/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 Test for convergence or divergen... [FREE SOLUTION] | 91Ó°ÊÓ

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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 $$ \sum_{n=0}^{\infty} 5\left(-\frac{1}{5}\right)^{n} $$

Short Answer

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

Step by step solution

01

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.
02

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.
03

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 \(\frac{5}{1 - (-\frac{1}{5})} = 4\).

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