∑ 
n=2
∞
​
  
2
3*n
 
5 
n+1
 
​


The given summation appears to be:

\[
\sum_{n=2}^{\infty} \frac{2}{3n} \cdot \frac{5}{n+1}
\]

We can break it down into a clearer form:

\[
\sum_{n=2}^{\infty} \frac{10}{3n(n+1)}
\]

### Step-by-Step Explanation:
1. The summation starts at \( n = 2 \) and continues indefinitely (\( \infty \)).
2. The expression inside the sum is \( \frac{10}{3n(n+1)} \).
   - This can be simplified further by partial fraction decomposition.

### Partial Fraction Decomposition:
We rewrite \( \frac{10}{3n(n+1)} \) as:

\[
\frac{10}{3n(n+1)} = \frac{A}{n} + \frac{B}{n+1}
\]

Solving for \( A \) and \( B \), we multiply both sides by \( 3n(n+1) \):

\[
10 = A(n+1) + Bn
\]

Expanding the right-hand side:

\[
10 = An + A + Bn
\]

Now, grouping terms with \( n \):

\[
10 = (A + B)n + A
\]

This gives the system of equations:
\[
A + B = 0 \quad \text{and} \quad A = 10
\]

Solving this system:
- \( A = 10 \)
- \( B = -10 \)

Thus, we have:

\[
\frac{10}{3n(n+1)} = \frac{10}{3} \left( \frac{1}{n} - \frac{1}{n+1} \right)
\]

### Simplified Summation:
Now, substituting this back into the sum:

\[
\sum_{n=2}^{\infty} \frac{10}{3n(n+1)} = \frac{10}{3} \sum_{n=2}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right)
\]

This is a telescoping series, meaning most terms cancel out. The remaining terms are:

\[
\frac{10}{3} \left( \frac{1}{2} - \lim_{n \to \infty} \frac{1}{n+1} \right)
\]

As \( n \to \infty \), \( \frac{1}{n+1} \to 0 \), so we are left with:

\[
\frac{10}{3} \times \frac{1}{2} = \frac{5}{3}
\]

### Final Answer:
\[
\sum_{n=2}^{\infty} \frac{10}{3n(n+1)} = \frac{5}{3}
\]

Would you like more details or have any questions about the steps?

Here are some related questions:
1. How does partial fraction decomposition work in summation problems?
2. What is a telescoping series and how does it simplify summations?
3. Can this method apply to finite sums, and if so, how does the result change?
4. What are other ways to simplify summations involving products of terms like \( n \) and \( n+1 \)?
5. How does the behavior of infinite sums change with different starting indices?

**Tip:** Partial fractions are powerful for simplifying rational functions in summations and integrals.

This guide explains how to solve the infinite series ∑_{n=2}^{∞} (10 / 3n(n+1)) using partial fraction decomposition and telescoping series techniques. Learn the detailed steps, including how terms cancel out, leading to the final answer of 5/3. This problem is suitable for undergraduate calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation