Determine whether the series is convergent or divergent by expressing sn as a telescoping sum. If it is convergent, find its sum. (If the quantity diverges, enter DIVERGES.)
∞
21
n(n + 3)

n = 1


We are given the infinite series:

\[
S = \sum_{n=1}^{\infty} \frac{21}{n(n+3)}
\]

### Step 1: Partial fraction decomposition

We want to decompose \(\frac{21}{n(n+3)}\) into partial fractions. We express it as:

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

Multiplying both sides by \(n(n+3)\), we get:

\[
21 = A(n+3) + Bn
\]

Expanding and simplifying

Learn how to determine convergence and find the sum of the infinite series \( \sum_{n=1}^{\infty} \frac{21}{n(n+3)} \) using partial fraction decomposition and telescoping series method. Understand when a series converges or diverges and apply these concepts to solve advanced mathematical problems. Suitable for advanced high school students and those interested in calculus.

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