91Ó°ÊÓ

In a geometric series, the series' convergence is heavily dependent on the common ratio, denoted as \( r \). For the series to be convergent, a pivotal rule must be met: the absolute value of the common ratio \( r \) must be less than 1. This means that \( |r| < 1 \). Convergence occurs when the terms of the series become smaller and smaller, approaching zero as they progress to infinity. If this condition holds, the series sum will approach a finite limit rather than growing indefinitely.For example, in the given series \( \sum_{n=1}^{\infty} \left(\frac{1}{7^{1/3}}\right)^n \), the common ratio \( r \) is \( \frac{1}{7^{1/3}} \). Calculating \( |r| = \left|\frac{1}{7^{1/3}}\right| = 7^{-1/3} \), we find that it is indeed less than 1. This verifies that the series converges and reassures us that we can compute its sum.

An infinite series is a sum of an endless sequence of numbers, which is often written as \( \sum_{n=1}^{\infty} a_n \). Unlike a finite series which contains a fixed number of terms, an infinite series continues indefinitely, adding terms one after another.The concept of infinity implies no end, so a natural curiosity arises about whether these sums can reach a limit, especially when involving endless addition. This is where convergence becomes significant. In our specific case, the series \( \sum_{n=1}^{\infty} 7^{-n/3} \) is transformed into a geometric form \( \sum_{n=1}^{\infty} \left(\frac{1}{7^{1/3}}\right)^n \). Despite being infinite, this series converges, allowing us to compute a definite sum due to its decreasing term pattern.

Once a geometric series is identified as convergent, finding its sum becomes straightforward using a specific formula. For an infinite geometric series \( \sum_{n=0}^{\infty} ar^n \), the sum \( S \) is calculated using:\[ S = \frac{a}{1 - r} \]where \( a \) is the first term and \( r \) is the common ratio.In the provided problem, we first recognized the equivalence \( a = \frac{1}{7^{1/3}} \) and \( r = \frac{1}{7^{1/3}} \). Employing the formula, we substitute these values to find the sum:\[ S = \frac{\frac{1}{7^{1/3}}}{1 - \frac{1}{7^{1/3}}} = \frac{7^{-1/3}}{1 - 7^{-1/3}} \]Simplifying the result gives \( \frac{1}{7^{1/3} - 1} \). This neat answer makes understanding the behavior and results of geometric series much more accessible when applying the sum formula.