Question

Show that ∑ₙ=1[infinity]2^(n²)/n ! diverges. Recall from the Laws of Exponents that 2^(n²)=(2ⁿ)ⁿ. Find the correct limit for the Ratio test, where

aₙ = 2(n²) / n!

Answer 1

To show that the series diverges, we can use the Ratio Test. By simplifying the limit of the ratio of consecutive terms, we find that the limit is 2, which is greater than 1, indicating that the series diverges.

Step-by-step explanation:

To show that the series ∑ₙ=1[infinity]2^(n²)/n ! diverges, we can use the Ratio Test. Let's consider the limit of the ratio of consecutive terms aₙ = 2(n²) / n!:

limₙ→∞ |(aₙ₊₁) / aₙ| = limₙ→∞ |(2(n+1)² / (n+1)!) / (2(n²) / n!)|

We can simplify this expression:

= limₙ→∞ |2(n+1)² / (n+1)(n!)| * |n! / 2(n²)|

= limₙ→∞ 2(n+1)² / (n+1)(n)

= limₙ→∞ 2(n+1) / n

Since the limit as n approaches infinity is 2, which is greater than 1, the series diverges.