Problem 49 - Essential Calculus

Question

Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} e^{-n^{2}} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=0}^{\infty} e^{-n^{2}} \) converges by the Direct Comparison Test.

Step 1: Understand the Direct Comparison Test

The Direct Comparison Test (or Comparison Test) is a method used to test for the convergence of an infinite series. It states that if 0 ≤ aₙ ≤ bₙ for all n, and the series of bₙ converges, then the series of aₙ also converges. If the series of bₙ diverges, the series of aₙ does too. In this case, a suitable bₙ to compare with the given series needs to be selected.

Step 2: Choose an Appropriate Series for Comparison

For the series \( \sum_{n=0}^{\infty} e^{-n^{2}} \), a logical choice would be the series \( \sum_{n=0}^{\infty} e^{-n} \). The reason being \( e^{-n^{2}} \) is always less than or equal to \( e^{-n} \) for all \( n \geq 0 \) since when you square a nonnegative number, you only increase or keep it the same (for zero), and the exponential function is decreasing for negative inputs, so \( e^{-n^{2}} \leq e^{-n} \) holds.

Step 3: Apply the Direct Comparison Test

It's known from the properties of geometric series that the series \( \sum_{n=0}^{\infty} e^{-n} \) converges because it's a geometric series with \( -1 < r = e^{-1}<1 \). Therefore, by the Direct Comparison Test, since the terms of the series \( \sum_{n=0}^{\infty} e^{-n^{2}} \) are nonnegative and less than or equal to the corresponding terms of a convergent series, the given series \( \sum_{n=0}^{\infty} e^{-n^{2}} \) also converges.