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Let's analyze and solve this problem step-by-step.

### Problem Breakdown
The series in question is:

\[
\sum_{n=1}^{\infty} \frac{3}{(3n - 2)(3n + 1)}
\]

This is a telescoping series, which typically simplifies when expressed as a difference of fractions.

#### Step 1: Express the General Term in Partial Fractions
Let's rewrite the general term in partial fractions. We want to express

\[
\frac{3}{(3n - 2)(3n + 1)}
\]

as:

\[
\frac{A}{3n - 2} + \frac{B}{3n + 1}
\]

Multiplying through by \((3n - 2)(3n + 1)\) gives:

\[
3 = A(3n + 1) + B(3n - 2)
\]

Expanding and collecting terms:

\[
3 = (3A + 3B)n + (A - 2B)
\]

By comparing coefficients, we get the system of equations:

1. \(3A + 3B = 0\)
2. \(A - 2B = 3\)

Solving this system:

1. From the first equation, \(A + B = 0\), so \(A = -B\).
2. Substitute \(A = -B\) into the second equation: \(-B - 2B = 3\), so \(-3B = 3\), hence \(B = -1\).
3. Then \(A = 1\).

So we can rewrite the term as:

\[
\frac{3}{(3n - 2)(3n + 1)} = \frac{1}{3n - 2} - \frac{1}{3n + 1}
\]

#### Step 2: Write Out the First Four Terms of the Series
Using this form, the series becomes:

\[
\sum_{n=1}^{\infty} \left( \frac{1}{3n - 2} - \frac{1}{3n + 1} \right)
\]

The first few terms are:

- For \(n=1\): \(\frac{1}{1} - \frac{1}{4} = 1 - \frac{1}{4} = \frac{3}{4}\)
- For \(n=2\): \(\frac{1}{4} - \frac{1}{7}\)
- For \(n=3\): \(\frac{1}{7} - \frac{1}{10}\)
- For \(n=4\): \(\frac{1}{10} - \frac{1}{13}\)

So the first four terms are:

\[
\frac{3}{4}, \quad \frac{1}{4} - \frac{1}{7}, \quad \frac{1}{7} - \frac{1}{10}, \quad \frac{1}{10} - \frac{1}{13}
\]

#### Step 3: Derive a Formula for the \(n\)-th Partial Sum \(S_n\)
The \(n\)-th partial sum \(S_n\) of a telescoping series is obtained by adding up the first \(n\) terms and canceling intermediate terms. For our series, most terms cancel out, leaving only:

\[
S_n = \left( 1 - \frac{1}{3n + 1} \right) = 1 - \frac{1}{3n + 1}
\]

#### Step 4: Determine if the Series Converges and Find Its Sum
To find the sum as \(n \to \infty\), we take the limit of \(S_n\):

\[
\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( 1 - \frac{1}{3n + 1} \right) = 1 - 0 = 1
\]

### Conclusion
The series converges, and its sum is:

\[
\sum_{n=1}^{\infty} \frac{3}{(3n - 2)(3n + 1)} = 1
\]

Would you like further details on any specific step, or do you have any questions?

### Related Questions
1. How do we determine if a series is telescoping?
2. What are partial fractions, and how are they used in series expansion?
3. How can we verify convergence for other types of series?
4. What happens if the limit in a partial sum does not exist?
5. Why do intermediate terms cancel in a telescoping series?

#### Tip
Telescoping series often simplify to a manageable form by partial fraction decomposition, making it easier to identify terms that cancel.

Consider the following telescoping series:

\(\sum_{n=1}^{\infty} \frac{3}{(3n - 2)(3n + 1)}\)

**i.** List out the first four terms of the series. Then derive a formula for the nth partial sum \(S_n\).

**ii.** Use the definition of infinite series to determine if the series converges or diverges. If the series converges, find its sum.

Explore the telescoping series \(\sum_{n=1}^{\infty} \frac{3}{(3n - 2)(3n + 1)}\) with a detailed solution. Learn how to list the first four terms, derive a partial sum formula, and determine the convergence and sum of the series. Ideal for students studying advanced calculus or mathematical series.

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