frac{1}{1 cdot 2} + frac{1}{2 cdot 3} + frac{ | vedclass

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(B) The given series is $S_n = \sum_{k=1}^{n} \frac{1}{k(k+1)}$.Using the method of partial fractions,we can write $\frac{1}{k(k+1)} = \frac{1}{k} - \

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$\frac{n}{n + 1}$

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(B) The given series is $S_n = \sum_{k=1}^{n} \frac{1}{k(k+1)}$.Using the method of partial fractions,we can write $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.Substituting this into the sum,we get:$S_n = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{n} - \frac{1}{n+1} \right)$.This is a telescoping series where all intermediate terms cancel out:$S_n = 1 - \frac{1}{n+1}$.Simplifying the expression,we get $S_n = \frac{n+1-1}{n+1} = \frac{n}{n+1}$.

$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n + 1)}$ equals

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