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+1 vote

The sum of the series $\dfrac{1}{1.2} + \dfrac{1}{2.3}+ \cdots + \dfrac{1}{n(n+1)} + \cdots $ is

  1. $1$
  2. $1/2$
  3. $0$
  4. non-existent
in Numerical Ability by Veteran (434k points)
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2 Answers

+6 votes

$=\sum _{k=1}^{\infty}\frac{1}{k(k+1)}$

$=\sum _{k=1}^{\infty}\frac{(k+1)-k}{k(k+1)}$

$=\sum _{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)$




$\therefore $ Option (A).
by Boss (13.4k points)
0 votes
the series can be expanded as 1-1/2+1/2-1/3...........1/n-1/n+1, all the terms will cancel out each other except first and last term so at the end we will get 1-1/n+1, and for larger N this term will be getting smaller and smaller so when n is very large 1/n+1 will be equal to 0, therefore, the final answer is 1.
by (299 points)
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