Question

Find the sum of the convergent series. $$ \sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^{n} $$

Short answer

The sum of the provided convergent series is \(\frac{2}{3}\).

Step-by-step solution

Identify the First Term and Common Ratio

In the given series \(\sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^{n}\), the first term \(a\) is when \(n = 0\) which gives \(a = (-\frac{1}{2})^{0} = 1\). The common ratio \(r\) is \(-\frac{1}{2}\).

Check if the Series is Convergent

For a series to be convergent, the absolute value of the common ratio must be less than 1. In this case: \(|-1/2| = 1/2 < 1\). Hence, the series is convergent.

Calculate the Sum of the Convergent Series

Using the formula for the sum of an infinite geometric series \(S = \frac{a}{1-r}\), substitute the found values for \(a\) and \(r\). The sum \(S\) of the series then is \(S = \frac{1}{1--1/2} = \frac{1}{1+1/2} = \frac{1}{1.5} = \frac{2}{3}\).