To determine whether the series $\sum_{n=7}^{\infty} \frac{n!}{n^n}$ is convergent, we can apply the Ratio Test.

The Ratio Test states that for a series $\sum a_n$, if we compute $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$ and find that:

1. $L < 1$, then the series converges absolutely.
2. $L > 1$, then the series diverges.
3. $L = 1$, the test is inconclusive.

Applying the Ratio Test

Let $a_n = \frac{n!}{n^n}$.

1. Calculate $a_{n+1}$: $a_{n+1} = \frac{(n+1)!}{(n+1)^{n+1}}$

2. Set up $\frac{a_{n+1}}{a_n}$: $\frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} = \frac{(n+1)! \cdot n^n}{n! \cdot (n+1)^{n+1}}$

3. Simplify $\frac{a_{n+1}}{a_n}$: Since $(n+1)! = (n+1) \cdot n!$, we have: $\frac{a_{n+1}}{a_n} = \frac{(n+1) \cdot n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n}$

4. Take the limit as $n \to \infty$: $L = \lim_{n \to \infty} \frac{n^n}{(n+1)^n} = \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^n$

5. Evaluate the limit: Note that $\frac{n}{n+1} = 1 - \frac{1}{n+1}$, so $L = \lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right)^n = \lim_{n \to \infty} e^{-\frac{n}{n+1}} = e^{-1} = \frac{1}{e}$

Since $L = \frac{1}{e} < 1$, the series converges by the Ratio Test.

Answer

• Behavior: Convergent
• Primary Reason: Ratio Test
• Secondary Reason: None needed

Would you like further details or have any questions?

Follow-Up Questions:

1. How is the Ratio Test generally applied to series with factorial terms?
2. Why does $\left( \frac{n}{n+1} \right)^n$ tend to $e^{-1}$ as $n \to \infty$?
3. What are the implications of the Ratio Test result if $L = 1$?
4. Can we use other convergence tests for this series, like the Root Test?
5. How would the result change if the series had a different growth rate in the denominator?

Tip: When using the Ratio Test, factorials in the terms often simplify the limit calculations due to their rapid growth.