<s>[INST] Is $\lim_{n \to \infty}{\left(\frac{3}{1+\frac{1}{n}}\right)^n} = \lim_{n \to \infty}{3^n}$? I started out with the following series:
$\sum_{n=1}^{\infty}{\left(\frac{3}{n}\right)^nn!}$
The goal is to find out if the sequence converges or diverges. It went quite well, hopefully, until I got stuck here:
$\lim_{n \to \infty}{\left(\frac{3}{1+\frac{1}{n}}\right)^n} = \lim_{n \to \infty}{3^n}$
The problem is that I cannot figure out if this is a valid equality. Based on the property of multiplication of limits, i would assume it is, but the fact that it is raised to $n$ makes me wonder. 
 [/INST] Using the Ratio Test (expounding):
$$\frac{a_{n+1}}{a_n}= \frac{3^{n+1} \cdot (n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{3^n \cdot n!} = \frac{3^n \cdot 3 \cdot(n+1)n! \cdot n^n}{(n+1)^n \cdot (n+1)^1 \cdot 3^n \cdot n!} = \frac{3 \cdot n^n}{(n+1)^n} = 3 \cdot \left(\frac{n}{n+1} \right)^n$$ $$= 3 \cdot \left(\frac{1}{1+1/n} \right)^n = \frac{3}{(1+1/n)^n}$$ 
$$\lim_{n\to\infty} \frac{3}{(1+1/n)^n} = \frac{3}{e} > 1$$
Since the $L > 1 \implies$ the series diverges.
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