91Ó°ÊÓ

Let's start by understanding the **general term** of the series. The general term of a sequence is a formula that enables you to find each term based on its position in the sequence (denoted as n). In our given series, the general term is: \(\frac{1}{2n} - \frac{1}{3n}\). This formula helps us calculate any term in the series using its position. For instance, when n = 1, you substitute 1 in place of n into the general term formula. Breaking down the general term is crucial for evaluating the whole series and determining its behavior.

Now let's analyze the **first four terms** of our series. We substitute n = 1, 2, 3, and 4 into the general term. Here's the process:

• When n = 1: \(\frac{1}{2 \times 1} - \frac{1}{3 \times 1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\)
• When n = 2: \(\frac{1}{2 \times 2} - \frac{1}{3 \times 2} = \frac{1}{4} - \frac{1}{6} = \frac{1}{12}\)
• When n = 3: \(\frac{1}{2 \times 3} - \frac{1}{3 \times 3} = \frac{1}{6} - \frac{1}{9} = \frac{1}{18}\)
• When n = 4: \(\frac{1}{2 \times 4} - \frac{1}{3 \times 4} = \frac{1}{8} - \frac{1}{12} = \frac{1}{24}\)

So, the first four terms of the series are \(\frac{1}{6}, \frac{1}{12}, \frac{1}{18}, \frac{1}{24}\). By breaking down the general term for these initial values, we gain insight into the pattern and behavior of the series.

Next, we need to determine if the series is a **convergent series**. A series converges if the sum of its terms approaches a finite value as n approaches infinity. For our series, we simplify the general term to: \(\frac{1}{2n} - \frac{1}{3n} = \frac{1}{6n}\). This resembles the harmonic series \(\frac{1}{n}\), but multiplied by \(\frac{1}{6}\). Since the harmonic series is known to diverge, its behavior is modified by the constant multiplier.To confirm convergence, we recognize that \(\frac{1}{6n}\) has a similar form but it behaves differently due to the constant factor. Applying principles from calculus, such as the integral test, allows us to assert that \(\frac{1}{6n}\) does indeed converge. Hence, our series \(\frac{1}{2n} - \frac{1}{3n}\) is a convergent series.

Finally, let's find the **sum of the series**. Since we established the convergence, we now need the sum. Recall our series: \(\frac{1}{6} \times \frac{1}{n}\). Using logarithmic properties, we know that for a convergent series resembling \(\frac{1}{n}\), its sum can be linked to logarithmic values. The sum of \(\frac{1}{6} \times \frac{1}{n}\) translates to \(\frac{1}{6} \times \text{ln}\big(\frac{n+1}{n}\big)\) evaluated from 1 to infinity. However, simplifying this leads us to: \(\frac{1}{6} \times 1 = \frac{1}{6}\). Thus, the sum of the given series is \(\frac{1}{6}\).This approach ensures we don't just rely on series addition but apply mathematical understanding to link to known series behaviors and sums.