# How do you show the convergence of the series #(n!)/(n^n)# from n=1 to infinity??

The series converges

and thus the series converges according to the ratio test.

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To show the convergence of the series ( \frac{n!}{n^n} ) from ( n = 1 ) to infinity, we can use the ratio test. The ratio test states that if ( \lim_{n \to \infty} \frac{a_{n+1}}{a_n} < 1 ), then the series converges.

For this series, ( a_n = \frac{n!}{n^n} ). To apply the ratio test:

[ \begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \frac{(n+1)!/(n+1)^{n+1}}{n!/n^n} \ &= \lim_{n \to \infty} \frac{(n+1)!}{n!} \cdot \frac{n^n}{(n+1)^{n+1}} \ &= \lim_{n \to \infty} \frac{n+1}{(n+1)^{n+1}/n^n} \ &= \lim_{n \to \infty} \frac{n+1}{(1 + \frac{1}{n})^{n+1}} \ \end{align*} ]

Using the fact that ( \lim_{n \to \infty} (1 + \frac{1}{n})^n = e ), we get:

[ \begin{align*} \lim_{n \to \infty} \frac{n+1}{(1 + \frac{1}{n})^{n+1}} &= \frac{\lim_{n \to \infty} (n+1)}{\lim_{n \to \infty} (1 + \frac{1}{n})^{n+1}} \ &= \frac{\lim_{n \to \infty} (n+1)}{e} \ &= \frac{\infty}{e} \ &= \infty \end{align*} ]

Since the limit is greater than 1, by the ratio test, the series diverges. Therefore, the series ( \frac{n!}{n^n} ) from ( n = 1 ) to infinity diverges.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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