# How do you apply the ratio test to determine if #Sigma (3^n(n!))/(n^n)# from #n=[1,oo)# is convergent to divergent?

It is divergent.

We can apply d'Alembert's ratio test:

Suppose that;

Then

if L < 1 then the series converges absolutely; if L > 1 then the series is divergent; if L = 1 or the limit fails to exist the test is inconclusive.

So our series is;

So our test limit is:

Hence the series is divergent.

In case you are wondering how the final limit was established; it is a standard limit in disguise:

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See below.

Using Strirling approximation

we have

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To apply the ratio test to the series ( \sum_{n=1}^{\infty} \frac{3^n(n!)}{n^n} ), we compute the limit of the ratio of consecutive terms:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{3^{n+1}((n+1)!)}{(n+1)^{n+1}} \cdot \frac{n^n}{3^n(n!)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{3^{n+1}(n+1)(n!)}{(n+1)^{n+1}} \cdot \frac{n^n}{3^n(n!)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{3(n+1)}{(n+1)^n} \right| ]

[ = \lim_{n \to \infty} \left| \frac{3}{(1+\frac{1}{n})^n} \right| ]

[ = \lim_{n \to \infty} \left| 3 \right| = 3 ]

Since the limit is greater than 1, the series is divergent 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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