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Question Number 93859 by M±th+et+s last updated on 15/May/20

\underset{n = 1}{\sum^{\infty}} \frac{1}{n (n + 1)}

Answered by Ar Brandon last updated on 15/May/20

\frac{1}{n (n + 1)} = \frac{1}{n} - \frac{1}{n + 1} \Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{1}{n (n + 1)} = \underset{n = 1}{\sum^{\infty}} [\frac{1}{n} - \frac{1}{n + 1}] = \lim_{k \to + \infty} \underset{n = 1}{\sum^{k}} [\frac{1}{n} - \frac{1}{n + 1}]

= \lim_{k \to + \infty} [1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdot \cdot \cdot + \frac{1}{k} - \frac{1}{k + 1}]

= \lim_{k \to + \infty} [1 - \frac{1}{k + 1}] = 1