Problem 64 - Essential Calculus

Question

Test for convergence or divergence, using each test at least once. Identify which test was used. (a) \(n\) th-Term Test (b) Geometric Series Test (c) \(p\) -Series Test (d) Telescoping Series Test (e) Integral Test (f) Direct Comparison Test (g) Limit Comparison Test $$ \sum_{n=0}^{\infty} 5\left(-\frac{1}{5}\right)^{n} $$

Step-by-Step Solution

Answer

The given series is a geometric series that satisfies the conditions for convergence. The sum of the series is \(4\)

Step 1: Identify Type of Series

Given series \(\sum_{n=0}^{\infty} 5\left(-\frac{1}{5}\right)^{n}\) is a Geometric series. A geometric series takes the form \(\sum ar^{n}\), where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.

Step 2: Apply Geometric Series Test

The given series has the first term \(a = 5\) and common ratio \(r = -\frac{1}{5}\). A geometric series \(\sum ar^{n}\) converges if \(-1 < r < 1\). For this series, \(-1 < -\frac{1}{5} < 1\), so the Geometric Series Test can be assumed.

Step 3: Determine Convergence or Divergence

Since the absolute value of the common ratio is less than 1, the series converges. For a converging geometric series, the sum is equal to \(\frac{a}{1 - r}\). Substituting \(a = 5\) and \(r = -\frac{1}{5}\) into this formula, we find that the sum of the series is equal to \(\frac{5}{1 - (-\frac{1}{5})} = 4\).