$\frac{1}{2} - \frac{1}{2 \cdot 2^2} + \frac{1}{3 \cdot 2^3} - \frac{1}{4 \cdot 2^4} + \ldots$ is equal to

  • A
    $\frac{1}{4}$
  • B
    $\log _3\left(\frac{3}{4}\right)$
  • C
    $\log _e\left(\frac{3}{2}\right)$
  • D
    $\log _e\left(\frac{2}{3}\right)$

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$\frac{2}{1} \cdot \frac{1}{3} + \frac{3}{2} \cdot \frac{1}{9} + \frac{4}{3} \cdot \frac{1}{27} + \frac{5}{4} \cdot \frac{1}{81} + \dots \infty = $

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}{n^2} + \frac{1}{2n^4} + \frac{1}{3n^6} + \dots \infty = $

If $S = \sum\limits_{n = 0}^\infty \frac{(\log x)^{2n}}{(2n)!}$,then $S$ =

For $|x| < 1$,the coefficient of $x^3$ in the expansion of $\log(1+x+x^2)$ in ascending powers of $x$ is (in $/3$)

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