1. **State the problem:** Find the sum of the series $$\sum_{n=1}^\infty \left( \frac{3}{n(n+1)} + \frac{1}{2^n} \right).$$ 2. **Split the series:** The series is the sum of two separate series: $$\sum_{n=1}^\infty \frac{3}{n(n+1)} + \sum_{n=1}^\infty \frac{1}{2^n}.$$ 3. **Sum the first series:** Use partial fraction decomposition: $$\frac{3}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}.$$ Multiply both sides by $n(n+1)$: $$3 = A(n+1) + Bn = An + A + Bn = (A+B)n + A.$$ Equate coefficients: - Coefficient of $n$: $A + B = 0$ - Constant term: $A = 3$ So, $A=3$ and $B=-3$. Thus, $$\frac{3}{n(n+1)} = \frac{3}{n} - \frac{3}{n+1}.$$ 4. **Write the first series as telescoping:** $$\sum_{n=1}^\infty \left( \frac{3}{n} - \frac{3}{n+1} \right) = 3 \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right).$$ 5. **Evaluate the telescoping sum:** $$3 \left( 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdots \right) = 3 \times 1 = 3.$$ 6. **Sum the second series:** This is a geometric series with first term $\frac{1}{2}$ and ratio $\frac{1}{2}$: $$\sum_{n=1}^\infty \frac{1}{2^n} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1.$$ 7. **Add both sums:** $$3 + 1 = 4.$$ **Final answer:** The sum of the series is $4$.