The sum to infinity of the following series f | vedclass

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(B) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{n(n+1)}$.Using partial fractions,we can write the general term as $\frac{1}{n(n+1)} = \frac{

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$1$

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(B) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{n(n+1)}$.Using partial fractions,we can write the general term as $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.Thus,the sum of the first $N$ terms is $S_N = \sum_{n=1}^{N} \left( \frac{1}{n} - \frac{1}{n+1} ight)$.Expanding this,we get $S_N = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{N} - \frac{1}{N+1})$.All intermediate terms cancel out,leaving $S_N = 1 - \frac{1}{N+1}$.Taking the limit as $N 	o \infty$,we get $S = \lim_{N 	o \infty} (1 - \frac{1}{N+1}) = 1 - 0 = 1$.

The sum to infinity of the following series $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots$ is:

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