How do you decide if this is absolutely convergent, conditionally convergent, or divergent using the ratio test? power series of n=1 to infinity N!/N^N and please thoroughly go over the simplification for the factorials. Thanks!

Absolutely convergent

Step-by-step 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.

step-by-step explanation: it is d