3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Rewrite the expression in terms of the natural logarithm and then use Stirling's Approximation. $\endgroup$ – aleden Jan 7 at 18:22
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.