To determine whether the given series converges or diverges, we first need to express the general term of the series as a telescoping sum.

Series:

$\sum_{n=1}^{\infty} \frac{21}{n(n+3)}$

Step 1: Decompose the general term

The general term can be decomposed into partial fractions:

$\frac{21}{n(n+3)} = \frac{A}{n} + \frac{B}{n+3}$

To find the constants $A$ and $B$, we write:

$\frac{21}{n(n+3)} = \frac{A(n+3) + Bn}{n(n+3)} = \frac{An + 3A + Bn}{n(n+3)}$

This simplifies to:

$21 = (A + B)n + 3A$

We can equate the coefficients of $n$ and the constant terms:

A + B = 0 \quad \text{(coefficient of n)} $3A = 21 \quad \text{(constant term)}$

Solving these equations:

$A = 7, \quad B = -7$

Thus, the partial fraction decomposition is:

$\frac{21}{n(n+3)} = \frac{7}{n} - \frac{7}{n+3}$

Step 2: Express the series as a telescoping sum

The series can now be written as:

$\sum_{n=1}^{\infty} \left(\frac{7}{n} - \frac{7}{n+3}\right)$

This is a telescoping series. To better see this, let's write out the first few terms:

$\left(\frac{7}{1} - \frac{7}{4}\right) + \left(\frac{7}{2} - \frac{7}{5}\right) + \left(\frac{7}{3} - \frac{7}{6}\right) + \left(\frac{7}{4} - \frac{7}{7}\right) + \dots$

Notice that many terms cancel out:

$\text{Remaining terms: } \left(\frac{7}{1} + \frac{7}{2} + \frac{7}{3}\right) - \lim_{n \to \infty} \left(\frac{7}{n+1} + \frac{7}{n+2} + \frac{7}{n+3}\right)$

As $n$ approaches infinity, the terms $\frac{7}{n+1}$, $\frac{7}{n+2}$, and $\frac{7}{n+3}$ approach zero. Thus, the series converges, and the sum $S$ is:

$S = 7\left(\frac{1}{1} + \frac{1}{2} + \frac{1}{3}\right) = 7\left(1 + \frac{1}{2} + \frac{1}{3}\right) = 7\left(\frac{6}{6} + \frac{3}{6} + \frac{2}{6}\right) = 7 \times \frac{11}{6} = \frac{77}{6}$

Conclusion:

The series converges, and the sum is $\frac{77}{6}$.

Would you like more details or have any questions?

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Here are five related questions:

1. What is the general form of a telescoping series?
2. How do you identify when a series is telescoping?
3. Can a divergent series be expressed as a telescoping sum?
4. How would the sum change if the series started at a different index?
5. How do you evaluate the convergence of a series that is not telescoping?

Tip: When dealing with telescoping series, always look for terms that cancel out over a long sum to simplify the process of finding convergence and the total sum.