# How do you apply the ratio test to determine if #Sigma (n!)/n^n# from #n=[1,oo)# is convergent to divergent?

The series:

is convergent.

Evaluate the ratio:

We can write this expression differently in order to find the limit:

So:

which proves the series to be convergent.

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To apply the ratio test to determine the convergence of the series ( \sum_{n=1}^{\infty} \frac{n!}{n^n} ), you evaluate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where ( a_n = \frac{n!}{n^n} ). If the limit is less than 1, the series converges; if it's greater than 1 or approaches infinity, the series diverges; if the limit equals 1, the test is inconclusive.

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(n+1)!/(n+1)^{n+1}}{n!/n^n} \right| = \lim_{n \to \infty} \left| \frac{(n+1)!}{n!} \times \frac{n^n}{(n+1)^n} \times \frac{(n+1)^n}{(n+1)^{n+1}} \right| ]

[ = \lim_{n \to \infty} \left| \frac{(n+1)}{1} \times \frac{n^n}{(n+1)^n} \times \frac{1}{n+1} \right| = \lim_{n \to \infty} \left| \frac{n^n}{(n+1)^{n-1}} \right| ]

[ = \lim_{n \to \infty} \frac{n^n}{n^{n-1}} = \lim_{n \to \infty} n = \infty ]

Since the limit is infinity, the series diverges by the ratio test.

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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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