Question

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

Short answer

The series \(\sum_{n=0}^{\infty} e^{-n}\) converges according to the Root Test.

Step-by-step solution

Identify the series terms

The series is given as \(\sum_{n=0}^{\infty} e^{-n}\). Therefore, the sequence \(a_n\) is \(e^{-n}\). Identify the terms of the sequence. Here, \(a_n = e^{-n}\).

Calculate the nth root of the absolute value of the terms

We need to calculate \(L=\lim_{n \to \infty} |a_n|^{\frac{1}{n}}\). Since the terms are already positive, we have \(L=\lim_{n \to \infty} (e^{-n})^{1/n}\). Applying the exponent rules, this simplifies to \(L=\lim_{n \to \infty} e^{-1}\).

Evaluate the limit

Since \(e^{-1}\) is a constant, the limit of the sequence as \(n\) approaches infinity is \(e^{-1}\).

Apply the Root Test

The Root test states that if \(L<1\), the series converges; and if \(L>1\), it diverges. Here, since \(L=e^{-1}\) which is less than 1, the series \(\sum_{n=0}^{\infty} e^{-n}\) converges according to the Root Test.