91Ó°ÊÓ

A geometric series is a series of numbers where each term after the first is found by multiplying the previous term by a constant called the 'common ratio' (denoted as \(r\)). For example, in the series \(1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \text{...} \), each term is multiplied by \(\frac{1}{2}\) to get the next term.

Key characteristics of geometric series include:

• The first term (denoted as \(a\)).
• The common ratio \(r\).

The general form of a geometric series is:

\[a + ar + ar^2 + ar^3 + \text{...}\]

In the given series \(\sum_{n=1}^{+\text{infinity}}\bigg(2^{-n} + 3^{-n}\bigg)\), we actually have two separate geometric series:

• \(2^{-n}\) with \(a = \frac{1}{2}\) and \(r = \frac{1}{2}\).
• \(3^{-n}\) with \(a = \frac{1}{3}\) and \(r = \frac{1}{3}\).

Understanding these individual series helps in assessing the behavior and sum of the original series.

Calculating the sum of an infinite series can be manageable if the series belongs to a well-understood type like geometric series. For a geometric series to be finite sum convergable, the common ratio \(r\) must satisfy \(|r| < 1\).

The sum \(S\) of an infinite geometric series is given by the formula:

\[S = \frac{a}{1 - r}\]

where \(a\) is the first term and \(r\) is the common ratio.

For the given series:

• For \(2^{-n}\), the sum is \(\frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1\).
• For \(3^{-n}\), the sum is \(\frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}\).

Consequently, the total sum of the given series is:
\[ 1 + \frac{1}{2} = \frac{3}{2}\] This implies the sum of \(\frac{3}{2}\) results when evaluated and confirms our series culminates into a finite value

Testing for convergence in series is crucial, especially with infinite series, to ensure they have a finite limit. For geometric series, the convergence condition is quite simple: the series converges if \(|r| < 1\).

For our exercise, we examined two separate geometric series:

• In \(2^{-n}\), \(|r| = |\frac{1}{2}| = 0.5 < 1\), so this series converges.
• In \(3^{-n}\), \(|r| = |\frac{1}{3}| = 0.333 < 1\), thus this series converges too.

This was because each geometric series follows the rule \(|r| < 1\).

Hence, within our combined series \(\text{\textsum_{n=1}^{+\textinfty}}(2^{-n} + 3^{-n})\), each separate series is convergent, thereby ensuring the composite also converges.

Therefore, our series converges, and we effectively determined its sum as \(\frac{3}{2}\).