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The series $\sum_{k=2}^{\infty} \frac{1}{k(k-1)}$ converges to

  1. $-1$
  2. $1$
  3. $0$
  4. does not converge
in Numerical Ability by Veteran (425k points)
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1 Answer

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Answer:(B)

$\displaystyle\sum _{k=2}^{\infty}\frac{1}{k(k-1)}$

$\displaystyle=\sum _{k=2}^{\infty}\frac{k-(k-1)}{k(k-1)}$

$\displaystyle=\sum _{k=2}^{\infty}\left(\frac{1}{(k-1)}-\frac{1}{k}\right)$

$=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots$

$=1-\frac{1}{\infty}$

$=1$

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