Calculate \( S_{3}, S_{4} \), and \( S_{5} \) and then find the sum for the telescoping series \[ S=\sum_{n=7}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+2}\right) \] where \( S_{k} \) is the partial sum using the first \( k \) values of \( n \). \( S_{3}= \) \( S_{4}= \) \( S_{5}= \) \( S= \)

To find the partial sums \( S_{3}, S_{4}, S_{5} \) and the sum of the telescoping series \( S \), we start by calculating \( S_{k} \) for \( k = 3, 4, 5 \). The series \( S = \sum_{n=7}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n+2}\right) \) can be broken down as follows: 1. Observe the series: \[ S_{k} = \sum_{n=7}^{n=k+6} \left(\frac{1}{n+1} - \frac{1}{n+2}\right) \] where \( k \) starts from 3, and we will calculate it for \( k=3, 4, 5 \). ### Calculate \( S_{3} \) For \( n = 7, 8, 9 \): \[ S_{3} = \left( \frac{1}{8} - \frac{1}{9} \right) + \left( \frac{1}{9} - \frac{1}{10} \right) + \left( \frac{1}{10} - \frac{1}{11} \right) \] Notice that this is telescoping, and most terms will cancel: \[ S_{3} = \frac{1}{8} - \frac{1}{11} \] Calculating this, \[ S_{3} = \frac{1}{8} - \frac{1}{11} = \frac{11 - 8}{88} = \frac{3}{88} \] ### Calculate \( S_{4} \) For \( n = 7, 8, 9, 10 \): \[ S_{4} = S_{3} + \left( \frac{1}{11} - \frac{1}{12} \right) = \frac{3}{88} + \left(\frac{1}{11} - \frac{1}{12}\right) \] Calculating \( \left(\frac{1}{11} - \frac{1}{12}\right) \): \[ \frac{1}{11} - \frac{1}{12} = \frac{12 - 11}{132} = \frac{1}{132} \] Finding a common denominator for \( \frac{3}{88} \) and \( \frac{1}{132} \): \[ \text{lcm}(88, 132) = 132 \] Converting \( \frac{3}{88} \) to the common denominator: \[ \frac{3}{88} = \frac{3 \cdot 3}{88 \cdot 3} = \frac{9}{264} \] Similarly, \[ \frac{1}{132} = \frac{1 \cdot 2}{132 \cdot 2} = \frac{2}{264} \] Thus, \[ S_{4} = \frac{9}{264} + \frac{2}{264} = \frac{11}{264} \] ### Calculate \( S_{5} \) Now for \( n = 7, 8, 9, 10, 11 \): \[ S_{5} = S_{4} + \left( \frac{1}{12} - \frac{1}{13} \right) \] Calculating \( \left( \frac{1}{12} - \frac{1}{13} \right) \): \[ \frac{1}{12} - \frac{1}{13} = \frac{13 - 12}{156} = \frac{1}{156} \] Finding a common denominator between \( \frac{11}{264} \) and \( \frac{1}{156} \): \[ \text{lcm}(264, 156) = 1320 \] Converting \( \frac{11}{264} \) to the common denominator: \[ \frac{11}{264} = \frac{11 \cdot 5}{264 \cdot 5} = \frac{55}{1320} \] Converting \( \frac{1}{156} \): \[ \frac{1}{156} = \frac{1 \cdot 8.5}{156 \cdot 8.5} = \frac{80}{1320} \] Thus, \[ S_{5} = \frac{55}{1320} + \frac{80}{1320