1. The problem asks to evaluate the infinite series $$\sum_{n=1}^\infty \frac{n!}{n^{n-1}}$$ and check its convergence. 2. The general term of the series is $$a_n = \frac{n!}{n^{n-1}}$$. 3. To check convergence, we can use the Ratio Test, which states that a series $$\sum a_n$$ converges if $$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$$. 4. Compute the ratio: $$\frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)!}{(n+1)^n}}{\frac{n!}{n^{n-1}}} = \frac{(n+1)!}{(n+1)^n} \cdot \frac{n^{n-1}}{n!} = \frac{(n+1) \cdot n^{n-1}}{(n+1)^n} = \frac{n^{n-1}}{(n+1)^{n-1}} = \left(\frac{n}{n+1}\right)^{n-1}$$ 5. Evaluate the limit: $$\lim_{n \to \infty} \left(\frac{n}{n+1}\right)^{n-1} = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right)^{n-1} = e^{-1} \approx 0.3679 < 1$$ 6. Since the limit is less than 1, the series converges by the Ratio Test. 7. To approximate the sum, calculate the first few terms: - For $n=1$: $\frac{1!}{1^{0}} = 1$ - For $n=2$: $\frac{2!}{2^{1}} = \frac{2}{2} = 1$ - For $n=3$: $\frac{6}{3^{2}} = \frac{6}{9} = \frac{2}{3} \approx 0.6667$ - For $n=4$: $\frac{24}{4^{3}} = \frac{24}{64} = 0.375$ - For $n=5$: $\frac{120}{5^{4}} = \frac{120}{625} = 0.192$ Adding these partial sums: $1 + 1 + 0.6667 + 0.375 + 0.192 = 3.2337$ approximately. 8. The terms decrease rapidly, so the series converges to a value near 3.3. **Final answer:** The series $$\sum_{n=1}^\infty \frac{n!}{n^{n-1}}$$ converges, and its approximate sum is about 3.3.