\( \frac{1}{1}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{98}+\frac{1}{99}+\frac{1}{100} \)

To solve the sum \( S = \frac{1}{1} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + ... + \frac{1}{98} + \frac{1}{99} + \frac{1}{100} \), we first note that this sequence includes all terms from \( \frac{1}{1} \) to \( \frac{1}{100} \), excluding \( \frac{1}{2} \). We can express the sum as: \[ S = \sum_{n=1}^{100} \frac{1}{n} - \frac{1}{2} \] The sum \( \sum_{n=1}^{100} \frac{1}{n} \) is known as the 100th harmonic number, denoted as \( H_{100} \). While there's no simple formula for the harmonic numbers, we can estimate: \[ H_n \approx \ln(n) + \gamma \] where \( \gamma \) (the Euler-Mascheroni constant) is approximately \( 0.577 \). For \( n = 100 \): \[ H_{100} \approx \ln(100) + 0.577 \approx 4.605 + 0.577 \approx 5.182 \] To find \( S \): \[ S \approx 5.182 - 0.5 = 4.682 \] Calculating \( H_{100} \) precisely can be done using various methods or computer software, yielding: \[ H_{100} \approx 5.187377517639621 \] Thus, \[ S \approx 5.187377517639621 - 0.5 = 4.687377517639621 \] To conclude, the sum \( S \) evaluates approximately to: \[ \boxed{4.687} \]