To solve this problem, let us first determine the general formula for the series and then analyze the convergence.

Step 1: General Term Simplification

The given series is:

$\frac{6}{1 \cdot 2} + \frac{6}{2 \cdot 3} + \frac{6}{3 \cdot 4} + \dots + \frac{6}{n(n+1)} + \dots$

The general term of the series is:

$a_n = \frac{6}{n(n+1)}$

We can simplify this term using partial fractions:

$\frac{6}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$

Solving for $A$ and $B$, we get:

$6 = A(n+1) + Bn$

By equating coefficients:

• $A + B = 0$
• $A = 6$

Thus: $A = 6, \, B = -6$

So: $\frac{6}{n(n+1)} = \frac{6}{n} - \frac{6}{n+1}$

Step 2: Partial Sum Formula

The partial sum $S_n$ up to the $n$-th term is:

$S_n = \sum_{k=1}^n \left( \frac{6}{k} - \frac{6}{k+1} \right)$

This is a telescoping series. Writing out the terms:

$S_n = \left(\frac{6}{1} - \frac{6}{2}\right) + \left(\frac{6}{2} - \frac{6}{3}\right) + \dots + \left(\frac{6}{n} - \frac{6}{n+1}\right)$

Most terms cancel, leaving:

$S_n = 6 - \frac{6}{n+1}$

Step 3: Convergence of the Series

To find the sum of the infinite series, we take the limit as $n \to \infty$:

$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(6 - \frac{6}{n+1}\right) = 6 - 0 = 6$

Final Answer

• The formula for the $n$-th partial sum is:

$S_n = 6 - \frac{6}{n+1}$

• The sum of the series, if it converges, is:

$\text{Sum} = 6$

Would you like further clarification on this, or do you have additional questions? Here are some related queries to consider:

1. How does the partial fraction decomposition work in other contexts?
2. What is the significance of a telescoping series in calculus?
3. Can we derive the convergence using a different approach (e.g., integral test)?
4. What other types of series converge to finite sums?
5. How do we verify if a series diverges?

Tip: Telescoping series are an excellent way to simplify complex expressions quickly by leveraging cancellation.