# 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.

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$$\lim _{n\to \infty}(1+\frac{1}{n})^n=e$$ – Suraj M S Dec 9 '13 at 19:36
Yes, it is in this case, as your expression's denominator converges to a positive limit. – DonAntonio Dec 9 '13 at 19:37

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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Hint: For the original problem, use the Ratio Test. After a little manipulation, you will find that if $a_k$ denotes the $k$-th term, then $$\frac{a_{n+1}}{a_n}=\frac{3}{\left(1+\frac{1}{n}\right)^n}.$$ The denominator has limit $e\lt 3$.

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