The sum in question is:

$\sum_{r=1}^{\infty} \frac{3}{r(r+1)}.$

Step-by-Step Solution

1. Partial Fraction Decomposition:

   Let's break down the expression $\frac{3}{r(r+1)}$ using partial fractions. We seek constants $A$ and $B$ such that:

   $\frac{3}{r(r+1)} = \frac{A}{r} + \frac{B}{r+1}.$

   Multiplying through by $r(r+1)$, we get:

   $3 = A(r + 1) + Br.$

   Simplifying this, we have:

   $3 = (A + B)r + A.$

   For this equation to hold for all $r$, the coefficients of $r$ and the constant term must match on both sides. Therefore, we obtain the system of equations:

   $A + B = 0, \quad A = 3.$

   Solving this, we get $A = 3$ and $B = -3$. So, the partial fraction decomposition is:

   $\frac{3}{r(r+1)} = \frac{3}{r} - \frac{3}{r+1}.$

2. Rewrite the Series:

   Substitute the partial fractions into the sum:

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

3. Telescoping Series:

   This is a telescoping series, where most terms cancel out. Let's write out the first few terms to see the pattern:

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

   Notice that almost every positive term $\frac{3}{r}$ cancels with the subsequent negative term $-\frac{3}{r}$ in the next fraction. Therefore, after cancellation, we are left with:

   $3 - \lim_{n \to \infty} \frac{3}{n+1}.$

4. Evaluate the Limit:

   Since $\lim_{n \to \infty} \frac{3}{n+1} = 0$, the remaining sum is:

   $3.$

Final Answer

$\sum_{r=1}^{\infty} \frac{3}{r(r+1)} = 3.$

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Related Questions

1. How would you find the sum