frac{1}{1 times 2} + frac{1}{2 times 3} + fra | vedclass

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

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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 partial fractions,we can write $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.Thus,the sum becomes:$S_n = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{n} - \frac{1}{n+1})$.This is a telescoping series where all intermediate terms cancel out:$S_n = 1 - \frac{1}{n+1} = \frac{n+1-1}{n+1} = \frac{n}{n+1}$.

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

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