To determine whether the series

$\sum_{n=1}^\infty \frac{n^n}{n!}$

converges or diverges, we analyze its behavior.

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Step 1: Apply the Ratio Test

The Ratio Test states that for a series $\sum a_n$, consider the limit: $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.$

• If $L < 1$, the series converges absolutely.
• If $L > 1$, the series diverges.
• If $L = 1$, the test is inconclusive.

Here, $a_n = \frac{n^n}{n!}$. Compute $\frac{a_{n+1}}{a_n}$: $\frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)^{n+1}}{(n+1)!}}{\frac{n^n}{n!}} = \frac{(n+1)^{n+1} n!}{n^n (n+1)!}.$

Simplify: $\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{(n+1) n^n}.$

Break it into parts: $\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{n^n (n+1)} = \frac{(n+1)^n (n+1)}{n^n (n+1)} = \left(\frac{n+1}{n}\right)^n.$

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Step 2: Analyze $\left(\frac{n+1}{n}\right)^n$

Simplify $\frac{n+1}{n} = 1 + \frac{1}{n}$. As $n \to \infty$, this becomes: $\left(1 + \frac{1}{n}\right)^n \to e,$ where $e$ is the base of the natural logarithm.

Thus: $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = e.$

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Step 3: Apply the Ratio Test Result

Since the limit $L = e > 1$, the series diverges.

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Final Answer:

$\sum_{n=1}^\infty \frac{n^n}{n!} \quad \text{diverges}.$

Would you like a detailed explanation of any step or related examples?

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Practice Questions:

1. Determine if $\sum_{n=1}^\infty \frac{n!}{n^n}$ converges or diverges.
2. Analyze the series (\sum_{n=1}^\infty \frac{n