$1 + \left( \frac{1}{2} + \frac{1}{3} \right) \frac{1}{4} + \left( \frac{1}{4} + \frac{1}{5} \right) \frac{1}{4^2} + \left( \frac{1}{6} + \frac{1}{7} \right) \frac{1}{4^3} + \dots \infty = $

  • A
    $\log_e (2\sqrt{3})$
  • B
    $2 \log_e 2$
  • C
    $\log_e 2$
  • D
    $\log_e \left( \frac{2}{\sqrt{3}} \right)$

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If $x, y, z$ are three consecutive positive integers,then $\frac{1}{2}\log_e x + \frac{1}{2}\log_e z + \frac{1}{2xz + 1} + \frac{1}{3}\left( \frac{1}{2xz + 1} \right)^3 + \dots = $

The sum of the infinite series $\frac{1}{1 \times 2} - \frac{1}{2 \times 3} + \frac{1}{3 \times 4} - \dots \infty$ is equal to:

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$\frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} - \frac{1}{4 \cdot 5} + \dots \infty = $

If $|a| < 1$ and $b = \sum_{k=1}^{\infty} \frac{a^k}{k}$,then $a$ is equal to

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