Using the ratio test, the series ∑ n! / n^n converges because the limit of the ratio a_{n+1}/a_n equals which value?

Using the ratio test, the series ∑ n! / n^n converges because the limit of the ratio a_{n+1}/a_n equals which value?

• A. e
• B. 1/e
• C. 0
• D. ∞

Answer: (B)

Using the ratio test, look at a_n = n!/n^n. The ratio a_{n+1}/a_n equals [(n+1)!/(n+1)^{n+1}] / [n!/n^n] = (n^n)/(n+1)^n = [n/(n+1)]^n. As n grows, this approaches e^{-1}, since (n/(n+1))^n = (1 - 1/(n+1))^n → 1/e. The ratio test says if this limit L is less than 1, the series converges. Here L = 1/e < 1, so the series converges.

Using the ratio test, look at a_n = n!/n^n. The ratio a_{n+1}/a_n equals [(n+1)!/(n+1)^{n+1}] / [n!/n^n] = (n^n)/(n+1)^n = [n/(n+1)]^n. As n grows, this approaches e^{-1}, since (n/(n+1))^n = (1 - 1/(n+1))^n → 1/e. The ratio test says if this limit L is less than 1, the series converges. Here L = 1/e < 1, so the series converges.