limits of sequences exponential and factorial: $a_n=e^{5\cos((\pi/6)^n)}$ and $a_n=\frac{n!}{n^n}$

Question

Compute the limit $\lim\limits_{n\to\infty}a_n$ for the following sequences:

(a) $a_n=e^{5\cos((\pi/6)^n)}$

(b) $a_n=\frac{n!}{n^n}$

For part (a) do I just take the limit of the exponent part and then the answer would be $e$ raised to whatever the limit is?

And would the limit be $1$ or $-1$? because $\cos$ goes between those two.

For part $b$ it is in the form of infinity over infinity but how do you take the derivative of $n!$? Will it ever break out of infinity over infinity?

Answer 1

(a) $$ \lim_{n \to \infty} e^{5 \cos((\frac{\pi}6))^n} = e^{\lim_{n \to \infty} 5 \cos((\frac{\pi}6))^n} = e^{5 \cos(\lim_{n \to \infty}(\frac{\pi}6)^n)} = e^{5 \cdot 1} = e^5 $$

(b) $$ \frac{n!}{n^n} = \frac{1\cdot 2 \cdots n}{n\cdot n \cdot n} \leq \frac{1}{n} $$ From here $$ \lim_{n \to \infty} \frac{n!}{n^n} = 0 $$ since $$ \lim_{n \to \infty} \frac{1}{n} = 0 $$

Answer 2

Note that $$\lim_{n\to\infty}\left(\frac{\pi}{6}\right)^n=0.$$ Since $\cos x$ is continuous, it follows that $$\lim_{n\to\infty}\cos\left(\left(\frac{\pi}{6}\right)^n\right)=\cos(0)=1.$$ Since the function $e^x$ is continuous, it follows that the required limit is $e^5$.

For the second problem, please see the answer by UrošSlovenija.

Answer 3

For the second one (I've done this before btw), $n! = \prod_{i=1}^n i =\prod_{i=1}^n (n+1-i) $ so

$\begin{align} n!^2 &= \prod_{i=1}^n i(n+1-i)\\ &= \prod_{i=1}^n (i(n+1)-i^2)\\ &= \prod_{i=1}^n ((n+1)/2)^2-(n+1)/2)^2+ i(n+1)-i^2)\\ &= \prod_{i=1}^n ((n+1)/2)^2-((n+1)/2-i)^2)\\ &\le \prod_{i=1}^n ((n+1)/2)^2)\\ &= ((n+1)/2)^{2n} \end{align} $

so $n! < (\frac{n+1}{2})^{n} $ and $\frac{n!}{n^n} \le (\frac{n+1}{2n})^{n} $ and this $\to 0$ as $n \to \infty$.