$\lim_{n\to\infty}y_n = \infty$$ Now I'm having difficulties translating it in a backward direction. We have that: $$\lim_{n\to\infty}y_n = \lim_{n\to\infty}a^{x_n} = \infty$$ Or: $$\log_a \lim_{n\to\infty}a^{x_n} = \log_a(\infty)$$

The answer suggests that: $$\lim_{n\to\infty}x_n = \begin{cases} +\infty,\ a > 1\\ -\infty,\ 0 < a < 1 \end{cases}$$

And I don't see where this appears when going backward from $$a^{x_n}$$ to $$x_n$$. Could you please explain that to me?

• Rewrite the expression in terms of the natural logarithm and then use Stirling's Approximation. – aleden Jan 7 at 18:22

Hint: Note that $$\log_a{x}=\frac{\ln(x)}{\ln(a)}$$