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\(S = \frac{1}{96} + \frac{1}{97} + \frac{1}{98} + \frac{1}{99} + \frac{1}{100} + \frac{1}{101} + \frac{1}{102} + \frac{1}{103} + \frac{1}{104} + \frac{1}{105}\)

This is a sum of fractions of the form \(\frac{1}{n}\) where n is between 96 and 105.
since each \(\frac{1}{n}\) , are close in value to one another. The average denominator is approximately: \(\frac{96+105}{2} = \frac{201}{2} = 100.5\)
Approximating each fraction with the average denominator:
\(\frac{1}{100.5} \approx{ 0.01 }\)
\(S = \frac{1}{96} + \frac{1}{97} + \frac{1}{98} + \frac{1}{99} + \frac{1}{100} + \frac{1}{101} + \frac{1}{102} + \frac{1}{103} + \frac{1}{104} + \frac{1}{105}\)

\(S = \frac{1}{96+4.5} + \frac{1}{97+3.5} + \frac{1}{98+2.5} + \frac{1}{99+1.5} + \frac{1}{100+0.5} + \frac{1}{101-0.5} + \frac{1}{102-1.5} + \frac{1}{103-2.5} + \frac{1}{104-3.5} + \frac{1}{105-4.5} \)

\(S \approx{ 10 * \frac{1}{(100)} = \frac{1}{(10)} =0.1}\)

=> A
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