Answer:
Absolutely convergent
Explanation:
∑ₙ₌₁°° n! / (nⁿ)
Using ratio test:
L = lim(n→∞)│aₙ₊₁ / aₙ│
L = lim(n→∞)│[(n+1)! / (n+1)ⁿ⁺¹] / [n! / (nⁿ)]│
Change division to multiplication of the reciprocal:
L = lim(n→∞)│[(n+1)! / (n+1)ⁿ⁺¹] × [nⁿ / n!]│
Rearrange the numerators and denominators:
L = lim(n→∞)│[(n+1)! / n!] × [nⁿ / (n+1)ⁿ⁺¹]│
The next step is to simplify the factorials. Remember:
n! = n × n−1 × n−2 × n−3 ... × 3 × 2 × 1
(n+1)! = n+1 × n × n−1 × n−2 × n−3 ... × 3 × 2 × 1
(n+1)! = (n+1) n!
L = lim(n→∞)│[(n+1) n! / n!] × [nⁿ / (n+1)ⁿ⁺¹]│
L = lim(n→∞)│(n+1) × [nⁿ / (n+1)ⁿ⁺¹]│
L = lim(n→∞)│nⁿ / (n+1)ⁿ│
L = lim(n→∞)│(n / (n+1))ⁿ│
L = lim(n→∞)│(n+1) / n)⁻ⁿ│
L = lim(n→∞)│(1 + 1/n)⁻ⁿ│
To evaluate this, we need to use natural log.
ln L = lim(n→∞) ln│(1 + 1/n)⁻ⁿ│
ln L = lim(n→∞) -n ln│1 + 1/n│
ln L = lim(n→∞) ln│1 + 1/n│/ (-1/n)
Apply L'Hopital's rule:
ln L = lim(n→∞) 1 / (1 + 1/n) (-1/n²) / (1/n²)
ln L = lim(n→∞) -1 / (1 + 1/n)
ln L = -1
L = e⁻¹
L < 1
The limit is less than 1, so the series converges.
Since the series is always positive, the absolute value of the series is the same as the series. Therefore, it also converges. So the series is absolutely convergent.