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The comparison test is a method to determine the convergence or divergence of an infinite series by comparing it to another series with known behavior. To apply the comparison test:

1. Identify the general form of the series you want to compare.
2. Choose a comparison series that closely resembles the original series at large values of n.
3. Check if the comparison series converges or diverges.

If the comparison series \(\bold{\text{b}_n}\) converges and \(\bold{\text{a}_n \text{≤} \text{b}_n}\) for all terms, then the original series also converges. Conversely, if the comparison series diverges and \(\bold{\text{a}_n \text{≥} \text{b}_n}\) for all terms, then the original series also diverges.

Example:
If we consider the series \(\bold{\text{∑} \frac{1}{3^n}}\), we know it converges since it is a geometric series with a ratio less than 1. Therefore, we use this series to compare with our original series \(\bold{\text{∑} \frac{1}{3^n - \text{cos} n}}\).

The general term in a series refers to the expression that defines the nth term of the series. Understanding the general term is crucial to determining the behavior of the series.

For the series \(\bold{\text{∑} a_n}\), where \(\bold{a_n = \frac{1}{3^n - \text{cos} n}}\), the general term is defined as \(\bold{a_n}\). As n becomes large, \(\bold{\text{cos} n}\) oscillates between -1 and 1, while \(\bold{3^n}\) grows exponentially. This implies that the effect of \(\bold{\text{cos} n}\) becomes negligible as compared to \(\bold{3^n}\). Thus, the general term can be approximated by \(\bold{\frac{1}{3^n}}\).

Identifying the approximation helps in simplification and in comparing the series with well-known convergent or divergent series.

The limit comparison test is another method to determine the convergence of a series. It's particularly useful when the direct comparison test is hard to apply. For the limit comparison test:

1. Choose another series \(\bold{\text{b}_n}\) that is easy to compare and whose behavior (convergence/divergence) is known.
2. Compute the limit \(\bold{\text{L} = \text{lim}_{{n \to \text{∞}}} \frac{a_n}{b_n}}\).
- If 0 < L < ∞ and \(\bold{\text{∑b}_n}\) converges, then \(\bold{\text{∑a}_n}\) converges.
- If L = 0 and \(\bold{\text{∑b}_n}\) converges, then \(\bold{\text{∑a}_n}\) converges.
- If L = ∞ and \(\bold{\text{∑b}_n}\) diverges, then \(\bold{\text{∑a}_n}\) diverges.

Example:
Using \(\bold{\text{a}_n = \frac{1}{3^n - \text{cos} n}}\) and comparison series \(\bold{\text{b}_n = \frac{1}{3^n}}\), we compute:
\(\text{L} = \text{lim}_{{n \to \text{∞}}} \frac{a_n}{b_n} = \text{lim}_{{n \to \text{∞}}} \frac{\frac{1}{3^n - \text{cos} n}}{\frac{1}{3^n}} = \text{lim}_{{n \to \text{∞}}} \frac{3^n}{3^n - \text{cos} n} = 1 \).
Since 1 is a finite positive number and \(\bold{\text{∑} \frac{1}{3^n}}\) converges, \(\bold{\text{∑} \frac{1}{3^n - \text{cos} n}}\) also converges.

A geometric series is a series of the form \(\bold{\text{∑} ar^n}\), where \(\bold{a}\) is the first term, and \(\bold{r}\) is the common ratio.

A geometric series converges if the absolute value of the ratio is less than 1 (|r|<1). Otherwise, it diverges.

The series \(\bold{\text{∑} \frac{1}{3^n}}\) is a classic example of a convergent geometric series with a = 1 and r = \(\frac{1}{3}\). Since \(\frac{1}{3}\) is less than 1, the series converges.

In the given exercise, we used the geometric series to compare with the series \(\bold{\text{∑} \frac{1}{3^n - \text{cos} n}}\). After simplifying our general term to an approximate geometric series and applying the limit comparison test, we confirmed that our given series also converges.