91Ó°ÊÓ

A convergent series is one where the sum of all its terms approaches a specific finite value. Essentially, as you add more terms, the series gets closer and closer to a particular number. This is an integral concept in calculus and mathematical analysis.
In formal mathematical terms, a series \( \sum_{n=1}^\infty a_n \) is said to converge if the sequence of its partial sums \( S_N = a_1 + a_2 + \cdots + a_N \) approaches a finite limit as \( N \to \infty \).
- If the limit exists and is finite, the series converges. - If not, the series is divergent.
In the context of our problem, identifying that a series is convergent involves using a comparison series that converges, which directly affects the behavior of the original series.

A geometric series is a special type of series where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio \( r \).
A geometric series can be represented as:
\[ a + ar + ar^2 + ar^3 + \cdots \]
where \( a \) is the initial term. - If the absolute value of the common ratio \( |r| < 1 \), the geometric series converges.
- If \( |r| \geq 1 \), it diverges.
In the solution process of our exercise, the series \( \sum_{n=1}^{\infty} \frac{1}{3^n} \) is chosen for comparison, known to be convergent because \( r = \frac{1}{3} \) and this satisfies the condition \( |r| < 1 \). This allows us to use it as a benchmark for the convergence of the series in question.

Inequalities play a crucial role when determining the convergence of series. The basic comparison test, which we use in this scenario, heavily depends on the application of inequalities.
By comparing term-by-term, we ensure that our series of interest behaves similar to a known series regarding convergence.
For example, in our current problem, we have the inequality:
\[ \frac{1}{n 3^{n}} \leq \frac{1}{3^{n}} \text{ for all } n \geq 1 \]
This inequality helps establish that each term of the series \( \frac{1}{n 3^{n}} \) is smaller than or equal to the terms in the geometric series \( \frac{1}{3^{n}} \). Since the latter is convergent, this key inequality supports the conclusion of convergence for our series as well.

The basic comparison test is a valuable tool for determining the behavior of series, especially in terms of convergence or divergence. It operates on a simple principle: compare the terms of your series to those of another series whose convergence properties are already known.
Here's how it generally works:

• Identify a comparison series \( \sum_{n=1}^\infty b_n \), which serves as a benchmark.
• Ensure that for all terms \( a_n \) in the original series, it's true that \( a_n \leq b_n \) if \( \, \sum_{n=1}^\infty a_n \, \) is to be checked for convergence.
• If \( \sum_{n=1}^\infty b_n \) converges, then by comparison, \( \sum_{n=1}^\infty a_n \) must also converge.

This approach was applied to our exercise, where the given series \( \sum_{n=1}^{\infty} \frac{1}{n 3^n} \) showed each term to be less than those in the known convergent series \( \sum_{n=1}^{\infty} \frac{1}{3^n} \). By confirming this, we successfully determined the convergence of the original series using the basic comparison test.